MCQ 11 Mark
If an $A.P.$ is $1,7,13, 19, ………$ Find the sum of $22$ terms.
- A
$127$
- B
$1204$
- ✓
$1408$
- D
$1604$
AnswerCorrect option: C. $1408$
From the given $A.P., a = 1$ and $d = 7 - 1 = 6.$
We know, $\text{s}_\text{n}=\frac{n}{2}(2\text{a}+(\text{n}-1)\text{d}$
$\text{s}_\text{22}=\frac{22}{2}(2\times1+(22-1)6)$
$= 11(2 + 126)$
$= 11 \times 128$
$= 1408.$
View full question & answer→MCQ 21 Mark
If the decreasing $\text{GP}$ is considered, then the sum of infinite terms is:
View full question & answer→MCQ 31 Mark
if an $A.P.$ is $3,5,7,9…….$ Find the $12^{th}$ term of the $A.P.$
AnswerFrom the given $A.P., a = 3$ and $d = 5 - 3 = 2.$
We know, $a_n= a + (n - 1) d$
$\Rightarrow a_{12}= a + 11d$
$= 3 + 11 \times 2$
$= 3 + 22$
$= 25.$
View full question & answer→MCQ 41 Mark
If in an infinite $G.P.,$ first term is equal to $10$ times the sum of all successive terms, the its common ratio is:
- A
$\frac{1}{10}$
- ✓
$\frac{1}{11}$
- C
$\frac{1}{9}$
- D
$\frac{1}{20}$
AnswerCorrect option: B. $\frac{1}{11}$
Let the first term of the $G.P.$ be a.
Let its common ratio be $r.$
According to the question, we have:
First term $= 10 [$Sum of all successive terms$]$
$\text{a}=10\Big(\frac{\text{ar}}{1-\text{r}}\Big)$
$\Rightarrow\text{a}-\text{ar}=10\text{ar}$
$\Rightarrow11\text{ar}=\text{a}$
$\Rightarrow\text{r}=\frac{\text{a}}{11\text{a}}=\frac{1}{11}$
View full question & answer→MCQ 51 Mark
If $a, b, c$ are in $A.P.$ and $x, y, z$ are in $G.P.,$ then the value of $x^{b-c}y^{c-a}z^{a-b}$ is:
- A
$0$
- ✓
$1$
- C
$x y z$
- D
$x^ay^bz^c$
Answer$a, b$ and $c$ are in $A.P.$
$\therefore2\text{b}=\text{a}+\text{c}\ \cdots(\text{i})$
And, $x, y$ and $z$ are in $G.P.$
$\therefore\text{y}^2=\text{zy}$
Now, $\text{x}^{\text{b}-\text{a}}\text{ y}^{\text{c}-\text{a}}\text{ z}^{\text{a}-\text{b}}$
$=\text{x}^{\text{b}+\text{a}-2\text{b}}\text{ y}^{2\text{b}-\text{a}-\text{a}}\text{ z}^{\text{a}-\text{b}} [$From $(i)]$
$=\text{x}^{\text{a}-\text{b}}\text{ y}^{2(\text{b}-\text{a})}\text{ z}^{\text{a}-\text{b}}$
$=(\text{xz})^{\text{a}-\text{b}}(\text{xz})^{\text{b}-\text{a}}$
$[$From $(\text{ii}),\text{y}^2=\text{xz}]$
$=(\text{xz})^0$
$=1$
View full question & answer→MCQ 61 Mark
The sum of first three terms of a $G.P.$ is $\frac{21}{2}$ and their product is $27.$ Find the common ratio.
AnswerCorrect option: C. $2$ or $\frac{1}{2}$
Let three terms be $\frac{a}{r} a, a \times r.$
Product $= 27 \Rightarrow(\frac{a}{r}) (a) (a \times r) = 27$
$\Rightarrow a^3 = 27$
$\Rightarrow a = 3.$
$\text{sum}=\frac{21}{2}$
$\Rightarrow\frac{(\text{a}{}}{\text{r}+\text{a}\times{\text{r)}}}=\frac{21}{2}$
$\Rightarrow\text{a}\Big(\frac{1}{\text{r+1+1}\times\text{r}}\Big)=\frac{21}{2}$
$\Rightarrow\Big(\frac{1}{\text{r+1+1}\times\text{r}}\Big)$
$\Rightarrow\Big({\text{r}^2}+\text{r}+1\Big)=\Big(\frac{7}{2}\Big)$
$\Rightarrow\text{r}^2-\Big(\frac{5}{2}\Big)\text{r}+1=0$
$\Rightarrow\text{r}=2$ and $\frac{1}{2}.$
View full question & answer→MCQ 71 Mark
Which of the following is true if A means arithmetic mean and b means geometric mean of two numbers?
- ✓
$\text{A}>\text{G}$
- B
$\text{A}\geq\text{G}$
- C
$\text{G}<\text{A}$
- D
$\text{A}\leq\text{G}$
AnswerCorrect option: A. $\text{A}>\text{G}$
Solution: (B) $\text{A}\geq\text{G}$
We know, A.M. of two numbers a and b is $\frac{(a+\text{b)}}{2}$
Also, G.M. of two numbers a and b is $\sqrt{ab}$
$\text{A}-\text{G}=\frac{\text{(a+2)}}{2}-=\frac{((\text{a}+b)-2\sqrt{\text{ab)}}}{2}=\frac{(\sqrt{\text{a}}-\sqrt{\text{b}})^2}{2\geq0}$
$\text{SO},\text{A}\geq\text{G}.$
View full question & answer→MCQ 81 Mark
If $a, b, c$ are in $G.P.$ and $\text{a}^{\frac{1}{\text{x}}}=\text{b}^{\frac{1}{\text{y}}}=\text{c}^{\frac{1}{\text{z}}},$ then $xyz$ are in:
Answer$a, b$ and $c$ are in $G.P.$
$\therefore\text{b}^2=\text{ac}$
Taking $\log$ on both the sides:
$2\log\text{b}=\log\text{a}+\log\text{c}\ \cdots(\text{i})$
Now, $\text{a}^{\frac{1}{\text{x}}}=\text{b}^{\frac{1}{\text{y}}}=\text{c}^{\frac{1}{\text{z}}}$
Taking $\log$ on both the sides:
$\frac{\log\text{a}}{\text{x}}=\frac{\log\text{b}}{\text{y}}=\frac{\log\text{c}}{\text{z}}\ \cdots(\text{ii} )$
Now, comparing $(i)$ and $(ii):$
$\frac{\log\text{a}}{\text{x}}=\frac{\log\text{a}+\log\text{c}}{2\text{y}}=\frac{\log\text{c}}{\text{z}}$
$\Rightarrow\frac{\log\text{a}}{\text{x}}=\frac{\log\text{a}+\log\text{c}}{2\text{y}}$ and $\frac{\log\text{a}}{\text{x}}=\frac{\log\text{c}}{\text{z}}$
$\Rightarrow\log\text{a}(2\text{y}-\text{x})=\text{x}\log\text{c}$ and $\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{\text{z}}$
$\Rightarrow\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{(2\text{y}-\text{x})}$ and $\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{\text{z}}$
$\Rightarrow\frac{\text{x}}{2\text{y}-\text{x}}=\frac{\text{x}}{\text{z}}$
$\Rightarrow2\text{y}=\text{x}+\text{z}$
Thus, $x, y$ and $z$ are in $A.P.$
View full question & answer→MCQ 91 Mark
If $3^{rd}$ term of an $A.P.$ is $6$ and $5^{th}$ term of that $A.P.$ is $12.$ Then find the $21^{st}$ term of that $A.P.$
AnswerGiven, $a_3 = 6$ and $a_5 = 12$.
$\Rightarrow a + 2d = 6$ and $a + 4d = 12$
$\Rightarrow 2d = 6$
$\Rightarrow d = 3$
View full question & answer→MCQ 101 Mark
If general term of an $A.P.$ is $3n$ then find common difference.
AnswerGiven, $a_n = 3n$.
We know, $d = a_n- a_{n-1}$
$= 3n – 3(n - 1)$
$= 3.$
View full question & answer→MCQ 111 Mark
Number of identical terms in the sequence $2, 5, 8, 11,…$ upto $100$ terms and $3, 5, 7, 9, 11, …$ upto $100$ terms, are:
View full question & answer→MCQ 121 Mark
The sum of an infinite $G.P.$ is $4$ and the sum of the cubes of its terms is $92.$ The common ratio of original $G.P.$ is:
- ✓
$\frac12$
- B
$\frac{2}{3}$
- C
$\frac13$
- D
$\frac{-1}{2}.$
AnswerCorrect option: A. $\frac12$
$\frac{\text{a}}{1-\text{r}}=3$
$\text{a}=3-3\text{r}$
Sum of square terms of $G.P.$ is $\frac{\text{a}^2}{1-\text{r}^2}=3$
$\Rightarrow\frac{\text{a}}{1-\text{r}}=\frac{\text{a}^2}{1-\text{r}^2}$
or $\text{a}=1+\text{r}\ \dots(2)$
Solving $(1)$ and $(2),$
$\text{a}=\frac32$ and $r=\frac12$
View full question & answer→MCQ 131 Mark
In $A.P. 171, 162, 153, ……….$ Find first negative term.
AnswerExplanation: $a = 171$ and $d = 162 - 171 = -9.$
$a_n < 0$
$\Rightarrow 171+(n - 1) (-9) < 0$
$\Rightarrow 180 - 9n < 0$
$\Rightarrow 9n > 180$
$\Rightarrow n > 20$
$\Rightarrow n = 21$ for first negative term.
First negative term is $171+(20) (-9) = 171 - 180 = -9$
View full question & answer→MCQ 141 Mark
If three positive numbers are inserted between $4$ and $512$ such that the resulting sequence is a $G.P.,$ which of the following is not among the numbers inserted?
AnswerLet $G.P.$ be $4, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3, 512$.
$ \Rightarrow a=4$ and $a_5=a \times r^4=512 \times 4 \times r^4=512$
$\Rightarrow r^4 $
$ =\frac{512}{4}=128$
$\Rightarrow r=4$
$ \mathrm{G}_1=\mathrm{a}_2=\mathrm{a} \times \mathrm{r}=4 \times 4=16 $
$ \mathrm{G}_2=\mathrm{G}_1 \times \mathrm{r}=16 \times 4=64 $
$ \mathrm{G}_3=\mathrm{G}_2 \times \mathrm{r}=64 \times 4=256$
View full question & answer→MCQ 151 Mark
Find the sum of series $6^2+ 7^2+…………………..+ 15^2$.
- A
$55$
- ✓
$1185$
- C
$1240$
- D
$1385$
AnswerCorrect option: B. $1185$
$6^2+7^2+\ldots \ldots \ldots \ldots \ldots \ldots+15^2 $
$=\left(1^2+2^2+3^2+\ldots \ldots \ldots+15^2\right)-\left(1^2+2^2+3^2+4^2+5^2\right) $
$=\frac{15\times16\times31}{6}-\frac{5\times6\times11}{6}$
$=1240-55$
$=1185.$
View full question & answer→MCQ 161 Mark
Find the sum of series $ 1^2+3^2+5^2+\ldots………….+ 11^2$.
Answer$1^2+3^2+5^2+\ldots…………..+ 11^2$
$=\left(1^2+2^2+3^2+\ldots \ldots+11^2\right)-\left(2^2+4^2+6^2+8^2+10^2\right) $
$=\left(1^2+2^2+3^2+\ldots . .11^2\right)-2^2\left(1^2+2^2+3^2+4^2+5^2\right)$
$=\frac{16\times12\times23}{6}-\frac{4\times5\times6\times11}{6}$
$=506-220$
$=286.$
View full question & answer→MCQ 171 Mark
If $\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}},$ then $S_n$ is equal to:
- A
$2^{\text{n}}-\text{n}-1$
- B
$1-\frac{1}{2^{\text{n}}}$
- ✓
$\text{n}-1-\frac{1}{2^{\text{n}}}$
- D
${2^{\text{n}}}-1$
AnswerCorrect option: C. $\text{n}-1-\frac{1}{2^{\text{n}}}$
We have,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1(2^{\text{r}}-1)}{2\text{r}}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\Big(1-\frac{1}{2^{\text{r}}}\Big)$
$\Rightarrow\text{S}_\text{n}=\text{n}-\sum\limits^{\text{n}}_{\text{r}=1}\Big(\frac{1}{2^{\text{r}}}\Big)$
$\Rightarrow\text{S}_\text{n}=\text{n}-\Bigg[\frac{\big(\frac{1}{2}\big)\big\{1-\big(\frac{1}{2}\big)^\text{n}\big\}}{1-\frac{1}{2}}\Bigg]$
$\Rightarrow\text{S}_\text{n}=\text{n}-\Big[1-\Big(\frac{1}{2}\Big)^\text{n}\Big]$
$\Rightarrow\text{S}_\text{n}=\text{n}-1+\frac{1}{2^\text{n}}$
View full question & answer→MCQ 181 Mark
Choose the correct answer. If $t_n$ denotes the $n^{th}$ term of the series $2 + 3 + 6 + 11 + 18 + ...$ then $t_{50}$ is:
- A
$49^2- 1$
- B
$49^2$
- C
$50^2+ 1$
- ✓
$49^2+ 2$
AnswerCorrect option: D. $49^2+ 2$
$S_n= 2 + 3 + 6 + 11 + 18 + .... + t_{50}$
Using method of difference, we get
$S_n= 2 + 3 + 6 + 11 + 18 + .... + t_{50}....(1)$
And $S_n= 0 + 2 + 3 + 6 + 11 + .... + t_{49}+ t_{50}....(2)$
Subtracting eq. $(2)$ from eq. $(2),$ we get
$0 = 2 + 1 + 3 + 5 + 7 + .... -t_{50}$ terms
$ \Rightarrow t_{50}= 2 + (1 + 3 + 5 + 7 + .... $ upto $49$ terms$)$
$\Rightarrow\text{t}_{50}=2+\frac{49}{2}[2\times1+(49-1)2]=2+\frac{49}{2}[2+96]$
$\Rightarrow2+\frac{49}{2}\times98=2+49\times49=49^2+2$
Hence, the correct option is $(d).$
View full question & answer→MCQ 191 Mark
If $\text{A.M.}$ of two numbers is $\frac{15}{2}$ and their $\text{G.M.}$ is $6,$ then find the two numbers.
- A
$6$ and $8$
- ✓
$12$ and $3$
- C
$24$ and $6$
- D
$27$ and $3$
AnswerCorrect option: B. $12$ and $3$
We know, $\text{A.M.}$ of two numbers $a$ and $b$ is
$\frac{(\text{a}+\text{b)}}{2}$
$\Rightarrow\frac{(\text{a}+\text{b)}}{2}=\frac{15}{2}$
$\Rightarrow\text{}a+\text{b}=15.$
Also, $\text{G.M.}$ of two numbers $a$ and $b$ is $\sqrt{ab}$
$\Rightarrow \sqrt{ab}=6$
$\Rightarrow\text{ab}=36.$
$\Rightarrow a(15-a) = 36$
$\Rightarrow a=3$ or $12.$
For $a=3, b=12.$
For $a=12, b=3.$
So, the two numbers are $3$ and $12.$
View full question & answer→MCQ 201 Mark
If first term of a $\text{G.P.}$ is $20$ and common ratio is $4$. Find the $5^{th}$ term.
- A
$10240$
- B
$40960$
- ✓
$5120$
- D
$2560$
AnswerCorrect option: C. $5120$
Given, $a = 20$ and $r = 4.$
We know, $a_n= ar^{n-1}$
$\Rightarrow a_5= 20 \times 4^4= 20 \times 256 = 5120.$
View full question & answer→MCQ 211 Mark
The $n^{th}$ term of a $\text{G.P.}$ is $128$ and the sum of its $n$ terms is $225.$ If its common ratio is $2,$ then its first term is:
AnswerLet the firt term of the geometric progression $= x$
Common ration $= 2$
$\therefore 2^{nd}$ term of the $\text{G.P. = 2x}$
$\therefore 3^{rd}$ term $ = (2^2)x ...$
$N^{th}$ term can be written as $= (2^\text{n})\text{x}$
Sum of the $n$ terms $S = 255$
as we can see, except $x,$ all other terms in the $\text{G.P.}$ are multiples of $2$
and sum of all the terms is an odd number.
$\therefore\ x$ must be an odd number.
now $n^{th}$ term
$(2^{\text{n}})\text{x}=128=\big(2^7\big)\times1$
There are no factors of odd numbers in $128,$ except $1$
$\therefore x = 1$
Series of $\text{G.P}.$ is:
$1, 2, 4, 8, 16, 32, 64, 128$
Checking the sum of the $n$ terms,
$1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$
$\therefore$ First term of the $\text{G.P}. = 1$
View full question & answer→MCQ 221 Mark
Find the sum of squares of first $n$ terms.
- A
$\frac{\text{n}\text{(n}+1)}{2}$
- B
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^3$
- ✓
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
- D
$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
Sum of squares of first $n$ terms $= 1^2+2^2+3^2+……………+n^2$
$k^3–(k – 1)^3=3k^2–3k + 1$
On substituting $k = 1, 2, 3, ……, n$ and adding we get,
$\text{n}^3=\sum\limits^\text{n}_\text{i}= 0 \text{ k}^2=\sum\limits^\text{n}_\text{i}=0\text{ k} +\text{n}$
$\text{n}^3=3\sum\limits^2_\text{i}=0 \text{ k}^2-3\frac{\text{n}(\text{n}+1)}{2}+\text{n}$
$\sum\limits^\text{n}_\text{i}=0\text{ k}^2=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}.$
View full question & answer→MCQ 231 Mark
A sequence is called $...............$ if $a_{n+1}= a_n\times r.$
AnswerExplanation: $A$ sequence is called geometric progression if $a_{n+1}$
$= a_n * r$ where $a_1$ is the first term and $r$ is common ratio.
View full question & answer→MCQ 241 Mark
If $a=3$ and $r=2$ then find the sum up $5^{th}$ term.
AnswerWe know, $\text{s}_\text{n}=a\frac{\text({r^\text{n}-1)}}{\text({r}-1)}$
Here $a = 3, r = 2$ and $n = 5$
$\text{s}_5=3\frac{(2^5-1)}{(2-1)}$
$=3(32-1)$
$=3\times31$
$=93.$
View full question & answer→MCQ 251 Mark
A sequence is called $..........$ if $a_{n+1}= a_n\times r.$
AnswerExplanation: A sequence is called geometric progression if $a_{n+1}$
$= a_n * r$ where $a_1$ is the first term and $r$ is common ratio.
View full question & answer→MCQ 261 Mark
The sum of the series $\frac{1}{\log_24}+\frac{1}{\log_44}+\frac{1}{\log_84}+\ ...\ +\frac{1}{\log_2{^\text{n}4}}\text{ is}:$
- A
$\frac{\text{n}(\text{n}+1)}{2}$
- B
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{12}$
- ✓
$\frac{\text{n}(\text{n}+1)}{4}$
- D
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)}{4}$
Let $\text{S}_\text{n}=\frac{1}{\log_24}+\frac{1}{\log_44}+\frac{1}{\log_84}+\ ...\ +\frac{1}{\log_2{^\text{n}4}}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log4}{\log4}+\frac{\log8}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log2^2}{\log4}+\frac{\log2^3}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{2\log2}{\log4}+\frac{3\log2}{\log4}+\ ...\ +\frac{\text{n}\log2}{\log4}$
$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{\log4^{\frac{1}{2}}}{\log4}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{\frac{1}{2}\log4}{\log4}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{1}{2}(1+2+3+\ ...\ +\text{n})$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}$
View full question & answer→MCQ 271 Mark
Let $\text{ so}=\frac{8}{5}+\frac{16}{65}.....+\frac{128}{2}^{18}+1$:
- ✓
$\text{s}=\frac{1088}{545}$
- B
$\text{s}=\frac{1088}{545}$
- C
$\text{s}=\frac{1056}{545}$
- D
$\text{s}=\frac{545}{1056}$
AnswerCorrect option: A. $\text{s}=\frac{1088}{545}$
View full question & answer→MCQ 281 Mark
Find sum of series $2 + 3 + 5 + 7.$
AnswerSum of the series $2 + 3 + 5 + 7$ is finite because given series has finite number of terms.
The sum of given $4$ terms i.e. $17.$
View full question & answer→MCQ 291 Mark
If $x$ is psitive, the sum to in $\frac{1}{1+\text{x}}-\frac{1-\text{x}}{(1+\text{x})^2}+\frac{(1-\text{x})^2}{(1+\text{x})^3}-\frac{(1-\text{x})^3}{(1+\text{x})^4}+\cdots\text{ is:}$
- ✓
$\frac{1}{2}$
- B
$\frac{3}{4}$
- C
$1$
- D
AnswerCorrect option: A. $\frac{1}{2}$
$\text{S}=\frac{1}{1+\text{x}}-\frac{1-\text{x}}{(1+\text{x})^2}+\frac{(1-\text{x})^2}{(1+\text{x})^3}-\frac{(1-\text{x})^3}{(1+\text{x})^4}+\cdots\infty$
It is clear that it is a $G.P.$ with $\text{a}=\frac{1}{1+\text{x}}$ and $\text{r}=-\frac{(1-\text{x})}{(1+\text{x})}.$
$\therefore\text{S}=\frac{\text{a}}{(1-\text{r})}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1-\big(-\frac{(1-\text{x})}{(1+\text{x})}\big)\Big]}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1+\frac{(1-\text{x})}{(1+\text{x})}\Big]}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[\frac{(1+\text{x})+(1-\text{x})}{(1+\text{x})}\Big]}$
$\Rightarrow\text{S}=\frac12$
View full question & answer→MCQ 301 Mark
If $a, b$ and $c$ are in $AP$ and $p, p¢$ are the $AM$ and $GM$ respectively between $a$ and $b$, while $q, q¢$ are the $AM$ and $GM$ respectively between $b$ and $c,$ then:
AnswerCorrect option: C. $p^2-q^2=p^2-q^2 $
View full question & answer→MCQ 311 Mark
Choose the correct answer. Let $S_n$ denote the sum of the cubes of the first $n$ natural numbers and $s_n$ denote the sum of the first $n$ natural numbers. Then $\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}$ equals to:
- ✓
$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
- B
$\frac{\text{n}(\text{n}+1)}{2}$
- C
$\frac{\text{n}^{2}+3\text{n}+2}{2}$
- D
AnswerCorrect option: A. $\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
$\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\sum\limits^\text{n}_{\text{r}=1}\frac{\big[\frac{\text{r}(\text{r}+1)}{2}\big]^2}{\frac{\text{r}(\text{r}+1)}{2}}$
$=\sum\limits^\text{n}_{\text{r}=1}\frac{\text{r}(\text{r}+1)}{2}$
$=\frac{1}{2}\bigg[\sum\limits^\text{n}_{\text{r}=1}\text{r}^2+\sum\limits^\text{n}_{\text{r}=1}\text{r}\bigg]$
$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}{+1})(2\text{n}{+1})}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=\frac{1}{2}.\frac{\text{n}({\text{n}+1})}{2}\Big[\frac{2\text{n}{+1}}{3}+1\Big]$
$=\frac{\text{n}(\text{n}+1)}{4}\Big[\frac{2\text{n}+4}{3}\Big]$
$=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
View full question & answer→MCQ 321 Mark
What is $n^{th}$ term of a $G.P.?$
- A
$a_n=a+(n-1) d $
- B
$a_n=a+(n) d $
- ✓
$a_n=a \times r^{n-1} $
- D
$a_n=a \times r^n $
AnswerCorrect option: C. $a_n=a \times r^{n-1} $
Since every term of $a_n\ G.P.$ is $r$ times the previous term.
i.e. $a_{n+1}= a_n* r = a_{n-1} * r^2= ….. = a_1* r^n$
or $a_n= a*r^{n-1}$
View full question & answer→MCQ 331 Mark
Choose the correct answer. Let $S_n$ denote the sum of the first $n$ terms of an $A.P.$ If $S_{2n}= 3S_n,$ then $S_{3n} : S_n$ is equal to:
AnswerLet the first term be a and common difference be $d.$
Then,
$\text{S}_{2\text{n}}=3\text{S}_{\text{n}}$
$\Rightarrow\frac{2\text{n}}{2}[2\text{a}+(2\text{n}-1)\text{d}]=\frac{3\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow2[2\text{a}+(2\text{n}-1)\text{d}]=3[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow4\text{a}+(4\text{n}-2)\text{d}=6\text{a}+(3\text{n}-3)\text{d}$
$\Rightarrow2\text{a}=(\text{n}+1)\text{d}$
Now,
$\frac{\text{S}_{3\text{n}}}{\text{S}_{\text{n}}}$
$=\frac{\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}$
$=\frac{3[2\text{a}+(3\text{n}-1)\text{d}]}{[2\text{a}+(\text{n}-1)\text{d}]}$
$=\frac{3[(\text{n}+1)\text{d}+(3\text{n}-1)\text{d}]}{[(\text{n}+1)\text{d}+(\text{n}-1)\text{d}]}$
$=\frac{3[4\text{nd}]}{2\text{nd}}$
$=6$
View full question & answer→MCQ 341 Mark
If in an $A.P.,$ first term is $20,$ common difference is $2$ and $n^{th}$ term is $42,$ then find sum up to $n$ terms.
AnswerWe know, $a = 20, d = 2, a_n= 42.$
$a + (n - 1) d = 42$
$\Rightarrow 20 + 2(n - 1) = 42$
$\Rightarrow 2 (n - 1) = 42 - 20 = 22$
$\Rightarrow n - 1 = 11$
$\Rightarrow n = 12.$
$\text{s}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{I})$
$\Rightarrow\text{s}=\frac{12}{2}(20+42)=6\times62=372.$
View full question & answer→MCQ 351 Mark
If a sequence is in the form $2 \times 5n$ then which of the following may be the sequence?
AnswerIf $a_n = 2 \times 5n$ then
$a_1= 10, a_2= 50, a_3= 250.$
This is a geometric progression with first term $10$ and common ratio $5.$
View full question & answer→MCQ 361 Mark
If general term of an $A.P.$ is $3n$ then find common difference.
AnswerGiven, $a_n= 3n.$
We know, $d = a_n-a_{n-1}= 3n – 3(n-1) = 3.$
View full question & answer→MCQ 371 Mark
If $a, b, c$ are in $G.P.$ is $2$ and $x, y$ are $AM's$ between $a, b$ and $b, c$ respectively, then:
- A
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=2$
- B
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{2}$
- C
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{a}}$
- ✓
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{b}}.$
AnswerCorrect option: D. $\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{b}}.$
$a, b$ and $c$ are in $G.P.$
$\therefore\text{b}^2=\text{ac}\cdots(\text{i})$
$a, x$ and $b$ are in $A.P.$
$\therefore2\text{x}=\text{a}+\text{b}\ \cdots(\text{ii})$
Also, $b, y$ and $c$ are in $A.P.$
$\therefore2\text{y}=\text{b}+\text{c}$
$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}^2}{\text{a}} [$Using $(i)]$
$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}^2}{(2\text{x}-\text{b})} [$Using $(ii)]$
$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}(2\text{x}-\text{b})+\text{b}^2}{(2\text{x}-\text{b})}$
$\Rightarrow2\text{y}=\frac{2\text{bx}-\text{b}^2+\text{b}^2}{(2\text{x}-\text{b})}$
$\Rightarrow2\text{y}=\frac{2\text{b}\text{x}}{(2\text{x}-\text{b})}$
$\Rightarrow\text{y}=\frac{\text{bx}}{(2\text{x}-\text{b})}$
$\Rightarrow\text{y}(2\text{x}-\text{b})=\text{bx}$
$\Rightarrow2\text{xy}-\text{by}=\text{bx}$
$\Rightarrow\text{bx}+\text{by}=2\text{xy}$
Dividing both the sides by $xy:$
$\Rightarrow\frac{1}{\text{y}}+\frac{1}{\text{x}}=\frac{2}{\text{b}}$
View full question & answer→MCQ 381 Mark
The sum of $n$ terms of the infinite series $1.3^2+ 2.5^2+ 3.7^2 …\infty $ is:
AnswerCorrect option: A. $ n / 6(n+1)\left(6 n^2+14 n+7\right) $
View full question & answer→MCQ 391 Mark
After striking the floor, a certain ball rebounds $\Big(\frac{4}{5}\Big)^{th}$ of height from which it has fallen. Then, the total distance that it travels before coming to rest, if it is gentlydropped from a height of $120 m$ is:
- A
$1260 m$
- B
$600 m$
- ✓
$1080 m$
- D
AnswerCorrect option: C. $1080 m$
View full question & answer→MCQ 401 Mark
The fractional value of $2.357$ is:
- A
$\frac{2355}{1001}$
- B
$\frac{2379}{997}$
- ✓
$\frac{2355}{999}$
- D
AnswerCorrect option: C. $\frac{2355}{999}$
$2.\overline{357}=2.0+0.357+0.000357+0.000000357+\dots\infty$
$\Rightarrow2.\overline{357}=2+\Big[\frac{357}{10^3}+\frac{357}{10^6}+\frac{357}{10^9}+\dots\infty\Big]$
$\Rightarrow2.\overline{357}=2+\frac{\frac{357}{10^3}}{1-\frac{1}{10^3}}$
$\Rightarrow2.\overline{357}=2+\frac{357}{999}$
$\Rightarrow2.\overline{357}=\frac{2355}{999}$
View full question & answer→MCQ 411 Mark
Find the sum of cubes of first $n$ terms
- A
$\frac{\text{n}(\text{n+1)}}{2}$
- B
$\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^3$
- ✓
$\frac{\text{n}(\text{n+1)}(2\text{n}+1}{6}$
- D
$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
AnswerCorrect option: C. $\frac{\text{n}(\text{n+1)}(2\text{n}+1}{6}$
Sum of cubes of first $n$ terms $=1^3+2^3+3^3+.................+n^3$
$(k+1)^4-k^4=4 k^3+6 k^2+4 k+1$
On substituting $k = 1, 2, 3, ……, n$ and adding we get,
$4\text{n}^3+\text{n}^4+6\text{n}^2+4\text{n}=4$
$\text{n}=4\sum\limits^\text{n}_\text{i}=0\text{ k }^3+6\sum\limits^\text{n}_\text{i}=0\text{ k}^2+4\sum\limits^\text{n}_\text{i}=0\text{ k}+\text{n}$
$4\text{n}^3+\text{n}^4+6\text{n}^2+4\text{n}=4$
$=4\sum\limits^\text{n}_\text{i}=0\text{ k }^3+6\frac{(\text{n}(\text{n}+1)(2\text{n+1))}}{6}+4\frac{\text{n}(\text{n}+1)}{2}+\text{n}\sum\limits^\text{n}_\text{i}=0\text{ k}^\text{n}_\text{i}=0\text{ k}^3=\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2.$
View full question & answer→MCQ 421 Mark
The sixth term of an $\text{AP}$ is equal to $2.$ The value of the common difference of the $\text{AP}$ which makes the product $\text{T1 T4 T5}$ least, is given by:
AnswerCorrect option: C. $\frac{2}{3}$
View full question & answer→MCQ 431 Mark
The first two terms of a geometric progression add upto $12.$ The sum of the third and the fourth terms is $48.$ If the terms of the geometric progression are alternately positive and negative, then first term is:
View full question & answer→MCQ 441 Mark
Which of the following is the geometric mean of $3$ and $12.$
AnswerWe know, geometric mean of two numbers $a$ and $b$ is given by
$\text{G}.\text{M}.=\sqrt{\text{a}\times\text{b}}$
SO, $\text{G}.\text{M}.$ of $3$ and $12$ is $\sqrt{3}\times12=\sqrt{36}=6.$
View full question & answer→MCQ 451 Mark
The solution of the equation $(x + 1) + (x + 4) + (x + 7) + …+ (x + 28) = 155$ is:
View full question & answer→MCQ 461 Mark
If an increasing $\text{GP}$ is considered, then the number of terms in $\text{GP}$ is:
View full question & answer→MCQ 471 Mark
$8, 24, 48, 80, 120, .....:$
AnswerDifference of two successive numbers are $16, 24, 32, 40$ etc Hence the next number is $120 + 48 = 168$
View full question & answer→MCQ 481 Mark
Find the sum of series $\frac{1+1}{2}+\frac{1}{4}+……….$ up to $6$ term
- ✓
$\frac{63}{32}$
- B
$\frac{32}{63}$
- C
$\frac{26}{53}$
- D
$\frac{53}{26}$
AnswerCorrect option: A. $\frac{63}{32}$
Given series is $\text{G.P.}$ with first term $1$ and common ratio $\frac{1}{2}.$
We know,$\text{s}_{\text{n}}=\text{a}\frac{(1-\text{r}_\text{n})}{(1-\text{r)}}$ for $r<1.$
${\text{s}}_6=1\frac{(1-(\frac{1}{2})^6}{(1-\frac{1}{2})}$
$=\frac{(1-\frac{1}{64})}{(1+2)}$
$=63\times\frac{2}{64}$
$=\frac{63}{32}.$
View full question & answer→MCQ 491 Mark
What is the first term of Fibonacci sequence?
Answer$a_1=1 $ and $a_2=1 $
$a_n=a_{n-1}+a_{n-2}, n>2 $
This is a recurrence relation which gives the Fibonacci sequence.
View full question & answer→MCQ 501 Mark
$1 + 2 + 3 + 4$ or $10$ is a series?
AnswerCorrect option: A. $1 + 2 + 3 + 4$ only
$1 + 2 + 3 + 4$ is a finite series of $4$ terms.
$10$ is sum of the terms of this series not a series itself.
View full question & answer→MCQ 511 Mark
In any case, the difference of the least and greatest term is:
MCQ 521 Mark
The product of $n$ positive numbers is unity. Their sum is:
AnswerCorrect option: D. never less than $n$
View full question & answer→MCQ 531 Mark
Choose the correct answer: The minimum value of $4^{\text{x}}+4^{1-\text{x}},\text{x}\in\text{R}$ is:
AnswerWe know that $\text{AM}\geq\text{GM}$
$\therefore\ \frac{4^\text{x}+4^{1-\text{x}}}{2}\geq\sqrt{4^\text{x}.4^{1-\text{x}}}$
$\Rightarrow4^\text{x}+4^{1-\text{x}}\geq2\sqrt{4^{\text{x}+1-\text{x}}}$
$4^\text{x}+4^{1-\text{x}}\geq2.2$
$\Rightarrow4^\text{x}+4^{1-\text{x}}\geq4$
Hence, the correct option is $(b).$
View full question & answer→MCQ 541 Mark
If $\log_{ax},\log_{bx}, \log_{cx}$ be in $HP,$ then $a, b, c$ are in:
View full question & answer→MCQ 551 Mark
If $a + 2b + 3c = 12 , (a, b, c \in R+),$ then $ab^2c^3$ is:
AnswerCorrect option: C. $\leq2^6$
View full question & answer→MCQ 561 Mark
A sequence is called $..........$ if $a_{n+1} = a_n + d$.
AnswerA sequence is called arithmetic progression if $a_{n+1} = a_n + d$ where $a_1$ is the first term and $d$ is common difference.
View full question & answer→MCQ 571 Mark
Sumof nterms of series $12 + 16 + 24 + 40 + …$ will be:
- A
$2(2n – 1) + 8n$
- B
$2(2n – 1) + 6 n$
- C
$3(2n – 1) + 8n$
- ✓
$4(2n – 1) + 8n$
AnswerCorrect option: D. $4(2n – 1) + 8n$
View full question & answer→MCQ 581 Mark
If $Tn$ denotes the $n^{th}$ term of the series $2 + 3 + 6 + 11 + 18 + . . . ,$ then $T_{50}$ is:
- A
$49 2 - 1$
- B
$492$
- C
$502 + 1$
- ✓
$492 + 2$
AnswerCorrect option: D. $492 + 2$
View full question & answer→MCQ 591 Mark
How many terms of $\text{G.P.}\ 2,4,8,16, ………$ are required to give sum $254?$
Answer$=2$ and
$r=\frac{4}{2}=2.$
We know $\text{s}_\text{n}=\text{a}\frac{(\text{r}^\text{n}-1)}{\text{(r}-1)}$
$2\frac{(2^\text{n-1)}}{(2-1)}=254$
$\Rightarrow 2^n- 1 = 127$
$\Rightarrow 2^n = 128 = 2^7$
$\Rightarrow n = 7.$
View full question & answer→MCQ 601 Mark
Find the sum of series $6^3+ 7^3+…..…..+ 20^3$.
- ✓
$43875$
- B
$83775$
- C
$43775$
- D
$43975$
AnswerCorrect option: A. $43875$
$6^3+ 7^3+………..+ 20^3$
$= (1^3+ 2^3+ 3^3+……..+ 20^3) – (1^3+ 2^3+ 3^3+ 4^3+ 5^3)$
$=\Big(\frac{20\times21}{2}\Big)^2-\Big(\frac{5\times6}{2}\Big)^2$
$=(210)^2-(15)^2$
$=225\times195$
$=43875$
View full question & answer→MCQ 611 Mark
$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day, $4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is:
View full question & answer→MCQ 621 Mark
For an increasing $A.P.\ a_1, a_2, a_3..... a_n,$ if $a_1, a_3, a_5 = – 12$ and $a_1. a_3. a_5 = 80,$ then which of the following is/are true?
- ✓
$a_1 = -10$
- B
$a_2 = -1$
- C
$a_3= – 4$
- D
$a_5 = – 2 (a,c,d)$
AnswerCorrect option: A. $a_1 = -10$
View full question & answer→MCQ 631 Mark
If in an $A.P.,$ first term is $20,$ common difference is $2$ and $n^{th}$ term is $42,$ then find $n.$
AnswerWe know, $a = 20, d = 2, a_n = 42.$
$a + (n - 1) d = 42 20$
$\Rightarrow + 2(n - 1) = 42$
$\Rightarrow 2 (n - 1) = 42 - 20 = 22$
$\Rightarrow n - 1 = 11$
$\Rightarrow n = 12.$
View full question & answer→MCQ 641 Mark
Let $S$ be the sum, $P$ be the product and $R$ be the sum of the reciprocals of $3$ terms of a $G.P.$ then $\ce{P^2R^3 : S^3}$ is equal to:
AnswerCorrect option: A. $1 : 1$
Let the three terms of the $G.P.$ be $\frac{\text{a}}{\text{r}},\text{a},\text{ar}.$ Then
$\text{S}=\frac{\text{a}}{\text{r}}+\text{a}+\text{ar}$
$=\text{a}\Big(\frac{1}{\text{r}}+1+\text{r}\Big)$
$=\text{a}\Big(\frac{1+\text{r}+\text{r}^2}{\text{r}}\Big)$
$=\frac{\text{a}(\text{r}^2+\text{r}+1)}{\text{r}}$
Also,
$\text{P}=\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=\text{a}^3$
And,
$\text{R}=\frac{\text{r}}{\text{a}}+\frac{1}{\text{a}}+\frac{1}{\text{ar}}$
$=\frac{1}{\text{a}}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)$
Now,
$\frac{\text{P}^2\text{R}^2}{\text{S}^3}=\frac{\big(\text{a}^3\big)^2\times\Big[\frac{1}{\text{a}}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)\Big]^3}{\Big[\text{a}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)\Big]^3}$
$=\frac{\text{a}^6\times\frac{1}{\text{a}^3}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)^3}{\text{a}^3\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)^3}$
$=\frac11$
So, the ratio is $1 : 1.$
Hence, the correct alternative is option $(a).$
View full question & answer→MCQ 651 Mark
If the first term of a $\ce{G.P. a_1, a_2, a_3, ...}$ is unity such that $4 a_2+ 5a_3$ is least, then common ratio of $\text{G.P.}$ is:
- ✓
$\frac{-2}{5}$
- B
$\frac{-3}{5}$
- C
$\frac25$
- D
AnswerCorrect option: A. $\frac{-2}{5}$
If the first term is $1,$ then, the $\text{G.P.}$ will be $1, \text{r},\text{r}^2,\text{r}^3,\ \dots$
Now, $5\text{r}^2+4\text{r}=5\Big(\text{r}^2+\frac45\text{r}\Big)$
$=5\Big(\text{r}^2+\frac45\text{r}+\frac{4}{25}-\frac{4}{25}\Big)$
$=5\Big(\text{r}+\frac25\Big)^2-\frac45$
This will be the least when $\text{r}+\frac25=0,$
i.e. $\text{r}=-\frac25.$
View full question & answer→MCQ 661 Mark
Find the sum of series $1^3+ 3^3+ 5^3+………….+ 11^3$.
- ✓
$2556$
- B
$5248$
- C
$6589$
- D
$9874$
AnswerCorrect option: A. $2556$
$1^3+ 3^3+ 5^3+……………..+ 11^3$
$= (1^3+ 2^3+ 3^3+……+ 11^3) – (2^3+ 4^3+ 6^3+ 8^3+ 10^3)$
$= (1^3+ 2^3+ 3^3+……11^3) – 2^3(1^3+ 23 + 3^3+ 4^3+ 5^3)$
$\Big(\frac{11\times12}{2}\Big)^2-8\Big(\frac{5\times6}{2}\Big)^2$
$=66^2-8\times15^2$
$4356-1800=2556.$
View full question & answer→MCQ 671 Mark
The ratio of the $A.M.$ and $G.M.$ of two positive numbers $a$ and $b$ is $5 : 3.$ Find the ratio of $a$ to $b.$
- ✓
$9 : 1$
- B
$3 : 5$
- C
$1 : 9$
- D
$3 : 1$
AnswerCorrect option: A. $9 : 1$
$\frac{(\text{A.}\text{M.})}{\text{(G.}\text{M.)}}=\frac{5}{3}$
$\Rightarrow\frac{\text{a}+\text{b}}{2\sqrt{a}\text{b}}=\frac{5}{3}$
Applying componendo and dividendo rule, we get
$\Rightarrow\frac{\text{a}+\text{b+2}\sqrt{\text{a}\text{b}}}{\text{a}+\text{b}-2\sqrt{a}\text{b}}=\frac{8}{2}$
$\Rightarrow\Big(\frac{\sqrt{\text{a}+\text{b}}}{\sqrt{\text{a}-\text{b}}}\Big)^2=4$
$\Rightarrow\Big(\frac{\sqrt{\text{a}+\text{b}}}{\sqrt{\text{a}-\text{b}}}\Big)^1=2$
$\Big(\frac{\sqrt{\text{a}}}{\sqrt{\text{b}}}\Big)^1=3$
Again applying componendo and dividendo rule, we get
$\frac{\text{a}}{\text{b}}=3\Big(\frac{3}{1}\Big)^2=9.$
$\text{ so},\text{a}:\text{b}=9:1$
View full question & answer→MCQ 681 Mark
Choose the correct answer. If x, 2y and 3z are in A.P. where the distinct numbers x, y and z are in G.P., then the common ratio of the G.P. is:
- A
$3$
- B
$\frac{1}{3}$
- C
$2$
- ✓
$\frac{1}{2}$
AnswerCorrect option: D. $\frac{1}{2}$
Solution: (D) $\frac{1}{2}.$
Since, x, 2y and 3z are in A. P., we get
$\text{2y}=\frac{\text{x}+3\text{z}}{2}$
⇒ 4y = x + 3z
Also, x, y and z are ibn G.P.
Therefore, y = xr and $z = xr^2$.
Where 'r' is the common ratio.
$\therefore$ $4xr = x + 3xr^2$ [Using (1)]
$⇒ 4r = 1 + 3r^2$
$⇒ 3r^2 - 4r + 1 = 0$
$⇒ (3r - 1)(r - 1) = .0$
$\Rightarrow\text{r}=\frac{1}{3}$
(For r = 1; x, y, z are not distinct)
View full question & answer→MCQ 691 Mark
Find the sum to $6$ terms of each of the series $2*3+4*6+6*11+8*18+………..$
AnswerGeneral term of above series is $a_k = 2k*(k^2+2) = 2k^3+4k$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=2\sum\limits^\text{n}_\text{i}=0\text{ k}^3+4\sum\limits^\text{n}_\text{i}=0\text{ k}$
$=2\Big(\frac{(\text{n}(\text{n+1)}}{2}\Big)^2+4\frac{\text{n}(\text{n+1)}}{2}$
$=\text{n}^2\frac{(\text{n}+1)}{2}+2\text{n}(\text{n+1)}$
$=\frac{36\times49}{2}+2\times6\times7$
$=966.$
View full question & answer→MCQ 701 Mark
Find the sum $1^3+2^3+3^3+………+8^3$.
- A
$1225$
- B
$1184$
- C
$1475$
- ✓
$1296$
AnswerCorrect option: D. $1296$
We know, sum of cubes of first $n$ terms is given by $\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)^2.$
$\text{Here},\text{n}=8$
$\text{ so},\text{sum}=\Big(\frac{8\times9}{2}\Big)^2=1296.$
View full question & answer→MCQ 711 Mark
Sum of $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ ...\text{ is}$
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
Let $T_N$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=\sqrt{2\times\text{n}^2}=\text{n}\sqrt{2}$
Now, let $S_n$ be the sum of $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sqrt{2}\sum\limits^{\text{n}}_{\text{k}=1}(\text{k})$
$\Rightarrow\text{S}_\text{n}=\sqrt{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
View full question & answer→MCQ 721 Mark
The sum of the infinity of the series $1+2/3 + 6/32 + 10/33 +14/34$ is:
View full question & answer→MCQ 731 Mark
Choose the correct answer. If $9$ times the $9^{th}$ term of an $A.P.$ is equal to $13$ times the $13^{th}$ term, then the $22^{nd}$ term of the $A.P.$ is:
AnswerLet the first term and common difference of given $A.P.$ be $a$ and $d$, respectively.
It is given that $9 \times t_9= 13 \times t_{13}$
$\Rightarrow 9(a + 8d) = 13(a + 12d)$
$\Rightarrow 9a + 72d = 13a + 156d$
$\Rightarrow 4a + 84d = 0$
$\Rightarrow 4(a + 21d) = 0$
$\Rightarrow t_{22}= 0$
View full question & answer→MCQ 741 Mark
Let $x$ be the $A.M.$ and $y, z$ be two $G.M.s$ between two positive numbers. Then, $\frac{\text{y}^3+\text{z}^3}{\text{xyz}}$ is equal to:
AnswerLet the two numbers be $a$ and $b.$
$a, x$ and $b$ are in $A.P.$
$\therefore2\text{x}=\text{a}+\text{b}\ \cdots(\text{i})$
Also, $a, y, z$ and $b$ are in $G.P.$
$\therefore\frac{\text{y}}{\text{a}}=\frac{\text{z}}{\text{y}}=\frac{\text{b}}{\text{z}}$
$\Rightarrow\text{y}^2=\text{az},\text{yz}=\text{ab},\text{z}^2=\text{by}\ \cdots(\text{ii})$
Now, $\frac{\text{y}^3+\text{z}^3}{\text{xyz}}$
$=\frac{\text{y}^2}{\text{xz}}+\frac{\text{z}^2}{\text{xy}}$
$=\frac{1}{\text{x}}\Big(\frac{\text{y}^2}{\text{z}}+\frac{\text{z}^2}{\text{y}}\Big)$
$=\frac{1}{\text{x}}\Big(\frac{\text{az}}{\text{z}}+\frac{\text{by}}{\text{y}}\Big) [$Using $(ii)]$
$=\frac{1}{\text{x}}(\text{a}+\text{b})$
$=\frac{2}{(\text{a}+\text{b})}(\text{a}+\text{b}) [$Using $(i)]$
$=2$
View full question & answer→MCQ 751 Mark
$a_1 = a_2= 2, a_n = a_n- 1 - 1, n > 2$. Find $a_5$.
Answer$ \Rightarrow a_n=a_n-1-1, n>2 $
$ \Rightarrow a_3=a_2-1=2-1=1 $
$ \Rightarrow a_4=a_3 1=11=0 $
$\Rightarrow a_5=a_4-1=0-1=-1 $
View full question & answer→MCQ 761 Mark
If a be $\text{A.M.}$ and $p, q$ be two $\text{G.M.'s}$ between two numbers, then $2A$ is equal to:
- ✓
$\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
- B
$\frac{\text{p}^3-\text{q}^3}{\text{pq}}$
- C
$\frac{\text{p}^2+\text{q}^2}{2}$
- D
$\frac{\text{pq}}{2}.$
AnswerCorrect option: A. $\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
Let the two positive numbers be $a$ and $b.$
$a, A $ and $b$ are in $A.P.$
$\therefore2\text{A}=\text{a}+\text{b}\cdots(\text{i})$
Also, $a, p, q$ and $b$ are in $G.P.$
$\therefore\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}$
Again, $p = ar$ and $q = ar^2........(ii)$
Now, $2A = a + b [$ From $(i)]$
$=\text{a}+\text{a}\Big(\frac{\text{b}}{\text{a}}\Big)$
$=\text{a}+\text{a}\Bigg(\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}\Bigg)^3$
$=\text{a}+\text{ar}^3$
$=\frac{(\text{ar})^2}{\text{ar}^2}+\frac{\big(\text{ar}^2\big)^2}{\text{ar}}$
$=\frac{\text{p}^2}{\text{q}}+\frac{\text{q}^2}{\text{p}} [$Using $(ii)]$
$=\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
View full question & answer→MCQ 771 Mark
Jairam purchased a house in $Rs. 15000$ and paid $Rs. 5000$ at once. Rest money he promised to pay in annual instalment of $Rs. 1000$ with $10\%$ per annum interest. How much money is to be paid by Jairam?
- A
$Rs. 21555$
- B
$Rs. 20475$
- ✓
$Rs. 20500$
- D
$Rs. 20700$
AnswerCorrect option: C. $Rs. 20500$
View full question & answer→MCQ 781 Mark
Choose the correct answer. If in an $A.P., S_n= qn^2$ and $S_m = qm^2$, where $S_r$ denotes the sum of $r$ terms of the $AP,$ then $S_q$ equals:
- A
$\frac{\text{q}^3}{2}$
- B
$\text{mnq}$
- ✓
$q^3$
- D
$(m + n)q^2$
AnswerGiven,
$S n=q n^2$ and $S m=q m^2 $
$\therefore S_1=q, S_2=4 q, S_3=9 q$ and $S_4=16 q$
Now, $\mathrm{t}_1=\mathrm{q}$
$\therefore \mathrm{t}_2=S_2-S_1=4 q-q=3 q $
$t_3=S_3-S_2=9 q-4 q=5 q $
$ t_4=S_4-S_3=16 q-9 q=7 q$
So, the $A.P.$ is: $q, 3q, 5q, 7q, ....$
Thus, first term is $q$ and common difference is $3q - q = 2q.$
$\therefore\ \text{S}_\text{q}=\frac{\text{q}}{2}[2\times\text{q}+(\text{q}-1)2\text{q}]=\frac{\text{q}}{2}\times[2\text{q}+2\text{q}^2-2\text{q}]$
$=\frac{\text{q}}{2}\times2\text{q}^2$
$=\text{q}^3$
View full question & answer→MCQ 791 Mark
Given that $x > 0,$ the sum $\sum\limits^\infty_{\text{n}=1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{\text{n}-1}$ equals:
AnswerCorrect option: B. $\text{x}+1$
$\sum\limits^\infty_{\text{x}+1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{(\text{n}-1)}$
$=1+\Big(\frac{\text{x}}{\text{x}+1}\Big)+\Big(\frac{\text{x}}{\text{x}+1}\Big)^2+\Big(\frac{\text{x}}{\text{x}+1}\Big)^3+\Big(\frac{\text{x}}{\text{x}+1}\Big)^4+\dots\infty$
$=\frac{1}{1-\big(\frac{\text{x}}{\text{x}+1}\big)}$ $\Big[\because$ it is a $G.P.$ with $a = 1$ and $\text{r}=\Big(\frac{\text{x}}{\text{x}+1}\Big)\Big]$
$=\frac{(\text{x}+1)}{(\text{x}+1-\text{x})}$
$=\frac{(\text{x}+1)}{1}$
$=(\text{x}+1)$
View full question & answer→MCQ 801 Mark
Find out next term of the series $2, 7, 28, 63, 126, ...:$
AnswerGiven the series' terms can be written as
$13 + 1, 23 - 1, 33 + 1, 43 - 1, 53 + 1, 63 - 1$ etc.
Hence the next number is $63 - 1 = 216 - 1 = 215$
View full question & answer→MCQ 811 Mark
If $a, b, c$ are in $AP,$ then the straight line $ax + by + c = 0$ will always pass through the point:
- A
$(-1, – 2)$
- ✓
$(1, – 2)$
- C
$(-1, 2)$
- D
$(1, 2)$
AnswerCorrect option: B. $(1, – 2)$
View full question & answer→MCQ 821 Mark
If $a_1, a_2, a-n$, are in $AP$ with common difference $d,$ then the sum of the series $\sin \ce{d(cosec{~a_1}\ cosec {~a_2} + \ cosec {~a_2}\ cosec {~a_3} +… + \ cosec {~a_{n-1}}\ cosec {~a_n})}$ is:
- A
$\sec {~a_1} \sec {~a_n} –$
- ✓
$\cot {~a_1}– \cot {~a_n}$
- C
$\tan {~a_1}– \tan {~a_n}$
- D
$\ce{cosec {~a_1} – cosec {~a_n}}$
AnswerCorrect option: B. $\cot {~a_1}– \cot {~a_n}$
View full question & answer→MCQ 831 Mark
In $G.P.\ 4, 8, 16, 32, …$ find the sum up to $5^{th}$ term.
AnswerIn the given $\text{G.P.}$
$=4$ and $\text{r}=\frac{8}{4}=2.$
We know, $\text{s}_\text{n}=a\frac{\text({r}_\text{n}-1)}{\text({r}-1)} $
$\Rightarrow{\text{s}}_5$
$=4\frac{(2^5-1)}{(2-1)}$
$=4\times31$
$=124.$
View full question & answer→MCQ 841 Mark
If the sum of the roots of the equation $\ce{ax^2bx + c} = 0$ is be equal to the sum of the reciprocals of their squares, then $\ce{bc^2,ca^2,ab^2}$ will be in:
View full question & answer→MCQ 851 Mark
The value of $\sum\limits^{\text{n}}_{\text{r}=1}\Big\{\big(2\text{r}-1\big)\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$ is equal to:
- A
$\text{an}^2+\frac{\text{b}^{\text{n}-1}-1}{\text{b}^{\text{n}-1}(\text{b}-1)}$
- ✓
$\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
- C
$\text{an}^3+\frac{\text{b}^{\text{n}-1}-1}{\text{b}^{\text{n}}(\text{b}-1)}$
- D
AnswerCorrect option: B. $\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
We have,
$\sum\limits^{\text{n}}_{\text{r}=1}\Big\{(2\text{r}-1)\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$
$=\sum\limits^{\text{n}}_{\text{r}=1}\Big\{2\text{ra}-\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$
$=\sum\limits^{\text{n}}_{\text{r}=1}2\text{ar}-\sum\limits^{\text{n}}_{\text{r}=1}\text{a}+\sum\limits^{\text{n}}_{\text{r}=1}\frac{1}{\text{b}^{\text{r}}}$
$=\text{an}(\text{n}+1)-\text{an}+\frac{(1-\text{b}^\text{n})}{(1-\text{b})\text{b}^\text{n}}$
$=\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
View full question & answer→MCQ 861 Mark
A sequence is called $............$ if $a_{n+1} = a_n + d.$
AnswerA sequence is called arithmetic progression if $a_{n+1} = a_n+ d$ where $a_1$ is the first term and $d$ is common difference.
View full question & answer→MCQ 871 Mark
If the altitudes of a triangle are in $AP,$ then the sides of the triangle are in:
View full question & answer→MCQ 881 Mark
If the sum of two numbers is $4$ times the geometric mean then find the ratio of numbers.
- ✓
$\frac{8\pm3\sqrt{5}}{1}$
- B
$\frac{8\pm3\sqrt{7}}{1}$
- C
$\frac{6\pm3\sqrt{5}}{1}$
- D
$\frac{6\pm3\sqrt{7}}{1}$
AnswerCorrect option: A. $\frac{8\pm3\sqrt{5}}{1}$
We know, $G.M.$ of two numbers $a$ and $b$ is $\sqrt{\text{ab.}}$
so, $\text{a}+\text{b}=\sqrt{\text{ab}}$
Squaring we get, $a^2+b^2= 16ab$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{a}}{\text{b}}\Big)$
Let $x=\frac{\text{a}}{\text{b}}.$
So $,x+\frac{1}{x}=16$
$\Rightarrow x^2 – 16x + 1 = 0$
$\Rightarrow x=\frac{16\pm\sqrt{256-4}}{4}$
$=\frac{16\pm\sqrt{252}}{2}$
$=\frac{16\pm6\sqrt{7}}{2}$
$=\frac{8\pm3\sqrt{7}}{1}.$
View full question & answer→MCQ 891 Mark
The first three of four given numbers are in $G.P.$ and their last three are $A.P.$ with common difference $6.$ If first and fourth numbers are equal, then the first number is:
AnswerThe first and the last numbers are equal.
Let the four given numbers be $p, q, r$ and $p.$
The first three of four given numbers are in $G.P.$
$\therefore\text{q}^2=\text{p}\cdot\text{r}\cdots(\text{i})$
And, the last three numbers are in $A.P.$ with common difference $6.$
We have:
First term $= q$
Second term $= r = q + 6$
Third term $= p = q + 12$
Also, $2r = q + p$
Now, putting the values of $p$ and $r$ in $(i):$
$q^2= (q + 12)(q + 6)$
$\Rightarrow q^2 = q^2 + 18q + 72$
$\Rightarrow 18q + 72 = 0$
$\Rightarrow q + 4 = 0$
$\Rightarrow q = -4$
Now, putting the value of $q$ in $p = q + 12:$
$p = -4 + 12 = 8$
View full question & answer→MCQ 901 Mark
Which of the following relation gives Fibonacci sequence?
- ✓
${a}_{n}={a}_{{n}-1}+{a}_{t{n}-2}$
- B
${a}_{{n}-1}={a}_{n}+{a}_{{n}-2}$
- C
${a}_{{n}-2}={a}_{n}+{a}_{{n}-1}$
- D
${a}_{n}={a}_{{n}+1}+{a}_{{n}-2}$
AnswerCorrect option: A. ${a}_{n}={a}_{{n}-1}+{a}_{t{n}-2}$
This is a recurrence relation which gives Fibonacci sequence.
View full question & answer→MCQ 911 Mark
length of a side of $S_n$ equals the length of a diagonal of $S_n+1$. If the length of a side of $S_1$ is $10 \ cm$, then for which of the following values of $n,$ the area of $S_n$ less than $1\ \text{sq cm} ?$
View full question & answer→MCQ 921 Mark
In a $G.P., 5^{th}$ term is $27$ and $8^{th}$ term is $729.$ Find its $11^{th}$ term.
- A
$729$
- B
$2187$
- C
$6561$
- ✓
$19683$
AnswerCorrect option: D. $19683$
Given, $a_5 = 27$ and $a_8 = 729.$
$\Rightarrow ar^4 = 27$ and $ar^7 = 729$
On dividing we get, $r^3 = 27$
$\Rightarrow r=3$
$\Rightarrow\text{a}=\frac{23}{(3^4)}=\frac{1}{3}$
$\Rightarrow\text{a}_{11}=\text{a}^{10}$
$=(\frac{1}{3})(3^{10})$
$=39$
$=19683$
View full question & answer→MCQ 931 Mark
The sum of $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ ...\text{ is}$
- ✓
$121\big(\sqrt{6}+\sqrt{2}\big)$
- B
$243\big(\sqrt{3}+1\big)$
- C
$\frac{121}{\sqrt{3}-1}$
- D
$242\big(\sqrt{3}-1\big)$
AnswerCorrect option: A. $121\big(\sqrt{6}+\sqrt{2}\big)$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{S}_{10}=\sqrt{2}\sum\limits^{10}_{\text{k}=1}\Big(\sqrt{3^{(\text{k}-1)}}\Big)$
$\Rightarrow\text{S}_{10}=\sqrt{2}\Big(1+\sqrt{3}+\sqrt{3^2}+\ ...\ +\sqrt{3^9}\Big)$
$\Rightarrow\text{S}_{10}=\sqrt{2}\bigg(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\bigg)$
$\Rightarrow\text{S}_{10\text{n}}=\sqrt{2}\Big(\frac{3^5-1}{\sqrt{3}-1}\Big)\Big(\frac{\sqrt{3}+1}{\sqrt{3}+1}\Big)$
$\Rightarrow\text{S}_{10}=\frac{\sqrt{2}}{2}\big(3^5-1\big)\big(\sqrt{3}+1\big)$
$\Rightarrow\text{S}_{10}=\frac{1}{2}(242)\Big(\sqrt{6}+\sqrt{2}\Big)$
$\Rightarrow\text{S}_{10}=121\big(\sqrt{6}+\sqrt{2}\big)$
View full question & answer→MCQ 941 Mark
Find the sum to $n$ terms of the series whose $n^{th}$ term is $n (n-2).$
- A
$\frac{\text{n}(\text{n-1)}(2\text{n}+4)}{6}$
- ✓
$\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{6}$
- C
$\frac{(\text{n-1)}(2\text{n}-5)}{3}$
- D
$\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{3}$
AnswerCorrect option: B. $\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{6}$
Given, $n^{th}$ term is $n(n-2)$
So, $a_k = k(k-2)$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=\sum\limits^\text{n}_\text{i}$
$=0\text{ k}^2-2\sum\limits^\text{n}_\text{i}=0\text{ k}$
$=\frac{\text{n}(\text{n+1)}(2\text{n+1}}{6}-2\frac{\text{n}(\text{n+1)}}{2}$
$=\frac{\text{n}(\text{n+1)(2}\text{n}-5)}{6}.$
View full question & answer→MCQ 951 Mark
The value of $9^{\frac13}.9^{\frac19}.9^{\frac{1}{27}}\dots\text{to }\infty,$ is:
Answer$9^{\frac13}\times9^{\frac19}\times9^{\frac{1}{27}}\times\dots\infty$
$=9^{\big(\frac13+\frac19+\frac{1}{27}+\dots\infty\big)}$
Here, it is a $G.P.$ with $\text{a}=\frac13$ and $\text{r}=\frac13.$
$\therefore9^{\Bigg(\frac{\frac{1}{3}}{1-\frac13}\Bigg)}$
$=9^{\big(\frac12\big)}$
$=3$
View full question & answer→MCQ 961 Mark
If $p^{th,} q^{th}$ and $r^{th}$ terms of an $A.P.$ are in $G.P.$, then the common ratio of this $G.P$. is:
AnswerCorrect option: B. $\frac{\text{q}-\text{r}}{\text{p}-\text{q}}$
Let a be the first term and $d$ be the common difference of the given $A.P.$
Then, we have:
$p^{th}$ term, $ap = a + (p−1)d$
$q^{th}$ term, $aq = a + (q−1)d$
$r^{th}$ term, $ar = a + (r−1)d$
Now, according to the question the $p^{th}$, the $q^{th}$ and the $r^{th}$ terms are in $G.P.$
$\therefore (\text{a} + (\text{q}−1)\text{d})^2=( \text{a} + (\text{p}−1)\text{d})\times(\text{a} + (\text{r}−1)\text{d})$
$\Rightarrow\text{a}^2+2\text{a} (\text{q}−1)\text{d}+( (\text{q}−1)\text{d})^2=\text{a}^2+\text{ad}(\text{r}−1+\text{p}−1)+(\text{p}−1) (\text{r}−1)\text{d}^2$
$\Rightarrow(2\text{q}−2−\text{r}−\text{p}+2)+\text{d}^2(\text{q}^2−2\text{q}+1−\text{pr}+\text{p}+\text{r}−1)=0$
$\Rightarrow(2\text{q}−\text{r}-\text{p})+\text{d}(\text{q}^2−2\text{q}−\text{pr}+\text{p}+\text{r})=0$
$\Rightarrow\text{a}=-\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(2\text{q}-\text{r}-\text{p})}$
$\therefore\text{Common ratio},\text{ r}=\frac{\text{a}_\text{q}}{\text{a}_\text{p}}$
$=\frac{\text{a}+(\text{q}-1)\text{d}}{\text{a}+(\text{p}-1)\text{d}}$
$=\frac{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{q}-1)\text{d}}{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{p}-1)\text{d}}$
$=\frac{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{pq}+\text{rq}+\text{rq}-2\text{q}^2-\text{p}-\text{r}+2\text{q}}{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{p}^2+\text{pr}-2\text{pq}-\text{p}-\text{r}+2\text{q}}$
$=\frac{\text{pq}-\text{pr}-\text{q}^2+\text{qr}}{\text{p}^2+\text{q}^2-2\text{pq}}$
$=\frac{\text{p}(\text{p}-\text{r})-\text{q}(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{p}-\text{q})(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{q}-\text{r})}{(\text{p}-\text{q})}$
View full question & answer→MCQ 971 Mark
Choose the correct answer. The lengths of three unequal edges of a rectangular solid block are in $G.P.$ If the volume of the block is $216\ cm^3$ and the total surface area is $252\ cm^2$, then the length of the longest edge is:
- ✓
$12\ cm$
- B
$6\ cm$
- C
$18\ cm$
- D
$3\ cm$
AnswerCorrect option: A. $12\ cm$
Let the length, breadth and height of rectangular solid block be $\frac{\text{a}}{\text{r}}, a$ and $ar,$ respectively.
$\therefore\ \text{Volume}=\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=216\text{cm}^3$
$\Rightarrow\text{a}^3=216=6^3$
$\Rightarrow\text{a}=6$
Also, Surface area $=2\Big(\frac{\text{a}}{\text{r}}.\text{a}+\text{a}.\text{ar}+\frac{\text{a}}{\text{r}}.\text{ar}\Big)=252$
$\Rightarrow2\text{a}^2\Big(\frac{1}{\text{r}}+\text{r}+1\Big)=252$
$\Rightarrow2\times36\Big(\frac{1+\text{r}^2+\text{r}}{\text{r}}\Big)=252$
$\Rightarrow2(1+\text{r}^2+\text{r})=7\text{r}$
$\Rightarrow2\text{r}^2-5\text{r}+2=0$
$\Rightarrow(2\text{r}-1)(\text{r}-2)=0$
$\therefore\ \text{r}=\frac{1}{2},2$
For $\text{r}=\frac{1}{2}:$ Length $=\frac{\text{a}}{\text{r}}=\frac{6\times2}{1}=12,$ Breadth $= a = 6$
Height $=\text{ar}=6\times\frac{1}{2}=3$
For $r = 2:$ Length $=\frac{\text{a}}{\text{r}}=\frac{6}{2}=3,$ Breadth $= a = 6$
Height $= ar = 6 \times 2 = 12$
View full question & answer→MCQ 981 Mark
If $2p + 3q + 4r = 15,$ then the maximum value of $p_3\ q_5\ r_7$ is:
AnswerCorrect option: C. $5^5\times\frac{7^7}{2^{17}}\times9$
View full question & answer→MCQ 991 Mark
A series can also be denoted by symbol $.........$
- A
$\pi\text{a}_\text{n}$
- ✓
$\sum\text{a}_{\text{n}}$
- C
$\phi\text{a}_{\text{n}}$
- D
$\theta\text{a}_{\text{n}}$
AnswerCorrect option: B. $\sum\text{a}_{\text{n}}$
When we use addition between the terms of sequence, it is said to be series.
We know that addition can also be written in the form of sigma
so, series can also be denoted by $\sum\text{a}_{\text{n}}.$
View full question & answer→MCQ 1001 Mark
If $r = 1$ in a $G.P.$ then what is the sum to $n$ terms?
AnswerCorrect option: A. $n\times a$
If a is the first term of $G.P.,$ then $G.P.$
look like $a, a, a, a, …………$
Then sum to n terms becomes $n \times a.$
View full question & answer→MCQ 1011 Mark
If in an $A.P.,$ first term is $20,$ common difference is $2$ and nth term is $42$, then find $n.$
AnswerWe know, $a = 20, d = 2, a_n = 42$
$a + (n - 1) d = 42 $
$\Rightarrow 20 + 2(n - 1) = 42$
$\Rightarrow 2 (n - 1) = 42 - 20 = 22 $
$\Rightarrow n - 1 = 11 $
$\Rightarrow n = 12.$
View full question & answer→MCQ 1021 Mark
Which of the following is not a series?
AnswerThe isometric series is not a series.
Rest all are series i.e. arithmetic series, geometric series and harmonic series.
View full question & answer→MCQ 1031 Mark
A person is to count $4500$ currency notes. Let an denotes the number of notes he counts in the nth minute. If $a_1 = a_2 = . . . .= a_{10}= 150$ and $a_{10} , b_{11}$, . . . are in $AP$ with common difference $2,$ then the time taken by him to count all notes, is:
- A
$24$ min
- ✓
$34$ min
- C
$125$ min
- D
$135$ min
AnswerCorrect option: B. $34$ min
View full question & answer→MCQ 1041 Mark
The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+\ ....$ to $n$ terms is:
- A
$\text{n}-\frac{1}{2}(3^{-\text{n}}-1)$
- ✓
$\text{n}-\frac{1}{3}(1-3^{-\text{n}})$
- C
$\text{n}+\frac{1}{2}(3^\text{n}-1)$
- D
$\text{n}-\frac{1}{1}(3^\text{n}-1)$
AnswerCorrect option: B. $\text{n}-\frac{1}{3}(1-3^{-\text{n}})$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=\frac{3^\text{n}-1}{3^\text{n}}=1-\frac{1}{3^\text{n}}$
Now,
Let $S_n$ be the sum of $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum^\limits\text{n}_{\text{k}=1}\text{T}_\text{k}$
$=\sum^\limits\text{n}_{\text{k}=1}\Big[1-\frac{1}{3^\text{k}}\Big]$
$=\sum^\limits\text{n}_{\text{k}=1}1-\sum^\limits\text{n}_{\text{k}=1}\frac{1}{3^\text{k}}$
$=\text{n}-\Big[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ ....\ +\frac{1}{3^\text{n}}\Big]$
$=\text{n}-\frac{1}{3}\Bigg[\frac{1-\big(\frac{1}{3}\big)^\text{n}}{1-\frac{1}{3}}\Bigg]$
$=\text{n}-\frac{1}{3}\Big[1-\Big(\frac{1}{3}\Big)^\text{n}\Big]$
$=\text{n}-\frac{1}{3}\big[1-3^{-\text{n}}\big]$
View full question & answer→MCQ 1051 Mark
Find the sum of first $n$ terms.
- ✓
$\frac{\text{n}(\text{n}+1)}{2}$
- B
$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
- C
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
- D
$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
AnswerCorrect option: A. $\frac{\text{n}(\text{n}+1)}{2}$
Sum of first $n$ terms $= 1+2+3+4+……+n$
$\Rightarrow\Big(\frac{\text{n}}{2}\Big)=\text{(a}+\text{b)}$
$=\Big(\frac{\text{n}}{2}\Big)(\text{1}+\text{n})$
$=\frac{\text{n}(\text{n}+1)}{2}.$
View full question & answer→MCQ 1061 Mark
If $a, b, c$ are in $AP,$ then $10^{ax+10}, 10^{bx+10}, 10^{cx}+10(\mathrm{x} \neq 0)$ are in:
- A
$AP$
- B
$GP$ only when $x > 0$
- ✓
$GP$ for all $x$
- D
$GP$ only when $x < 0$
AnswerCorrect option: C. $GP$ for all $x$
View full question & answer→MCQ 1071 Mark
If $a, b, c, d$ are any four consecutive coefficients of any expanded binomial, then $\frac{\text{a}+\text{b}}{a},\frac{\text{b}+\text{c}}{b},\frac{\text{c }+\text{d}}{c}$ are in:
View full question & answer→MCQ 1081 Mark
The sum of $n$ terms of two arithmetic progressions are in the ratio $(2n + 3) : (7n + 5).$ Find the ratio of their $9^{th}$ terms.
- A
$4 : 5$
- B
$5 : 4$
- ✓
$9 : 31$
- D
$31 : 9$
AnswerCorrect option: C. $9 : 31$
Let $\text{a, a’}$ be the first terms and $\text{d, d’}$ be the common differences of $\text{2 A.P.’s}$ respectively.
Given, $\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}2[2a'+(\text{n}-1){\text{d}]}}=\frac{2\text{n}+3}{7\text{n}+5}$
$\Rightarrow\frac{\text{a}+(\text{n}-1){\frac{d}2{}}}{\text{a'}+(\text{n}-\frac{\text{d'}}{2}}{}=\frac{2\text{n}+3}{7\text{n}+5}$
If we have to find ratio of $9^{th}$ terms then
$\frac{(\text{n}-1)}{2}=8$
$\Rightarrow\text{n}=17$
$\Rightarrow\frac{\text{a}+8\text{d}}{\text{a'}+8\text{d'}}$
$=\frac{2\times17+3}{3\times17+5}$
$=\frac{34+3}{119+5}$
$=\frac{36}{124}$
$=\frac{9}{31}.$
View full question & answer→MCQ 1091 Mark
In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then, the common ratio of this progression is equal to:
AnswerCorrect option: C. $\sqrt{5}$
View full question & answer→MCQ 1101 Mark
If second term of a $\text{G.P.}$ is $2$ and the sum of its infinite terms is $8,$ then its first terms is:
- A
$\frac{1}{4}$
- B
$\frac{1}{2}$
- C
$2$
- ✓
$4.$
Answer$\text{a}_2=2$
$\therefore\text{ar}=2\ \cdots(\text{i})$
Also, $\text{S}_\infty=8$
$\Rightarrow\frac{\text{a}}{(1-\text{r})}=8$
$\Rightarrow\frac{\text{a}}{\Big(1-\frac{2}{\text{a}}\Big)}=8 [$Using $(i)]$
$\Rightarrow\text{a}^2=8(\text{a}-2)$
$\Rightarrow\text{a}^2-8\text{a}+16=0$
$\Rightarrow(\text{a}-4)^2=0$
$\Rightarrow\text{a}=4$
View full question & answer→MCQ 1111 Mark
Find the sum of first $5$ terms of series $2 + 4 + 6 +...........$
AnswerSince $2, 4$ and $6$ all are even numbers
so, given series involve all even number terms.
The next two terms will be $8$ and $10$
so, sum will be $2 + 4 + 6 + 8 + 10 = 30.$
View full question & answer→MCQ 1121 Mark
Choose the correct answer. If the third term of $\text{G.P.}$ is $4,$ then the product of its first $5$ terms is:
AnswerGiven that:
$T_3 = 4$
$\Rightarrow ar^{3-1}= 4$
$\big[\because\ \text{T}_\text{n}=\text{ar}^{\text{n}-1}\big]$
$\Rightarrow ar^2 = 4$
Product of first $5$ terms $= \ce{a. ar. ar^2. ar^3. ar^4}$
$= a^5r^{10}= (ar^2)^5= (4)^5$
Hence, the corrrect option is $(c).$
View full question & answer→MCQ 1131 Mark
The AM,HMandGMbetween two numbers are $\frac{144}{15},15$ and 12, but not necessarily in thisorder. Then, HM, GM and AM respectively are:
AnswerCorrect option: B. $ \frac{144}{15},12, 15$
$ \frac{144}{15},12, 15$
View full question & answer→MCQ 1141 Mark
If $100$ times the $100^{th}$ term of an $AP$ with non$-$zero common difference equals the $50$ times its $50^{th}$ term, then the $150^{th}$ term of this $AP$ is:
View full question & answer→MCQ 1151 Mark
The sum of first three terms of a $G.P.$ is $\frac{21}{2}$ and their product is $27.$ Which of the following is not a term of the $G.P.$ if the numbers are positive?
- A
$3$
- ✓
$\frac{2}{3}$
- C
$\frac{3}{2}$
- D
$6$
AnswerCorrect option: B. $\frac{2}{3}$
Let three terms be $\frac{\text{a}}{\text{r}}, a, \times ar.$
$\text{product}=27\Rightarrow\Big(\frac{\text{a}}{\text{r}}\Big)\Big(\text{a}\Big)\Big(\text{a}\times\text{r}\Big)=27$
$\Rightarrow\text{a}^3=27$
$\Rightarrow\text{a}=3.$
$\text{sum}=\frac{21}{2}$
$\Rightarrow\Big(\frac{a}{r+\text{a}+\text{a}\times\text{r}}\Big)=\frac{21}{2}$
$\Rightarrow\text{a}\Big( \frac{1 }{\text{r+1+1}\times\text{r}}\Big) =\frac{21}{2} $
$\Rightarrow\Big(\frac{1}{\text{r+1+1}\times\text{r}}\Big)=\Big(\frac{\frac{21}{2}}{3}\Big)=\frac{7}{2}$
$\Rightarrow\Big(\text{r}^2+\text{r}+1=\Big)\Big(\frac{7}{2}\Big)$
$\Rightarrow\text{r}^2 -\Big(\frac{5}{2}\Big)\text{r}+1=0$
$\Rightarrow\text{r}=2$ and $\frac{1}{2}.$
Terms are $\frac{3}{2}, 3, 3 \times 2 $
i.e. $\frac{3}{2} , 3, 6.$
View full question & answer→MCQ 1161 Mark
The sum of $n$ terms of an $AP$ is a $n(n – 1).$ The sum of the squares of these terms is:
- A
$\text{n}^2-\text{n}^2(\text{n-1)}^2$
- B
$\frac{\text{a}^2}{6}-\text{n}(\text{n-1})(2\text{n-1)}$
- ✓
$\frac{2\text{a}^2}{3}-\text{n}(\text{n-1})(2\text{n-1)}$
- D
$\frac{2\text{a}^2}{3}-\text{n}(\text{n+1})(2\text{n+1)}$
AnswerCorrect option: C. $\frac{2\text{a}^2}{3}-\text{n}(\text{n-1})(2\text{n-1)}$
View full question & answer→MCQ 1171 Mark
Complete $2, 4, 6, 8, .............$
AnswerSince sequence $2, 4, 6, 8, 10$ contains limited number of terms
so, it is finite sequence.
Rest all are infinite sequences.
View full question & answer→MCQ 1181 Mark
If the non$-$zero numbers $\text{x, y, z}$ are in $\text{AP}$ and $\ce{\tan - 1 , \tan - 1 , \tan - 1 x y z}$ are in $\text{AP,}$ then:
- ✓
$x = y = z$
- B
$yzx 2 =$
- C
$xyz 2 =$
- D
$zxy2 =$
AnswerCorrect option: A. $x = y = z$
View full question & answer→MCQ 1191 Mark
In a $\text{G.P.}$ of ever number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the $\text{G.P.}$ is:
- A
$-\frac{4}{5}$
- B
$\frac{1}{5}$
- ✓
$4$
- D
AnswerLet there be $2n$ terms in a $G.P.$
Let a be the first term and $r$ be the common ratio.
$\because\text{ S}_{2\text{n}}=5(\text{S}_{\text{odd terms}})$
$\Rightarrow\frac{\text{a}\big(\text{r}^{2\text{n}-1}\big)}{(\text{r}-1)}=5\big(\text{a}+\text{ar}^2+\text{ar}^4+\text{ar}^6+\dots\text{ar}^{(2\text{n}-1)}\big)$
$\Rightarrow\frac{\text{a}\big(\text{r}^{2\text{n}}-1\big)}{(\text{r}-1)}=5\Bigg(\frac{\text{a}\big(\big(\text{r}^2\big)^\text{n}\big)}{\big(\text{r}^2-1\big)}\Bigg)$
$\Rightarrow\frac{\big(\text{r}^{2\text{n}}-1\big)}{(\text{r}-1)}=5\frac{\big(\big(\text{r}^2\big)^\text{n}-1\big)}{\big(\text{r}^2-1\big)}$
$\Rightarrow\frac{\big(\big(\text{r}^\text{n}\big)^2-1^2\big)}{(\text{r}-1)}=5\frac{\big(\big(\text{r}^\text{n}\big)^2-1^2\big)}{\big(\text{r}^2-1\big)}$
$\Rightarrow\frac{\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)}{(\text{r}-1)}=5\frac{\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)}{\big(\text{r}-1\big)\big(\text{r}+1\big)}$
$\Rightarrow\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)\big(\text{r}-1\big)\big(\text{r}+1\big) -5\big(\text{r}-1\big)\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)=0$
$\Rightarrow\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)\big(\text{r}-1\big)\big(\text{r}+1-5\big)=0$
But, $r = 1$ or $−1$ is not possible.
$\therefore\text{r}=4$
View full question & answer→MCQ 1201 Mark
If sum of $n$ terms of an $A.P.$ is $n^2+5n$ then find general term.
- A
$n + 1$
- ✓
$2n$
- C
$3n$
- D
$n^2+ 3n$
AnswerGiven, $S_n=n^2+5 n$
We know, $a_n=S_n-S_{n-1}$
$=\left(n^2+5 n\right)-\left((n-1)^2+5(n-1)\right)$
$=\left(n^2+5 n\right)-\left(n^2+1-2 n+5 n-1\right)$
$=2 n$.
View full question & answer→MCQ 1211 Mark
The product $(32),(32)^{\frac{1}{6}}(32)^{\frac{1}{36}}\ \dots\text{ to }\infty$ is equal to:
Answer$32\times32^{\frac{1}{6}}\times32^{\frac{1}{36}}\times\ \cdots\infty$
$=32^{\big(1+\frac{1}{6}+\frac{1}{36}+\ \cdots\infty\big)}$
$=32^{\Bigg(\frac{1}{1-\frac{1}{6}}\Bigg)}$
$[\because$ it is a $G.P. ]$
$=32^{\big(\frac65\big)}$
$=\big(2^5\big)^{\big(\frac65\big)}$
$=2^6$
$=64$
View full question & answer→MCQ 1221 Mark
In a $\text{G.P.}$ if the $(m + n)^{th}$ terms is $p$ and $(m - n)^{th}$ term is $q,$ then its $m^{th}$ term is:
AnswerCorrect option: C. $\sqrt{\text{pq}}$
Here, $\text{a}_{(\text{m}+\text{n})}=\text{p}$
$\Rightarrow\text{ar}^{(\text{m}+\text{n}-1)}=\text{p}\ \cdots(\text{i})$
Also, $\text{a}_{(\text{m}-\text{n})}=\text{q}$
$\Rightarrow\text{ar}^{(\text{m}-\text{n}-1)}=\text{q}\ \cdots(\text{ii})$
Mutliplying $(i)$ and $(ii):$
$\Rightarrow\text{ar}^{(\text{m}+\text{n}-1)}\text{ar}^{(\text{m}-\text{n}-1)}=\text{pq}$
$\Rightarrow\text{a}^2\text{r}^{(2\text{m}-2)}=\text{pq}$
$\Rightarrow\Big(\text{ar}^{(\text{m}-1)}\Big)^2=\text{pq}$
$\Rightarrow\text{ar}^{(\text{m}-1)}=\sqrt{\text{pq}}$
$\Rightarrow\text{a}_\text{m}=\sqrt{\text{pq}}$
Thus, the $m^{th}$ term is $\sqrt{\text{pq}}.$
View full question & answer→MCQ 1231 Mark
The sum of first $20$ terms of the sequence $0.7, 0.7 7, 0.7 7 7, … ,$ is:
- A
$7/81 (179 – 10^{-20})$
- B
$7/9 (99 – 10^{-20})$
- ✓
$7/81 (179 + 10^{-20})$
- D
$7/9 (99 + 10^{-20})$
AnswerCorrect option: C. $7/81 (179 + 10^{-20})$
View full question & answer→MCQ 1241 Mark
If $1+\frac{1+2}{2}+\frac{1+2+3}{3}+\ ....$ to $n$ terms is $S,$ then $S$ is equal to:
- ✓
$\frac{\text{n}(\text{n}+3)}{4}$
- B
$\frac{\text{n}(\text{n}+2)}{4}$
- C
$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
- D
$\text{n}^2$
AnswerCorrect option: A. $\frac{\text{n}(\text{n}+3)}{4}$
Let $T_n$ be the nth term of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{\text{k}}{2}+\frac{1}{2}\Big)$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\frac{\text{k}}{2}+\frac{\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}+\frac{\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+1}{2}+1\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+3}{2}\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+3)}{4}$
View full question & answer→MCQ 1251 Mark
The sum of the series $1^2 + 3^2 + 5^2 + ...$ to $n$ terms is:
- A
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{2}$
- ✓
$\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
- C
$\frac{(\text{n}-1)^2(2\text{n}+1)}{6}$
- D
$\frac{(2\text{n}+1)^3}{3}$
AnswerCorrect option: B. $\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=(2\text{n}-1)^2$
$=4\text{n}^2+1-4\text{n}$
Now, let $S_n$ be the sum of $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_\text{k=1}(4\text{k}^2+1-4\text{k})$
$\Rightarrow\text{S}_\text{n}=4\sum\limits^{\text{n}}_\text{k=1}\text{k}^2+\sum\limits^{\text{n}}_\text{k=1}1-4\sum\limits^{\text{n}}_\text{k=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{4\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}-\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{2\text{n}(\text{n}+1)(2\text{n}+1)}{3}+\text{n}-2\text{n}(\text{n}+1)$
$\Rightarrow\text{S}_\text{n}=\text{n}\Big[\frac{2(\text{n}+1)(2\text{n}+1)}{3}+1-2(\text{n}+1)\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(2\text{n}+2)(2\text{n}+1)+3-6(\text{n}+1)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(4\text{n}^2-1)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
View full question & answer→MCQ 1261 Mark
If in an $A.P.,$ first term is $20$ and $12^{th}$ term is $120.$ Find the sum up to $12^{th}$ term.
AnswerGiven, $a = 20, a_{12}= 120, n = 12$
$\text{s}_{\text{n}}=\frac{\text{n}}{2}(\text{a}+\text{I})$
$\Rightarrow\text{s}_{\text{12}}=\frac{12}{2}(20+120)$
$= 6 \times 140$
$= 840$
View full question & answer→MCQ 1271 Mark
The sum of the integers from $1$ to $100$ which are not divisible by $3$ or $5$ is:
- A
$2489$
- B
$4735$
- ✓
$2317$
- D
$2632$
AnswerCorrect option: C. $2317$
View full question & answer→MCQ 1281 Mark
The consecutive digits of a three digit number are in $GP.$ If the middle digit be increased by $2$, then they form an $AP.$ If $792$ is subtracted from this, then we get the number constituting of same three digits but in reverse order. Then, number is divisible by:
View full question & answer→MCQ 1291 Mark
In numbers from $1$ to $100$ the digit $"0"$ appears $...........$ times:
Answer$10, 20, 30, 40, 50, 60, 70, 80, 90, 100$
Thus the digit $0$ appears $11$ times.
View full question & answer→MCQ 1301 Mark
Concentric circles of radii $1, 2, 3, … , 100 \ cm$ are drawn. The interior of the smallest circle is coloured red and the angular regions are coloured alternately green and red, so that no two adjacent regions are of the same colour. Then, the total area of the green regions in $\text{sq \ cm}$ is equal to:
- A
$1000\pi$
- ✓
$5050\pi$
- C
$4950\pi$
- D
$5151\pi$
AnswerCorrect option: B. $5050\pi$
View full question & answer→MCQ 1311 Mark
$\sum\limits^{4}_{\text{i = 1}}2\text{n}+3=............$
Answer$a_1 = 2 \times 1 + 3 = 5,$
$a_2 = 2 \times 2 + 3 = 7,$
$a_3 = 2 \times 3 + 3 = 9,$
$a_4= 2 \times 4 + 3 = 11.$
Sum $= 5 + 7 + 9 + 11 = 32.$
View full question & answer→MCQ 1321 Mark
If $\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54},$ the value of x is:
Answer
Solution: (B) 8
$\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54}$
$\Rightarrow4^{(3+6+9+12+\ \dots+3\text{x})}=\Big(\frac{625}{10000}\Big)^{-54}$
$\Rightarrow4^{3(1+2+3+4+\dots+\text{x})}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(4^{-2}\Big)^{-54}$
Comparing both the sides:
$\Rightarrow3\Big(\frac{\text{x}(\text{x}+1)}{2}\Big)=108$
$\Rightarrow\text{x}(\text{x}+1)=72$
$\Rightarrow\text{x}^2+\text{x}-72=0$
$\Rightarrow\text{x}^2+9\text{x}-8\text{x}-72=0$
$\Rightarrow\text{x}(\text{x}+9)-8(\text{x}+9)=0$
$\Rightarrow(\text{x}+9)(\text{x}-8)=0$
$\Rightarrow\text{x}=8,-9$
$\Rightarrow\text{x}=8$ $[\because\text{ x}\text{ is psitive}]$
View full question & answer→MCQ 1331 Mark
Which term of $G.P. 25, 125, 625, ………….$ is $390625?$
AnswerIn the given $G.P.$, In the given $G.P., a=25$ and
$\text{r}=\frac{125}{25}=5$
Given, $a_n= 390625$
$\Rightarrow ar^{n-1}= 390625$
$\Rightarrow 25 \times 5^{n-1} = 390625$
$\Rightarrow 5^\text{n-1}=\frac{390625}{25}=15625=5^6$
$\Rightarrow n-1 = 6$
$\Rightarrow n=7.$
View full question & answer→MCQ 1341 Mark
If $\sum\text{n}=210,$ then $\sum\text{n}^2=$
AnswerCorrect option: A. $2870$
Given,
$\sum\text{n}=210$
$\Rightarrow\text{n}\Big(\frac{\text{n}+1}{2}\Big)=210$
$\Rightarrow\text{n}^2+\text{n}-420=0$
$\Rightarrow(\text{n}-20)(\text{n}+21)=0$
$\Rightarrow\text{n}=20$
$(\because\ \text{n}>0)$
Now,
$\sum\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\Rightarrow\frac{\text{n}(\text{n}+1)}{2}\times\frac{(2\text{n}+1)}{3}$
$\Rightarrow(210)\times\Big(\frac{41}{3}\Big)$
$\Rightarrow(70)\times(41)$
$\Rightarrow2870$
View full question & answer→MCQ 1351 Mark
Complete $2, 3, 5, 7, ..........$
AnswerSince $2, 3, 5$ and $7$ all are consecutive prime numbers
so, it is a sequence of prime numbers.
Prime number next to $7$ is $11.$
So, $2, 3, 5, 7, 11.$
View full question & answer→MCQ 1361 Mark
Find the sum up to $7^{th}$ term of series $2+3+5+8+12+…………$
Answer$S_n = 2+3+5+8+12+……………+ a_n$
$S_n = 2+3+5+8+12+ ……. + a_{n-1}+ a_n$
Subtracting we get, $0 = 2+1+2+3+4+………… – a_n$
$\Rightarrow\text{a}^\text{n}=2+1+2+3+4+...........+(\text{n}+1)$
$=2+(\text{n}+1)\frac{\text{n}}{2}=\Big(\frac{1}{2}\Big)\text{(n}^2-\text{n}+4)\text{n}^\text{th}\text{terms}\text{ is}\Big(\frac{1}{2}\Big)(\text{n}^2-\text{n}+4$ so $\text{ a}_\text{k} $
$=2+(\text{n}+1)\frac{\text{n}}{2}$
$=\Big(\frac{1}{2}\Big)(\text{k}^2-\text{k+4)}$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=\Big(\frac{1}{2}\Big)\sum\limits^\text{n}_\text{i}=0\text{ k}^2-\Big(\frac{1}{2}\Big)\sum\limits^\text{n}_\text{i}=0\text{ k}+2\text{n}$
$=\frac{\text{n}(\text{n+1)}(2\text{n}+1)}{(2\times6)}$
Here $,\text{n}=7.$
so, $\sum\limits^\text{n}_\text{i}=0\text{ a}\text{k}=\frac{(7\times8\times15)}{12}-\frac{(7\times8)}{4}+2\times7=70.$
View full question & answer→MCQ 1371 Mark
If $a_n = 4n + 6$, find $15^{th}$ term of the sequence.
Answer$a_n = 4n + 6$ and $n = 15$
$\Rightarrow a_{15}= 4 \times 15 + 6$
$= 60 + 6$
$= 66.$
View full question & answer→MCQ 1381 Mark
Insert $4$ numbers between $2$ and $22$ such that the resulting sequence is an $A.P.$
- A
$4, 8, 12, 16$
- B
$5, 9, 13, 17$
- C
$4, 10, 15, 19$
- ✓
$6, 10, 14, 18$
AnswerCorrect option: D. $6, 10, 14, 18$
Let $A.P.$ be $2, A_1, A_2, A_3, A_4, 22$.
$\Rightarrow a=2$ and $a_6=a+5 d=22$
$\Rightarrow 2+5 \times d=22$
$\Rightarrow d=4$
$ A_1=a_2=a+d=2+4=6 $
$ A_2=A_1+d=6+4=10 $
$ A_3=10+4=14 $
$ A_4=14+4=18$
View full question & answer→MCQ 1391 Mark
If $a_1, a_2, \ldots a_n$ are in $HP,$ then the expression $a_1 a_2+a_2 a_3+\ldots a_n-1$ is equal to:
- A
$(n-1) (a_1 – a_n )$
- B
$na_1 a_n$
- ✓
$(n-1) a_1 a_n$
- D
$n (a_1- a_n)$
AnswerCorrect option: C. $(n-1) a_1 a_n$
View full question & answer→MCQ 1401 Mark
If $p, q$ be two $A.M.'s$ and $G$ be one $G.M.$ between two numbers, then $G^2 =$
- ✓
$(2\text{p}-\text{q})(\text{p}-2\text{q})$
- B
$(2\text{p}-\text{q})(2\text{q}-\text{p})$
- C
$(2\text{p}-\text{q})(\text{p}+2\text{q})$
- D
AnswerCorrect option: A. $(2\text{p}-\text{q})(\text{p}-2\text{q})$
Let the two numbers be $a$ and $b.$
$a, p, q$ and $b$ are in $A.P.$
$\therefore\text{ p}-\text{a}=\text{q}-\text{q}=\text{b}-\text{q}$
$\Rightarrow\text{ p}-\text{a}=\text{q}-\text{p}$ and $\text{ q}-\text{p}=\text{b}-\text{q}$
$\Rightarrow\text{ a}=2\text{p}-\text{q}$ and $\text{ b}=2\text{q}-\text{p}\cdots(\text{i})$
Also, $a, G$ and $b$ are in $G.P.$
$\therefore\text{G}^2=\text{ab}$
$\Rightarrow\text{G}^2=(2\text{p}-\text{q})(2\text{q}-\text{p})$
View full question & answer→MCQ 1411 Mark
A man saves $Rs. 200$ in each of the first three months of his service. In each of the subsequent months his saving increases by $Rs. 40$ more than the saving of immediately previous month. His total saving from the start of service will be $Rs. 11040$ after:
- A
$19$ months
- B
$20$ months
- ✓
$21$ months
- D
$18$ months
AnswerCorrect option: C. $21$ months
View full question & answer→MCQ 1421 Mark
The sum to $n$ terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ ....\text{ is}:$
AnswerCorrect option: D. $\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=\frac{1}{\sqrt{2\text{n}-1}+\sqrt{2\text{n}+1}}$
$=\frac{\sqrt{2\text{n}+1}-\sqrt{2\text{n}-1}}{2}$
Now,
Let $S_n$ be the sum $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$=\sum\limits^{\text{n}}_{\text{k}=1}\bigg(\frac{\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}}{2}\bigg)$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\big(\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}\big)$
$=\frac{1}{2}\Big[\big(\sqrt{3}-\sqrt{1}\big)+\big(\sqrt{5}-\sqrt{3}\big)+\big(\sqrt{7}-\sqrt{5}\big)+\ ...\ +\big(\sqrt{2\text{n}+1}-\sqrt{2\text{n}-1}\big)\Big]$
$=\frac{1}{2}\Big\{(-1)+\sqrt{2\text{n}+1}\Big\}$
$=\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$
View full question & answer→MCQ 1431 Mark
Find the sum of series up to $6^{th}$ term whose $n^{th}$ term is given by $n^2 + 3^n$.
- A
$91$
- B
$1284$
- ✓
$1183$
- D
$1092$
AnswerCorrect option: C. $1183$
Given, $n^{th}$ term is $n^2+ 3^n$
So, $a_k= k^2 + 3^k$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}\text{k}=\sum\limits^\text{n}_\text{i}=0\text{ k}^2+\sum\limits^\text{n}_\text{i}=0 3 ^\text{k}$
$\sum\limits^\text{n}_\text{i}=0\text{ k}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\sum\limits^\text{n}_\text{i}=0 3^\text{ k}=\frac{3\times(3^\text{n}-1)}{(3-1)}=\Big(\frac{3}{2}\Big)(3^\text{n}-1)$
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=\frac{\text{n}(\text{n+1}(2\text{n+1)}}{6}+\Big(\frac{3}{2}\Big)(3^\text{n}-1) $
Sum up to $6^{th}$ term
$=\frac{6\times7\times13}{6}+\Big(\frac{3}{2}\Big)(3^6-1)$
$=91+1092$
$=1183.$
View full question & answer→MCQ 1441 Mark
If two numbers are $2$ and $6$ then find their arithmetic mean.
AnswerExplanation: We know that arithmetic mean of two numbers is given by the average of two numbers
i.e. $A.M.$
$\frac{(2+6)}{2}=\frac{8}{2}=4.$
View full question & answer→MCQ 1451 Mark
Complete $2, 4, 6, 8, ...........$
AnswerSince $2, 4, 6$ and $8$ are even numbers so it is a sequence of even numbers.
Even number next to $8$ is $10.$
So, $2, 4, 6, 8, 10.$
View full question & answer→MCQ 1461 Mark
If the sum of first two terms of an infinite $G.P.$ is $1$ and every term is twice the sum of all the successive terms, then its first term is:
- A
$\frac13$
- B
$\frac23$
- C
$\frac14$
- ✓
$\frac34.$
AnswerCorrect option: D. $\frac34.$
Let the terms of the $G.P.$ be $\text{a},\text{a}_2,\text{a}_3,\text{a}_4.\text{a}_5,\ \dots,\infty.$
And, let the common ratio be $r.$
Now, $\text{a}+\text{a}_2=1$
$\therefore\text{a}+\text{ar}=1\dots(\text{i})$
Also, $\text{a}=2(\text{a}_2+\text{a}_3+\text{a}_4+\text{a}_5+\dots\infty)$
$\Rightarrow\text{a}=2\big(\text{ar}+\text{ar}^2+\text{ar}^3+\text{ar}^4+\ \dots\infty\big)$
$\Rightarrow\text{a}=2\Big(\frac{\text{ar}}{1-\text{r}}\Big)$
$\Rightarrow1-\text{r}=2\text{r}$
$\Rightarrow3\text{r}=1$
$\Rightarrow\text{r}=\frac13$
Putting the value of $r$ in $(i):$
$\text{a}+\frac{a}{3}=1$
$\Rightarrow\frac{4\text{a}}{3}=1$
$\Rightarrow4\text{a}=3$
$\Rightarrow\text{a}=\frac34$
View full question & answer→MCQ 1471 Mark
If in an $A.P.,$ first term is $20,$ common difference is $2$ and nth term is $42,$ then find sum up to $n$ terms.
AnswerWe know, $a = 20, d = 2, a_n= 42.$
$a + (n - 1) d = 42$
$\Rightarrow 20 + 2(n - 1) = 42$
$\Rightarrow 2(n - 1) = 42 - 20 = 22$
$\Rightarrow n - 1 = 11$
$\Rightarrow n = 12$
$\text{s}_\text{a}=\frac{\text{n}}{2}(a+I)$
$\Rightarrow\text{s}_{\text{12}=\frac{12}{2}}(20+42)=6\times62=372$
View full question & answer→MCQ 1481 Mark
Choose the correct answer. If the sum of $n$ terms of an $A.P.$ is given by $S_n = 3n + 2n^2$, then the common difference of the $A.P.$ is:
AnswerGiven that:
$ S_n=3 n+2 n^2 $
$ S_1=3(1)+2(1) 2=5 $
$ S_2=3(2)+2(4)=14 $
$ S_1=a_1=5 $
$ S_2-S_1=a_2=14-5=9$
$\therefore$ Common difference $\mathrm{d}=\mathrm{a}_2-\mathrm{a}_1=9-5=4$
Hence, the correct option is $(d).$
View full question & answer→MCQ 1491 Mark
If $|x| < 1,$ then the sum of the series $1 + 2x + 3 \times 2 + 4 \times 3…\infty $ will be:
- A
$1/1 - x$
- B
$1/1 + x$
- C
$1(1 + x^2)$
- ✓
$1(1 - x^2)$
AnswerCorrect option: D. $1(1 - x^2)$
View full question & answer→MCQ 1501 Mark
If $S$ be the sum, $P$ the product and $R$ be the sum of the reciprocals of n terms of a $\text{G.P.}$ then $P^2$ is equal to:
- A
$\frac{\text{S}}{\text{R}}$
- B
$\frac{\text{R}}{\text{S}}$
- C
$\Big(\frac{\text{R}}{\text{S}}\Big)^\text{n}$
- ✓
$\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$
AnswerCorrect option: D. $\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$
Sum of $n$ terms of the $\text{G.P.}, \text{S}=\frac{\text{a}(\text{r}^{\text{n}}-1)}{(\text{r}-1)}$
Product of $n$ terms of the $\text{G.P.}, \text{P}=\text{a}^{\text{n}}\text{r}^{\big[\frac{\text{n}(\text{n}-1)}{2}\big]}$
Sum of the reciprocals of $n$ terms of the $\text{G.P.}, \text{R}=\frac{\Big[\frac{1}{\text{r}^\text{n}}-1\Big]}{\text{a}\big(\frac{1}{\text{r}}-1\big)}=\frac{(\text{r}^{\text{n}}-1)}{\text{ar}^{(\text{n}-1)}(\text{r}-1)}$
$\therefore\text{P}^2=\bigg\{\text{a}^2\text{r}^\frac{2(\text{n}-1)}{2}\bigg\}^\text{n}$
$\Rightarrow\text{P}^2=\Bigg\{\frac{\frac{\text{a}(\text{r}^\text{n}-1)}{(\text{r}-1)}}{\frac{(\text{r}^\text{n}-1)}{\text{ar}^{(\text{n}-1)(\text{r}-1)}}}\Bigg\}^\text{n}$
$\Rightarrow\text{P}^2=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$
Let the first term of the $\text{G.P.}$ be a and the common ratio be $r.$
Sum of $n$ terms, $\text{S}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}$
Product of the $\text{G.P.,}$ $\text{P}=\text{a}^{\text{n}}\text{r}^{\frac{\text{n}(\text{n}+1)}{2}}$
Sum of the reciprocals of $n$ terms,
$\Rightarrow\text{R}=\frac{\big(\frac{1}{\text{r}^{\text{n}-1}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}$
$=\frac{\big(\frac{1-\text{r}^{\text{n}}}{\text{r}^\text{n}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}$
$\text{P}^2=\bigg\{\text{a}^2\text{r}^{\frac{(\text{n}+1)}{2}}\bigg\}^{\text{n}}$
$\text{P}^2=\begin{Bmatrix} \frac{\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}}{\frac{\Big(\frac{1-\text{r}^\text{n}}{\text{r}^\text{n}}\Big)}{\text{a}\Big(\frac{1-\text{r}}{\text{r}}\Big)}}\end{Bmatrix}$
$=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$
View full question & answer→MCQ 1511 Mark
If $a_1,a_2,a_2,….., a_{20}$ are $\text{AM’s}$ between $13$ and $67,$ then the maximum value of $a_1 a_2 a_3.... a_{20} …$ is equal to:
- A
$(20)^{20}$
- ✓
$(40)^{20}$
- C
$(60)^{20}$
- D
$(80)^{20}$
AnswerCorrect option: B. $(40)^{20}$
View full question & answer→MCQ 1521 Mark
What is the third term of Fibonacci sequence?
Answer$a_1=1$ and $a_2=1$.
$a_n=a_{n-1}+a_{n-2}, n>2$
This is a recurrence relation which gives Fibonacci sequence.
$\Rightarrow a_3=a_1+a_2=1+1=2$
View full question & answer→MCQ 1531 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ If $5^{th}$ and $8^{th}$ term of a $\text{GP}$ be $48$ and $384$ respectively, then the common ratio of $\text{GP}$ is $2.$
Reason $(R):$ If $18, x, 14$ are in $AP,$ then $x = 16.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- ✓
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: B. $A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
Assertion Let a be the first term and $r$ be the common ratio of the given $GP.$
According to the question,
$T_5= 48$
$\Rightarrow ar^4 = 48 ...(i)$
and $T8 = 384$
$\Rightarrow ar^7= 384 ...(ii)$
On dividing Eq. $(ii)$ by Eq. $(i)$, we get
$=\frac{\text{ar}^7}{\text{ar}^4}=\frac{384}{48}$
$\Rightarrow\text{r}^3=8$
$\Rightarrow\text{r}=2$
Reason $18, x, 14$ are in $\text{AP.}$
$\Rightarrow x - 18 = 14 - x$
$\Rightarrow 2x = 32$
$\Rightarrow x = 16$
View full question & answer→MCQ 1541 Mark
The two geometric means between the numbers $1$ and $64$ are:
- A
$1$ and $64$
- ✓
$4$ and $16$
- C
$2$ and $16$
- D
$8$ and $16.$
AnswerCorrect option: B. $4$ and $16$
Let the two $\text{G.M.s}$ between $1$ and $64$ be $G_1$ and $G_2$.
Thus, $1, \mathrm{G}_1, \mathrm{G}_2$ and $64$ are in $\text{G.P.}$
$64=1\times\text{r}^3$
$\Rightarrow\text{r}=\sqrt[3]{64}$
$\Rightarrow\text{r}=4$
$\Rightarrow\text{G}_1=\text{ar}=1\times4=4$
And, $\text{G}_2=\text{ar}^2=1\times4^2=16$
Thus, $4$ and $16$ are the required $\text{G.M.s.}$
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