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→