MCQ 11 Mark
If an A.P. is 1,7,13, 19, ……… Find the sum of 22 terms.
AnswerFrom 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 × 128 = 1408.
View full question & answer→MCQ 21 Mark
If the decreasing GP is considered, then the sum of infinite terms is:
MCQ 31 Mark
if an A.P. is 3,5,7,9……. Find the 12th term of the A.P.
Answer
- 25
Solution:
From the given A.P., $\mathrm{a}=3$ and $\mathrm{d}=5-3=2$.
We know, $a_n=a+(n-1) d=>a_{12}=a+11 d=3+11 \times 2=3+22=25$. 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:
Answer
- 1
Solution:
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)]
$=\text{x}^{\text{a}-\text{b}}\text{ y}^{2(\text{b}-\text{a})}\text{ z}^{\text{a}-\text{b}}$
$=(\text{xz})^{\text{a}-\text{b}}(\text{xz})^{\text{b}-\text{a}}$ $[\text{From}(\text{ii}),\text{y}^2=\text{xz}]$
$=(\text{xz})^0$
$=1$ 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. $2\text{ or}\frac{1}{2}$
- $2\text{ or}\frac{1}{2}$
Solution:
Let three terms be $\frac{a}{r}$ a, a × r.
Product = 27 $\Rightarrow(\frac{a}{r}) (a) (a \times r) = 27 \Rightarrow a^3 = 27$
⇒ a = 3.
$\text{sum}=\frac{21}{2}\Rightarrow\frac{(\text{a}{}}{\text{r}+\text{a}\times{\text{r)}}}=\frac{21}{2}\Rightarrow\text{a}\Big(\frac{1}{\text{r+1+1}\times\text{r}}\Big)=\frac{21}{2}$
$\Rightarrow\Big(\frac{1}{\text{r+1+1}\times\text{r}}\Big)$
$\Rightarrow\Big({\text{r}^2}+\text{r}+1\Big)=\Big(\frac{7}{2}\Big)\Rightarrow\text{r}^2-\Big(\frac{5}{2}\Big)\text{r}+1=0$
$\Rightarrow\text{r}=2\text{ and }\frac{1}{2}.$ 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?
- A
$\text{A}>\text{G}$
- ✓
$\text{A}\geq\text{G}$
- C
$\text{G}<\text{A}$
- D
$\text{A}\leq\text{G}$
AnswerCorrect option: 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:
Answera, 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):
$\frac{\log\text{a}}{\text{x}}=\frac{\log\text{a}+\log\text{c}}{2\text{y}}=\frac{\log\text{c}}{\text{z}}$
$\Rightarrow\frac{\log\text{a}}{\text{x}}=\frac{\log\text{a}+\log\text{c}}{2\text{y}}\text{ and }\frac{\log\text{a}}{\text{x}}=\frac{\log\text{c}}{\text{z}}$
$\Rightarrow\log\text{a}(2\text{y}-\text{x})=\text{x}\log\text{c}\text{ and }\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{\text{z}}$
$\Rightarrow\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{(2\text{y}-\text{x})}\text{ and }\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{\text{z}}$
$\Rightarrow\frac{\text{x}}{2\text{y}-\text{x}}=\frac{\text{x}}{\text{z}}$
$\Rightarrow2\text{y}=\text{x}+\text{z}$
Thus, x, y and z are in A.P.
View full question & answer→MCQ 91 Mark
If 3rd term of an A.P. is 6 and 5th term of that A.P. is 12. Then find the 21st term of that A.P.
Answer
- 60
Solution:
$\text { Given, } a_3=6 \text { and } a_5=12$
$\Rightarrow a+2 d=6 \text { and } a+4 d=12$
$\Rightarrow 2 d=6 \Rightarrow d=3$ View full question & answer→MCQ 101 Mark
If general term of an A.P. is 3n then find common difference.
Answer
- 3
Solution:
Given, $a_n=3 n$.
We know, $d=a_n-a_{n-1}=3 n-3(n-1)=3$. 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:
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),
$\text{a}=\frac32\text{ and r}=\frac12$
View full question & answer→MCQ 131 Mark
In A.P. 171, 162, 153, ………. Find first negative term.
Answer
- -9
Solution:
Explanation: $\mathrm{a}=171$ and $\mathrm{d}=162-171=-9$.
$a_n<0$
$\Rightarrow 171+(n-1)(-9)<0$
$\Rightarrow 180-9 n<0$
$\Rightarrow 9 n>180$
$\Rightarrow \mathrm{n}>20 \Rightarrow \mathrm{n}=21 \text { for first negative term. }$
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?
Answer
- 128
Solution:
Let G.P. be $4, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3, 512$.
$\Rightarrow a=4 \text { and } a_5=a \times r^4=512 \times 4 \times r^4=512 \Rightarrow r^4$
$=\frac{512}{4}=128 \Rightarrow r=4 .$
$G_1=a_2=a \times r=4 \times 4=16 .$
$G_2=G_1 \times r=16 \times 4=64 .$
$G_3=G_2 \times r=64 \times 4=256 .$ View full question & answer→MCQ 151 Mark
Find the sum of series $6^2+ 7^2+…………………..+ 15^2$.
Answer
- 1185
Solution:
$6^2+ 7^2+………………..….. + 15^2$
$= (1^2+ 2^2+ 3^2+ …….. +15^2) – (1^2+ 2^2+ 3^2+ 4^2+ 5^2)$
$=\frac{15\times16\times31}{6}-\frac{5\times6\times11}{6}$
$=1240-55=1185.$ View full question & answer→MCQ 161 Mark
Find the sum of series $1^2+ 3^2+ 5^2+$…………………………..$+ 11^2$.
Answer
- 286
solution:
$1^2+ 3^2+ 5^2+…………………………..+ 11^2$
$= (1^2+ 2^2+ 3^2+……+ 11^2) – (2^2+ 4^2+ 6^2+ 8^2+ 10^2)$
$= (1^2+ 2^2+ 3^2+……11^2) – 2^2(1^2+ 2^2+ 3^2+ 4^2+5^2)$
$=\frac{16\times12\times23}{6}-\frac{4\times5\times6\times11}{6}$
$=506-220=286.$ 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. $\text{n}-1-\frac{1}{2^{\text{n}}}$
- $\text{n}-1-\frac{1}{2^{\text{n}}}$
Solution:
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 $\mathrm{t}_{\mathrm{n}}$ denotes the $\mathrm{n}^{\text {th }}$ term of the series $2+3+6+11+18+\ldots$ then $\mathrm{t}_{50}$ is:
- A
$49^2-1$
- B
$49^2$
- C
$50^2+1$
- ✓
$49^2+2$
AnswerCorrect option: D. $49^2+2$
- $49^2+2$
Solution:
$S_n=2+3+6+11+18+\ldots .+t_{50}$
Using method of difference, we get
$S_n=2+3+6+11+18+\ldots .+t_{50} \ldots$ (1)
And $S_n=0+2+3+6+11+\ldots .+t_{49}+t_{50} \ldots$ (2)
Subtracting eq. (2) from eq. (2), we get
$0=2+1+3+5+7+\ldots .-t_{50} \text { terms }$
$\Rightarrow \mathrm{t}_{50}=2+(1+3+5+7+\ldots$. 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 A.M. of two numbers is $\frac{15}{2}$ and their G.M. is 6, then find the two numbers.
AnswerWe know, 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, G.M. of two numbers a and b is $\sqrt{ab}$
$\Rightarrow\sqrt{ab}=6\Rightarrow\text{ab}=36.$
=> a(15-a) = 36 => 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 G.P. is 20 and common ratio is 4. Find the $5^{th}$ term.
Answer
- 5120
solution:
Given, $a=20$ and $\mathrm{r}=4$.
We know, $a_n=a r^{n-1}$
$\Rightarrow a_5=20 \times 4^4=20 \times 256=5120$ View full question & answer→MCQ 211 Mark
The $n$th term of a G.P. is 128 and the sum of its $n$ terms is 225 . If its common ratio is 2 , then its first term is:
Answer
- 1
Solution:
Let the firt term of the geometric progression $=\mathrm{x}$
Common ration $=2$
$\therefore 2$ nd term of the G.P. $=2 \mathrm{x}$
$\therefore 3$ rd term $=\left(2^2\right) \mathrm{x} \ldots$
N th term can be written as $=\left(2^{\mathrm{n}}\right) \mathrm{x}$
Sum of the $n$ terms $S=255$
as we can see, except $x$, all other terms in the G.P. are multiples of 2
and sum of all the terms is an odd number.
$\therefore \mathrm{x}$ must be an odd number.
now $\mathrm{n}^{\text {th }}$ term
$\left(2^{\mathrm{n}}\right) \mathrm{x}=128=\left(2^7\right) \times 1$
There are no factors of odd numbers in 128 , except 1
$\therefore x=1$
Series of 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 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. $\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
- $\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
Solution:
Sum of squares of first $n$ terms $=1^2+2^2+3^2+$......$+\mathrm{n}^2$
$k^3-(k-1)^3=3 k^2-3 k+1$
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 $a_{n+1}=a_n \times r$.
Answer
- geometric Progression
Solution:
Explanation: 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 5th 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 $a_{n+1}=a_n \times r$.
Answer
- geometric Progression
Solution:
Explanation: 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
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 281 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
- 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 291 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 301 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. $\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
- $\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}.$
Solution:
$\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\sum\limits^\text{n}_{\text{r}=1}\frac{\big[\frac{\text{r}(\text{r}+1)}{2}\big]^2}{\frac{\text{r}(\text{r}+1)}{2}}=\sum\limits^\text{n}_{\text{r}=1}\frac{\text{r}(\text{r}+1)}{2}=\frac{1}{2}\bigg[\sum\limits^\text{n}_{\text{r}=1}\text{r}^2+\sum\limits^\text{n}_{\text{r}=1}\text{r}\bigg]$
$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}{+1})(2\text{n}{+1})}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=\frac{1}{2}.\frac{\text{n}({\text{n}+1})}{2}\Big[\frac{2\text{n}{+1}}{3}+1\Big]=\frac{\text{n}(\text{n}+1)}{4}\Big[\frac{2\text{n}+4}{3}\Big]$
$=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$ View full question & answer→MCQ 311 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. $a_n=a \times r^{n-1}$
- $a_n=a \times r^{n-1}$
Solution:
Since every term of $a_n$ G.P. is $r$ times the previous term.
i.e. $a_{n+1}=a_n{ }^* r=a_{n-1}{ }^* r^2=\ldots . . .=a_1{ }^* r^n$
or $a_n=a^* r^{n-1}$ View full question & answer→MCQ 321 Mark
Choose the correct answer. Let $S_n$ denote the sum of the first $n$ terms of an A.P. If $S_{2 n}=3 S_n$, then $S_{3 n}: S_n$ is equal to:
Answer
- 6
Solution:
Let 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 331 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.
Answer
- 372
Solution:
We know, a = 20, d = 2, $a_n$ = 42.
a + (n - 1) d = 42 ⇒ 20 + 2(n - 1) = 42
⇒ 2 (n - 1) = 42 - 20 = 22 ⇒ n - 1 = 11 ⇒ 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 341 Mark
If a sequence is in the form 2 × 5n then which of the following may be the sequence?
Answer
- Geometric Progressio
Solution:
If $a_n=2 \times 5 n$ 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 351 Mark
If general term of an A.P. is 3n then find common difference.
Answer
- 3
Solution:
Given, $a_n=3 n$.
We know, $d=a_n-a_{n-1}=3 n-3(n-1)=3$. View full question & answer→MCQ 361 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 371 Mark
The sum of $n$ terms of the infinite series $1.3^2+2.5^2+3.72 \ldots \infty$ is:
AnswerCorrect option: A. $n / 6(n+1)\left(6 n^2+14 n+7\right)$
- $n / 6(n+1)\left(6 n^2+14 n+7\right)$
View full question & answer→MCQ 381 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:
View full question & answer→MCQ 391 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 401 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. $\frac{\text{n}(\text{n+1)}(2\text{n}+1}{6}$
- $\frac{\text{n}(\text{n+1)}(2\text{n}+1}{6}$
Solution:
Sum of cubes of first $n$ terms $=1^3+2^3+3^3+\ldots . . . . . . . . . . .+n^3$
$(k+1)^4-k^4=4 k^3+6 k^2+4 k+1$
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 411 Mark
The sixth term of an AP is equal to 2. The value of the common difference of the AP which makes the product T1 T4 T5 least, is given by:
AnswerCorrect option: C. $\frac{2}{3}$
View full question & answer→MCQ 421 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:
MCQ 431 Mark
Which of the following is the geometric mean of 3 and 12.
AnswerExplanation: We know, geometric mean of two numbers a and b is given by
$\text{G}.\text{M}.=\sqrt{\text{a}\times\text{b}}$
$\text{SO},\text{G}.\text{M}.\text{ of }3\text{ and }12\text{ is }\sqrt{3}\times12=\sqrt{36}=6.$
View full question & answer→MCQ 441 Mark
The solution of the equation (x + 1) + (x + 4) + (x + 7) + …+ (x + 28) = 155 is:
MCQ 451 Mark
If an increasing GP is considered, then the number of terms in GP is:
MCQ 461 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 471 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 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 481 Mark
What is the first term of Fibonacci sequence?
Answer
- 1
Solution:
$a_1= 1$ 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 491 Mark
1 + 2 + 3 + 4 or 10 is a series?
- ✓
- B
- C
- D
Neither 1 + 2 + 3 + 4 nor 10
Answer1 + 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 501 Mark
In any case, the difference of the least and greatest term is:
MCQ 511 Mark
The product of n positive numbers is unity. Their sum is:
MCQ 521 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 531 Mark
If $log_{ax}, log_{bx}, log_{cx}$ be in HP, then a, b, c are in:
View full question & answer→MCQ 541 Mark
If $a + 2b + 3c = 12 , (a, b, c ∈ R+)$, then $ab^2c^3$ is:
AnswerCorrect option: C. $\leq2^6$
View full question & answer→MCQ 551 Mark
A sequence is called ___________________ if $a_{n+1} = a_n + d$.
Answer
- Rithmetic progression.
Solution:
A 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 561 Mark
Sumof nterms of series12 + 16 + 24 + 40 + …will be:
MCQ 571 Mark
If Tn denotes the nth term of the series 2 + 3 + 6 + 11 + 18 + . . . , then $T_{50}$ is:
View full question & answer→MCQ 581 Mark
How many terms of G.P. 2,4,8,16, …………… are required to give sum 254?
Answer
- 7
Solution:
$=2$ 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 591 Mark
Find the sum of series $6^3+ 7^3+$………………..…..$+ 20^3$.
Answer
- 43875
Solution:
$6^3+ 7^3+………………..…..+ 20^3$
$= (1^3+ 2^3+ 3^3+……..+ 20^3) – (1^3+ 2^3+ 3^3+ 4^3+ 5^3)$
$=\Big(\frac{20\times21}{2}\Big)^2-\Big(\frac{5\times6}{2}\Big)^2$
$=(210)^2-(15)^2$
$=225\times195$
$=43875$ View full question & answer→MCQ 601 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:
MCQ 611 Mark
For an increasing A.P. a1, a2, a3..... an, if a1, a3, a5 = -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 621 Mark
If in an A.P., first term is 20, common difference is 2 and $n^{th}$ term is 42, then find n.
Answer
- 12
Solution:
$\text { We know, } a=20, d=2, a_n=42$
$a+(n-1) d=4220 \Rightarrow+2(n-1)=42$
$\Rightarrow 2(n-1)=42-20=22 \Rightarrow n-1=11 \Rightarrow n=12$ View full question & answer→MCQ 631 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 $P^2R^3 : S^3$ is equal to:
Answer
- 1 : 1
Solution:
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 641 Mark
If the first term of a G.P. $a_1, a_2, a_3, \ldots$ is unity such that $4 a_2+5 a_3$ is least, then common ratio of G.P. is:
- ✓
$\frac{-2}{5}$
- B
$\frac{-3}{5}$
- C
$\frac25$
- D
AnswerCorrect option: A. $\frac{-2}{5}$
- $-\frac{2}{5}$
Solution:
If the first term is 1, then, the 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 $\text{r}+\frac25=0,\text{ i.e. }\text{r}=-\frac25.$ View full question & answer→MCQ 651 Mark
Find the sum of series $1^3+3^3+5^3+$…………………………..$+11^3$.
Answer
- 2556
Solution:
$1^3+3^3+5^3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+11^3$
$=\left(1^3+2^3+3^3+\ldots . .+11^3\right)-\left(2^3+4^3+6^3+8^3+10^3\right)$
$=\left(1^3+2^3+3^3+\ldots . . .11^3\right)-2^3\left(1^3+23+3^3+4^3+5^3\right)$
$\Big(\frac{11\times12}{2}\Big)^2-8\Big(\frac{5\times6}{2}\Big)^2$
$=66^2-8\times15^2$
$4356-1800=2556.$ View full question & answer→MCQ 661 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.
Answer$\frac{(\text{A.}\text{M.})}{\text{(G.}\text{M.)}}=\frac{5}{3}$
$\Rightarrow\frac{\text{a}+\text{b}}{2\sqrt{a}\text{b}}=\frac{5}{3}$
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 671 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
$3$
- ✓
$\frac{1}{3}$
- C
$2$
- D
$\frac{1}{2}$
AnswerCorrect option: B. $\frac{1}{3}$
- $\frac{1}{2}.$
Solution:
Since, x, 2y and 3z are in A. P., we get
$\text{2y}=\frac{\text{x}+3\text{z}}{2}$
$\Rightarrow 4y = x + 3z$
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)]
$\Rightarrow 4r = 1 + 3r^2$
$\Rightarrow 3r^2 - 4r + 1 = 0$
$\Rightarrow (3r - 1)(r - 1) = .0$
$\Rightarrow\text{r}=\frac{1}{3}$
(For r = 1; x, y, z are not distinct) View full question & answer→MCQ 681 Mark
Find the sum to 6 terms of each of the series 2*3+4*6+6*11+8*18+………………..
Answer
- 966
Solution:
General 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 691 Mark
Find the sum $1^3+2^3+3^3+……………+8^3$.
Answer
- 1296
Solution:
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 701 Mark
Sum of n terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ ...\text{ is}$
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
- $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
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 711 Mark
The sum of the infinity of the series 1+2/3 + 6/32 + 10/33 +14/34 is:
MCQ 721 Mark
Choose the correct answer. If 9 times the 9th term of an A.P. is equal to 13 times the $13^{th}$ term, then the 22nd term of the A.P. is:
Answer
- 0
Solution:
Let the first term and common difference of given A.P. be a and d, respectively.
It is given that $9 \times \mathrm{t}_9=13 \times \mathrm{t}_{13}$
$\Rightarrow 9(a+8 d)=13(a+12 d)$
$\Rightarrow 9 a+72 d=13 a+156 d$
$\Rightarrow 4 a+84 d=0$
$\Rightarrow 4(a+21 d)=0$
$\Rightarrow t_{22}=0$ View full question & answer→MCQ 731 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 741 Mark
$a_1=a_2=2, a_n=a_n-1-1, n>2 \text {. Find } a_5$
Answer
- -1
Solution:
$\Rightarrow a_n=a_n-1-1, n>2$
$\Rightarrow a_3=a_2-1=2-1=1$
$\Rightarrow a_4=a_3 1=11=0$
$\Rightarrow a_5=a_4-1=0-1=-1$. View full question & answer→MCQ 751 Mark
If a be A.M. and p, q be two 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. $\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
- $\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
Solution:
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 761 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?
MCQ 771 Mark
Choose the correct answer. If in an A.P., $\mathrm{S}_{\mathrm{n}}=\mathrm{qn}^2$ and $\mathrm{S}_{\mathrm{m}}=\mathrm{qm}^2$, where $\mathrm{S}_{\mathrm{r}}$ denotes the sum of r terms of the AP , then $\mathrm{S}_{\mathrm{q}}$ equals:
- A
$\frac{\text{q}^3}{2}$
- B
- ✓
$q^3$
- D
$(m+n) q^2$
Answer
- $q^3$.
Solution:
Given,
$S n=q n^2 \text { and } S m=q m^2$
$\therefore S_1=q, S_2=4 q, S_3=9 q \text { and } S_4=16 q$
Now, $\mathrm{t}_1=\mathrm{q}$
$\therefore t_2=S_2-S_1=4 q-q=3 q$
$t_3=S_3-S_2=9 q-4 q=5 q$
$t_4=S_4-S_3=16 q-9 q=7 q$
So, the A.P. is: $q, 3 q, 5 q, 7 q, \ldots$.
Thus, first term is $q$ and common difference is $3 q-q=2 q$.
$\therefore\ \text{S}_\text{q}=\frac{\text{q}}{2}[2\times\text{q}+(\text{q}-1)2\text{q}]=\frac{\text{q}}{2}\times[2\text{q}+2\text{q}^2-2\text{q}]$
$=\frac{\text{q}}{2}\times2\text{q}^2=\text{q}^3$ View full question & answer→MCQ 781 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 791 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 801 Mark
If a, b, c are in AP, then the straight line ax + by + c = 0 will always pass through the point:
MCQ 811 Mark
If $a_1, a_2, a_n$, are in AP with common difference $d$, then the sum of the series $\sin d\left(\operatorname{cosec} a_1 \operatorname{cosec} a_2+\operatorname{cosec} a_2 \operatorname{cosec}\right.$ $a_3+\ldots+\operatorname{cosec} a_{n-1} \operatorname{cosec} a_n$ ) is:
AnswerCorrect option: B. $\cot a_1-\cot a_n$
View full question & answer→MCQ 821 Mark
In G.P. 4, 8, 16, 32, ………… find the sum up to 5th term.
AnswerIn the given 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 831 Mark
If the sum of the roots of the equation $a x^2 b x+c=0$ is be equal to the sum of the reciprocals of their squares, then $b c^2, c a^2, a b^2$ will be in:
View full question & answer→MCQ 841 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. $\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
- $\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
Solution:
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 851 Mark
A sequence is called ___________________ if $a_{n+1} = a_n + d$.
Answer
- Arithmetic progression
Solution:
A 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 861 Mark
If the altitudes of a triangle are in AP, then the sides of the triangle are in:
- A
- ✓
- C
- D
arithmetico-geometric progression
View full question & answer→MCQ 871 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. $\frac{8\pm3\sqrt{5}}{1}$
- $\frac{8\pm3\sqrt{7}}{1}$
Solution:
We know, G.M. of two numbers a and b is $\sqrt{\text{ab.}}$
$\text{so},\text{a}+\text{b}=\sqrt{\text{ab}}$
Squaring we get, $a^2+b^2 = 16ab$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{a}}{\text{b}}\Big)$
$\text{Let}\times=\frac{\text{a}}{\text{b}}.$
$\text{So},\times+\frac{1}{\times}=16$
$\Rightarrow x^2 – 16x + 1 = 0$
$\Rightarrow\times=\frac{16\pm\sqrt{256-4}}{4}=\frac{16\pm\sqrt{252}}{2}=\frac{16\pm6\sqrt{7}}{2}=\frac{8\pm3\sqrt{7}}{1}.$ View full question & answer→MCQ 881 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:
Answer
- 8
Solution:
The 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.
$\therefore \mathrm{q}^2=\mathrm{p} \cdot \mathrm{r} \cdots(\mathrm{i})$
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, $2 r=q+p$
Now, putting the values of $p$ and $r$ in (i):
$q^2=(q+12)(q+6)$
$\Rightarrow q^2=q^2+18 q+72$
$\Rightarrow 18 q+72=0$
$\Rightarrow q+4=0$
$\Rightarrow q=-4$
Now, putting the value of $q$ in $p=q+12$ :
$p=-4+12=8$ View full question & answer→MCQ 891 Mark
Which of the following relation gives Fibonacci sequence?
- ✓
$\text{a}_\text{n}=\text{a}_{\text{n}-1}+\text{a}_{\text{n}-2}$
- B
$\text{a}_{\text{n}-1}=\text{a}_\text{n}+\text{a}_{\text{n}-2}$
- C
$\text{a}_{\text{n}-2}=\text{a}_\text{n}+\text{a}_{\text{n}-1}$
- D
$\text{a}_\text{n}=\text{a}_{\text{n}+1}+\text{a}_{\text{n}-2}$
AnswerCorrect option: A. $\text{a}_\text{n}=\text{a}_{\text{n}-1}+\text{a}_{\text{n}-2}$
This is a recurrence relation which gives Fibonacci sequence.
View full question & answer→MCQ 901 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 $\mathrm{S}_{\mathrm{n}}$ less than 1 sq cm ?
View full question & answer→MCQ 911 Mark
In a G.P., 5th term is 27 and 8th term is 729. Find its 11th term.
Answer
- 19683
Solution:
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 921 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. $121\big(\sqrt{6}+\sqrt{2}\big)$
- $121\big(\sqrt{6}+\sqrt{2}\big)$
Solution:
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 931 Mark
Find the sum to n terms of the series whose nth 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. $\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{6}$
- $\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{6}$
Solution:
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 941 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 951 Mark
If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is:
AnswerCorrect option: B. $\frac{\text{q}-\text{r}}{\text{p}-\text{q}}$
- $\frac{\text{q}-\text{r}}{\text{p}-\text{q}}$
Solution:
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 961 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 $216cm^3$ and the total surface area is $252cm^2$, then the length of the longest edge is:
Answer
- 12cm.
Solution:
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 × 2 = 12 View full question & answer→MCQ 971 Mark
If 2p + 3q + 4r = 15, then the maximum value of $p_3q5 r_7$ is:
AnswerCorrect option: C. $5^5\times\frac{7^7}{2^{17}}\times9$
- $5^5\times\frac{7^7}{2^{17}}\times9$
View full question & answer→MCQ 981 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 991 Mark
If r = 1 in a G.P. then what is the sum to n terms?
AnswerIf a is the first term of G.P., then G.P.
look like a, a, a, a, …………
Then sum to n terms becomes n × a.
View full question & answer→MCQ 1001 Mark
If in an A.P., first term is 20, common difference is 2 and nth term is 42, then find n.
Answer
- 12
solution:
We know, a = 20, d = 2, $a_n$ = 42.
a + (n - 1) d = 42 => 20 + 2(n - 1) = 42
⇒ 2 (n - 1) = 42 - 20 = 22 ⇒ n - 1 = 11 ⇒ n = 12. View full question & answer→MCQ 1011 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 1021 Mark
A person is to count 4500 currency notes. Let an denotes the number of notes he counts in the $n$th minute. If $a_1=a 2$ $=\ldots$. $a_{10}=150$ and $a_{10}, b_{11}, \ldots$ are in AP with common difference 2 , then the time taken by him to count all notes, is:
View full question & answer→MCQ 1031 Mark
The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+\ ....$ to n terms is:
- A
$\text{n}-\frac{1}{2}(3^{-\text{n}}-1)$
- ✓
$\text{n}-\frac{1}{2}(1-3^{-\text{n}})$
- C
$\text{n}+\frac{1}{2}(3^\text{n}-1)$
- D
$\text{n}-\frac{1}{1}(3^\text{n}-1)$
AnswerCorrect option: B. $\text{n}-\frac{1}{2}(1-3^{-\text{n}})$
- $\text{n}-\frac{1}{2}(1-3^{-\text{n}})$
Solution:
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
$\text{S}_\text{n}=\sum^\limits\text{n}_{\text{k}=1}\text{T}_\text{k}$
$=\sum^\limits\text{n}_{\text{k}=1}\Big[1-\frac{1}{3^\text{k}}\Big]$
$=\sum^\limits\text{n}_{\text{k}=1}1-\sum^\limits\text{n}_{\text{k}=1}\frac{1}{3^\text{k}}$
$=\text{n}-\Big[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ ....\ +\frac{1}{3^\text{n}}\Big]$
$=\text{n}-\frac{1}{3}\Bigg[\frac{1-\big(\frac{1}{3}\big)^\text{n}}{1-\frac{1}{3}}\Bigg]$
$=\text{n}-\frac{1}{2}\Big[1-\Big(\frac{1}{3}\Big)^\text{n}\Big]$
$=\text{n}-\frac{1}{3}\big[1-3^{-\text{n}}\big]$ View full question & answer→MCQ 1041 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 1051 Mark
If $a, b, c$ are in $A P$, then $10^{a x+10}, 10^{b x+10}, 10^{c x}+10(x \neq 0)$ are in:
View full question & answer→MCQ 1061 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 1071 Mark
The sum of n terms of two arithmetic progressions are in the ratio (2n + 3) : (7n + 5). Find the ratio of their 9th terms.
Answer
- 9 : 31
Solution:
xplanation: Let a, a’ be the first terms and d, d’ be the common differences of 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 1081 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 1091 Mark
If second term of a 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 1101 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 1111 Mark
Choose the correct answer. If the third term of G.P. is 4, then the product of its first 5 terms is:
Answer
- $4^5$
Solution:
Given that:
$T_3=4$
$\Rightarrow ar^{3-1}=4\left[\because T_{n}= a r ^{n-1}\right]$
$\Rightarrow ar^2=4$
Product of first 5 terms $=a \cdot a r . a r^2 . a r^3 . a r^4$
$=a^5 r^{10}=\left(a r^2\right)^5=(4)^5$
Hence, the corrrect option is (c). View full question & answer→MCQ 1121 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$
b. $ \frac{144}{15},12, 15$
View full question & answer→MCQ 1131 Mark
If 100 times the 100th term of an AP with non-zero common difference equals the 50 times its 50th term, then the 150th term of this AP is:
MCQ 1141 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
- ✓
$\frac{2}{3}$
- C
$\frac{3}{2}$
- D
AnswerCorrect option: B. $\frac{2}{3}$
Let three terms be $\frac{\text{a}}{\text{r}},$ a, × 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} $
$\Rightarrow\Big(\frac{1}{\text{r+1+1}\times\text{r}}\Big)=\Big(\frac{\frac{21}{2}}{3}\Big)=\frac{7}{2}$
$\Rightarrow\Big(\text{r}^2+\text{r}+1=\Big)\Big(\frac{7}{2}\Big)\Rightarrow\text{r}^2 -\Big(\frac{5}{2}\Big)\text{r}+1=0$
$\Rightarrow\text{r}=2\text{ and }\frac{1}{2}.$
Terms are $\frac{3}{2}$, 3, 3 × 2 i.
e.$\frac{3}{2}$ , 3, 6.
View full question & answer→MCQ 1151 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. $\frac{2\text{a}^2}{3}-\text{n}(\text{n-1})(2\text{n-1)}$
- $\frac{2\text{a}^2}{3}-\text{n}(\text{n-1})(2\text{n-1)}$
View full question & answer→MCQ 1161 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 1171 Mark
If the non-zero numbers x, y, z are in AP and tan - 1 , tan - 1 , tan - 1 x y z are in AP, then:
MCQ 1181 Mark
In a 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 G.P. is:
- A
$-\frac{4}{5}$
- B
$\frac{1}{5}$
- ✓
- 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 1191 Mark
If sum of n terms of an A.P. is $n^2+5n$ then find general term.
Answer
- 2n
Solution:
Given, $S_n=n^2+5 n$
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 1201 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 1211 Mark
In a 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. $\sqrt{\text{pq}}$
- $\sqrt{\text{pq}}$
Solution:
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 1221 Mark
The sum of first 20 terms of the sequence 0.7, 0.7 7, 0.7 7 7, … , is:
- A
$7 / 81\left(179-10^{-20}\right)$
- B
$7 / 9\left(99-10^{-20}\right)$
- ✓
$7 / 81\left(179+10^{-20}\right)$
- D
$7 / 9\left(99+10^{-20}\right)$
AnswerCorrect option: C. $7 / 81\left(179+10^{-20}\right)$
- $7 / 81\left(179+10^{-20}\right)$
View full question & answer→MCQ 1231 Mark
Find the sum $1^2+2^2+3^2+……………+10^2$.
Answer
- 435
Solution:
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 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. $\frac{\text{n}(\text{n}+3)}{4}$
- $\frac{\text{n}(\text{n}+3)}{4}$
Solution:
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. $\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
- $\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
Solution:
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 12th term is 120. Find the sum up to 12th term.
Answer
- 840
Solution:
Given, $a = 20, a_{12}= 120, n = 12$
$\text{s}_{\text{n}}=\frac{\text{n}}{2}(\text{a}+\text{I})$
$\Rightarrow\text{s}_{\text{12}}=\frac{12}{2}(20+120)$
$= 6 \times 140 = 840$ 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:
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:
MCQ 1291 Mark
In numbers from 1 to 100 the digit "0" appears ____________times:
Answer10, 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 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
$\sum\limits^{4}_{\text{i = 1}}2\text{n}+3=\text{________}$
Answer
- 32
Solution:
$a_1=2 \times 1+3=5, a_2=2 \times 2+3=7, a_3=2 \times 3+3=9, a_4=2 \times 4+3=11$
$\text { Sum }=5+7+9+11=32$ 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$\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54}$
$\Rightarrow4^{(3+6+9+12+\ \dots+3\text{x})}=\Big(\frac{625}{10000}\Big)^{-54}$
$\Rightarrow4^{3(1+2+3+4+\dots+\text{x})}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(4^{-2}\Big)^{-54}$
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?
Answer
- 7
Solution:
In the given G.P., In the given G.P., a=25 and
$\text{r}=\frac{125}{25}=5$
$\text { Given, } a_n=390625 \Rightarrow a r^{n-1}=390625$
$\Rightarrow 25 \times 5^{n-1}=390625$
$\Rightarrow 5^{n-1}=\frac{390625}{25}=15625=5^6$
$\Rightarrow{n-1}=6=>n=7 .$ View full question & answer→MCQ 1341 Mark
If $\sum\text{n}=210,$ then $\sum\text{n}^2=$
AnswerGiven,
$\sum\text{n}=210$
$\Rightarrow\text{n}\Big(\frac{\text{n}+1}{2}\Big)=210$
$\Rightarrow\text{n}^2+\text{n}-420=0$
$\Rightarrow(\text{n}-20)(\text{n}+21)=0$
$\Rightarrow\text{n}=20$ $(\because\ \text{n}>0)$
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
- 70
solution:
$S_n = 2+3+5+8+12+……………………………+ a_n$
$S_n = 2+3+5+8+12+ ……. + a_{n-1} + a_n$
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{tems}\text{ is}\Big(\frac{1}{2}\Big)(\text{n}^2-\text{n}+4\text{ 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
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=\Big(\frac{1}{2}\Big)\sum\limits^\text{n}_\text{i}=0\text{ k}^2-\Big(\frac{1}{2}\Big)\sum\limits^\text{n}_\text{i}=0\text{ k}+2\text{n}$
$=\frac{\text{n}(\text{n+1)}(2\text{n}+1)}{(2\times6)}$
$\text{Here},\text{n}=7.\text{ so},$
$\sum\limits^\text{n}_\text{i}=0\text{ a}\text{k}=\frac{(7\times8\times15)}{12}-\frac{(7\times8)}{4}+2\times7=70.$ View full question & answer→MCQ 1371 Mark
If $a_n=4 n+6$, find 15 th term of the sequence.
Answer
- 66
Solution:
$a_n=4 n+6 \text { and } n=15$
$\Rightarrow a_{15}=4 \times 15+6=60+6=66$. View full question & answer→MCQ 1381 Mark
Insert 4 numbers between 2 and 22 such that the resulting sequence is an A.P.
Answer
- 6, 10, 14, 18
Solution:
$\text { Let A.P. be } 2, A_1, A_2, A_3, A_4, 22$
$\Rightarrow a=2 \text { and } a_6=a+5 d=22 \Rightarrow 2+5 \times d=22 \Rightarrow d=4$
$A_1=a_2=a+d=2+4=6$
$A_2=A_1+d=6+4=10$
$A_3=10+4=14$
$A_4=14+4=18$ 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:
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. $(2\text{p}-\text{q})(\text{p}-2\text{q})$
- $(2\text{p}-\text{q})(\text{p}-2\text{q})$
Solution:
Let the two numbers be a and b.
a, p, q and b are in A.P.
$\therefore\text{ p}-\text{a}=\text{q}-\text{q}=\text{b}-\text{q}$
$\Rightarrow\text{ p}-\text{a}=\text{q}-\text{p}\text{ and}\text{ q}-\text{p}=\text{b}-\text{q}$
$\Rightarrow\text{ a}=2\text{p}-\text{q}\text{ and}\text{ b}=2\text{q}-\text{p}\cdots(\text{i})$
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:
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. $\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$
- $\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$
Solution:
Let $T_n$ be the nth 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 6th term whose nth term is given by $n^2 + 3^n$.
Answer
- 1183
Solution:
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 6th 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.
Answer
- 372
Solution:
We know, $a=20, d=2, a_n=42$.
$a+(n-1) d=42$
$\Rightarrow 20+2(n-1)=42$
$\Rightarrow 2(n-1)=42-20=22$
$\Rightarrow n-1=11$
$\Rightarrow n=12$
$s_a=\frac{n}{2}(a+I) \Rightarrow s_{12=\frac{12}{2}}(20+42)=6 \times 62=372$ View full question & answer→MCQ 1481 Mark
Choose the correct answer. If the sum of $n$ terms of an A.P. is given by $S_n=3 n+2 n^2$, then the common difference of the A.P. is:
Answer
- 4.
Solution:
Given 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 $d=a_2-a_1=9-5=4$
Hence, the correct option is (d). View full question & answer→MCQ 1491 Mark
If $|x|<1$, then the sum of the series $1+2 x+3 \times 2+4 \times 3 \ldots \infty$ will be:
- A
$1 / 1- x$
- B
$1 / 1+x$
- C
$1\left(1+x^2\right)$
- ✓
$1\left(1-x^2\right)$
AnswerCorrect option: D. $1\left(1-x^2\right)$
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 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. $\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$
- $\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}$
Solution:
Sum of n terms of the G.P., $\text{S}=\frac{\text{a}(\text{r}^{\text{n}}-1)}{(\text{r}-1)}$
Product of n terms of the 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 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 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 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 $a_1, a_2, a_3, \ldots . ., a_{20}$ are AM's between 13 and 67 , then the maximum value of a1 a2 a3 $\ldots$ a $20 \ldots$ 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
- 2
Solution:
$a_1=1 \text { and } a_2=1$
$a_n=a_{n-1}+a_{n-2}, n>2 .$
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 5th and 8th term of a GP be 48 and 384 respectively, then the common ratio of 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
- D
AnswerCorrect option: B. A is true, R is true; R is not a correct explanation of A.
- A is true, R is true; R is not a correct explanation of A.
Solution:
Assertion Let a be the first term and r be the common ratio of the given GP. According to the question,
$T_5=48 \Rightarrow a r^4=48 \ldots \text {...i) }$
$\text { and } T 8=384 \Rightarrow a r^7=384 \ldots \text { (ii) }$
On dividing Eq. (ii) by Eq. (i), we get
$=\frac{\mathrm{ar}^7}{\mathrm{ar}^4}=\frac{384}{48}$
$\Rightarrow \mathrm{r}^3=8$
$\Rightarrow \mathrm{r}=2$
Reason $18, x, 14$ are in AP.
$\Rightarrow x-18=14-x$
$\Rightarrow 2 x=32$
$\Rightarrow x=16$ View full question & answer→MCQ 1541 Mark
The two geometric means between the numbers 1 and 64 are:
Answer
- 4 and 16
Solution:
Let the two G.M.s between 1 and 64 be $G_1$ and $G_2$.
Thus, 1, $G_1, G_2$ and 64 are in 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 G.M.s. View full question & answer→