MCQ 511 Mark
The common difference of the A.P. $\frac{1}{2 x}, \frac{1-4 x}{2 x}, \frac{1-8 x}{2 x}, \ldots$, is
View full question & answer→MCQ 521 Mark
The $14^{\text {th }}$ term from the end of the A.P. $-11,-8,-5, \ldots, 49$ is
View full question & answer→MCQ 531 Mark
Three numbers in A.P. have the sum 30 . What is the middle term?
MCQ 541 Mark
Which term of the A.P. $-29,-26,-23, \ldots, 61$ is 16 ?
- A
$11^{ th }$
- B
$16^{\text {th }}$
- C
$10^{\text {th }}$
- ✓
$31^{\text {st }}$
AnswerCorrect option: D. $31^{\text {st }}$
View full question & answer→MCQ 551 Mark
The common difference of an A.P. in which $a_{15}-a_{11}=48$, is
View full question & answer→MCQ 561 Mark
If $11^{\text {th }}$ term from the end of the A.P.: $10,7,4, \ldots,-62$ is
Answer(C)-32
We find that $10,7,4, \ldots,-62$ is an A.P. with common difference $d=-3$.
$
\therefore \quad 11^{\text {th }} \text { term from the end }=a_n+(11-1) \times-d=a_n-10 d=-62-10 \times-3=-32
$
View full question & answer→MCQ 571 Mark
If the $n^{\text {th }}$ term of an AP is $3 n+7$, then its common difference is
Answer(B)3
A sequence is an AP with common difference ' $A^{\prime}$ if and only if its $n^{\text {th }}$ term is of the form $A n+B$. Here, $a_n=3 n+7$. So, it is $n$th term of an A.P. with common difference 3 .
ALITER $d=a_n-a_{n-1}=(3 n+7)-(3(n-1)+7)=3$
View full question & answer→MCQ 581 Mark
If $p-1, p+1$ and $2 p+3$ are in A.P., then the value of $p$ is
Answer(c)0
$p-1, p+1$ and $2 p+3$ will be in A.P., iff
$
2(p+1)=p-1+2 p+3 \Rightarrow 2 p+2=3 p+2 \Rightarrow p=0
$
View full question & answer→MCQ 591 Mark
The next term of the A.P. $\sqrt{6}, \sqrt{24}, \sqrt{54}, \ldots$, is
- A
$\sqrt{60}$
- ✓
$\sqrt{96}$
- C
$\sqrt{72}$
- D
$\sqrt{216}$
AnswerCorrect option: B. $\sqrt{96}$
View full question & answer→MCQ 601 Mark
If the sum of $n$ terms of an A.P. be $3 n^2+n$ and its common difference is 6 , then its first term is
View full question & answer→MCQ 611 Mark
The next $\left(4^{ th }\right)$ term of the A.P. $\sqrt{18}, \sqrt{50}, \sqrt{98}, \ldots$. is
- A
$\sqrt{128}$
- B
$\sqrt{140}$
- ✓
$\sqrt{162}$
- D
$\sqrt{200}$
AnswerCorrect option: C. $\sqrt{162}$
View full question & answer→MCQ 621 Mark
The value of $x$ for which $2 x, x+10$ and $3 x+2$ are the three consecutive terms of an A.P. is
View full question & answer→MCQ 631 Mark
The first term of an A.P. is $p$ and the common difference is $q$, then its $10^{\text {th }}$ term is
- A
$q+9 p$
- B
$p-9 q$
- ✓
$p+9 q$
- D
$2 p+9 q$
AnswerCorrect option: C. $p+9 q$
View full question & answer→MCQ 641 Mark
The first three terms of an A.P. respectively are $3 y-1,3 y+5$ and $5 y+1$. Then, $y$ equals
View full question & answer→MCQ 651 Mark
The next term of the A.P. $\sqrt{7}, \sqrt{28}, \sqrt{63}, \ldots$
- A
$\sqrt{70}$
- B
$\sqrt{84}$
- C
$\sqrt{97}$
- ✓
$\sqrt{112}$
AnswerCorrect option: D. $\sqrt{112}$
View full question & answer→MCQ 661 Mark
If $k, 2 k-1$ and $2 k+1$ are three consecutive terms of an AP, the value of $k$ is
View full question & answer→MCQ 671 Mark
The common difference of the A.P. $\frac{1}{2 b}, \frac{1-6 b}{2 b}, \frac{1-12 b}{2 b}, \cdots$ is
View full question & answer→MCQ 681 Mark
The common difference of the A.P. $\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}, \cdots$ is
- A
$\frac{1}{3}$
- B
$-\frac{1}{3}$
- ✓
$-b$
- D
$b$
View full question & answer→MCQ 691 Mark
The common difference of the A.P. is $\frac{1}{2 q}, \frac{1-2 q}{2 q}, \frac{1-4 q}{2 q}, \cdots$ is
View full question & answer→MCQ 701 Mark
The next term of the A.P. $\sqrt{18}, \sqrt{50}, \sqrt{98}, \ldots$ is
- A
$\sqrt{146}$
- B
$\sqrt{128}$
- ✓
$\sqrt{162}$
- D
$\sqrt{200}$
AnswerCorrect option: C. $\sqrt{162}$
(C)$\sqrt{162}$
Given A.P. is $3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots$. Clearly, first term $a=3 \sqrt{2}$ and common difference $d=2 \sqrt{2}$. So, the next term is $2 \sqrt{2}$ more than the preceding term $7 \sqrt{2}$. Hence, it is $9 \sqrt{2}=\sqrt{162}$.
View full question & answer→MCQ 711 Mark
The sum of first 20 odd natural numbers is
AnswerCorrect option: C. $4 0 0$
View full question & answer→MCQ 721 Mark
If the sum of $n$ terms of two A.P.'s are in the ratio $(2 n+3):(3 n+2)$, then the ratio of then $m^{\text {th }}$ terms is
- A
$(4 m-1):(6 m+1)$
- B
$(6 m+1):(4 m+1)$
- ✓
$(4 m+1):(6 m-1)$
- D
$(4 m+1):(m+6)$
AnswerCorrect option: C. $(4 m+1):(6 m-1)$
(C)$(4 m+1):(6 m-1)$
Let $a_1, a_2$ be the first terms and $d_1, d_2$ be the common differences of two A.P.'s. It it given that
$
\frac{\frac{n}{2}\left\{2 a_1+(n-1) d_1\right\}}{\frac{n}{2}\left\{2 a_2+(n-1) d_2\right\}}=\frac{2 n+3}{3 n+2} \Rightarrow \frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2}=\frac{2 n+3}{3 n+2} \Rightarrow \frac{a_1+\left(\frac{n-1}{2}\right) d_1}{a_2+\left(\frac{n-1}{2}\right) d_2}=\frac{2 n+3}{3 n+2}
$
Replacing $\frac{n-1}{2}$ by $m-1$ i.e. $n$ by $(2 m-1)$, we obtain
$
\frac{a_1+(m-1) d_1}{a_2+(m-1) d_2}=\frac{2(2 m-1)+3}{3(2 m-1)+2} \Rightarrow \frac{a_1+(m-1) d_1}{a_2+(m-1) d_2}=\frac{4 m+1}{6 m-1}
$
Hence, $m^{\text {th }}$ terms of two A.P.'s are in the ratio $(4 m+1):(6 m-1)$.
View full question & answer→MCQ 731 Mark
If the ratio of $18^{\text {th }}$ term to $11^{\text {th }}$ term of an A.P. is $3: 2$, then the ratio of the $21^{\text {st }}$ term to $5^{\text {t }}$ term is
- A
$3: 2$
- ✓
$3: 1$
- C
$1: 3$
- D
$2: 3$
AnswerCorrect option: B. $3: 1$
(B)$3: 1$
Let $a$ be the first term and $d$ be the common difference of the given A.P. It is given the
$
\frac{a_{18}}{a_{11}}=\frac{3}{2} \Rightarrow \frac{a+17 d}{a+10 d}=\frac{3}{2} \Rightarrow a=4 d \therefore \frac{a_{21}}{a_5}=\frac{a+20 d}{a+4 d}=\frac{4 d+20 d}{4 d+4 d}=\frac{3}{1}
$
View full question & answer→MCQ 741 Mark
If $n$th term of an A.P. is $(2 n+1)$, then the sum of first $n$ terms of the A.P. is
- A
$n(n-2)$
- ✓
$n(n+2)$
- C
$n(n+1)$
- D
$n(n-1)$
AnswerCorrect option: B. $n(n+2)$
(B)$n(n+2)$
We have, $a_n=2 n+1$. Therefore, $a_1=2 \times 1+1=3$
Hence, $\quad S_n=\frac{n}{2}\left(a_1+a_n\right)=\frac{n}{2}(3+2 n+1)=n(n+2)$
View full question & answer→MCQ 751 Mark
The sum of $n$ terms of the series $\sqrt{3}+\sqrt{12}+\sqrt{27}+\sqrt{48}+\ldots \ldots$, is
- A
$\frac{2 n(n+1)}{\sqrt{3}}$
- B
$\frac{\sqrt{3} n(n-1)}{2}$
- ✓
$\frac{\sqrt{3} n(n+1)}{2}$
- D
$\frac{2 n(n-1)}{\sqrt{3}}$
AnswerCorrect option: C. $\frac{\sqrt{3} n(n+1)}{2}$
(C)$\frac{\sqrt{3} n(n+1)}{2}$
Let $S_n=\sqrt{3}+\sqrt{12}+\sqrt{27}+\sqrt{48}+\ldots$ to $n$ terms. Then $S_n=\sqrt{3}+2 \sqrt{3}+3 \sqrt{3}+4 \sqrt{3}+\ldots$ to $n$ terms
$
\Rightarrow \quad S_n=\sqrt{3}(1+2+3+4+\ldots \text { to } n \text { terms })=\frac{\sqrt{3} n(n+1)}{2} \quad\left[\because 1+2+3+\ldots+n=\frac{n(n+1)}{2}\right].
$
View full question & answer→MCQ 761 Mark
If the sum of first $n$ odd natural numbers is 225 , then the value of $n$ is
Answer(A)15
We have, $n^2=225 \Rightarrow n=15$
View full question & answer→MCQ 771 Mark
The sum of first $n$ even natural numbers is
- A
$2 n$
- B
$n^2$
- ✓
$n^2+n$
- D
$n^2-1$
AnswerCorrect option: C. $n^2+n$
(C)$n^2+n$
Let $S_n$ be the sum of first $n$ even natural numbers. Then,
View full question & answer→MCQ 781 Mark
The sum of the first $n$ odd natural numbers is
- A
$2 n$
- B
$2 n+1$
- ✓
$n^2$
- D
$n^2-1$
Answer(C)$n^2$
First $n$ odd natural numbers are $1,3,5,7, \ldots,(2 n-1)$. Clearly, these numbers form an A.P. with first term 1, last term $=2 n-1$ and common difference $=2$. Let $S_n$ denote the sum. Then, $S_n=\frac{n}{2}(1+2 n-1)=n^2$
View full question & answer→MCQ 791 Mark
If $a_p$, be the $p$ th term of A.P. $3,15,27, \ldots$, such that $a_p-a_{50}=180$, then $p=$
Answer(B)65
The common difference of the given A.P. is 12 .
$
\begin{array}{ll}
\text { Now, } & a_p-a_{50}=180 \\
\Rightarrow & (p-50) d=180 \qquad \left[\because a_m-a_n=(m-n) d\right]\\
\Rightarrow & 12(p-50)=180 \Rightarrow p-50=15 \Rightarrow p=65
\end{array}
$
View full question & answer→MCQ 801 Mark
If the sum of $n$ terms of an A.P. is $S_n=3 n^2+4 n$, then common difference of the A.P. is
Answer(D)6
We know that a sequence is an A.P. with common difference $2 A$ iff the sum $S_n$ of its $n$ terms is $S_n=A n^2+B n$. Hence, $S_n=3 n^2+4 n$. So the common difference of the A.P. is $2 \times 3=6$.
View full question & answer→MCQ 811 Mark
If $a_1, a_2, a_3, \ldots, a_n, \ldots$ is an AP such that $a_{20}-a_{12}=-32$ then the common difference of the A.P. is
Answer(B)-4
$
\begin{array}{l}
\text { } a_m-a_n=(m-n) d \\
\Rightarrow \quad a_{20}-a_{12}=(20-12) d \Rightarrow-32=8 d \Rightarrow d=-4
\end{array}
$
View full question & answer→MCQ 821 Mark
If the common difference of an $A P$ is 7 , then $a_{25}-a_{21}$ is equal to
Answer(C)28
If $a_m$ and $a_n$ are $m^{\text {th }}$ and $n^{\text {th }}$ terms of an AP with common difference ' $d$ ', then
$
a_m-a_n=\{a+(m-1) d\}-\{a+(n-1) d\}=(m-n) d
$
Hence, for the given A.P., we obtain
$
a_{25}-a_{21}=(25-21) \times 7=28
$
ALITER
$
a_{25}-a_{21}=(25-21) d=4 d=4 \times 7=28
$
View full question & answer→MCQ 831 Mark
The $7^{\text {th }}$ term from the end of the A.P. 7, 11, 15, ..., 107, is
Answer(B)83
The common difference and the last term of the given A.P. are 7 and 107 respectively. If $a_n$ is the last term of an AP with common difference ' $d$ '. Then,
$
\begin{array}{l}
\quad k^{\text {th }} \text { term from the end }=a_n-(k-1) d . \\
\therefore \quad 7^{\text {th }} \text { term from the end }=107-(7-1) \times 4=107-24=83
\end{array}
$
View full question & answer→MCQ 841 Mark
The $12^{\text {th }}$ term of an AP whose first two terms are -3 and 4 is
Answer(B)74
Let $a$ be the first term and $d$ be the common difference of the given A.P. It is given that $a=-3$ and $a+d=4 \Rightarrow a=-3$ and $d=7$
$
\therefore \quad 12^{\text {th }} \text { term }=a+(12-1) d=a+11 d=-3+11 \times 7=74
$
View full question & answer→MCQ 851 Mark
The sum of $n$ terms of two A.P.'s are in the ratio $5 n+9: 9 n+6$. Then, the ratio of their $18^{\text {th }}$ term is
- ✓
$\frac{184}{321}$
- B
$\frac{178}{321}$
- C
$\frac{175}{321}$
- D
$\frac{176}{321}$
AnswerCorrect option: A. $\frac{184}{321}$
View full question & answer→MCQ 861 Mark
If $S_n$ denote the sum of $n$ terms of an A.P. with first term $a$ and common difference $d$ such that $\frac{S_x}{S_{k x}}$ is independent of $x$, then
- A
$d=a$
- ✓
$d=2 a$
- C
$a=2 d$
- D
$d=-a$
AnswerCorrect option: B. $d=2 a$
View full question & answer→MCQ 871 Mark
If the sums of $n$ terms of two arithmetic progressions are in the ratio $\frac{3 n+5}{5 n+7}$, then their $n^{\text {th }}$ terms are in the ratio
- A
$\frac{3 n-1}{5 n-1}$
- ✓
$\frac{3 n+1}{5 n+1}$
- C
$\frac{5 n+1}{3 n+1}$
- D
$\frac{5 n-1}{3 n-1}$
AnswerCorrect option: B. $\frac{3 n+1}{5 n+1}$
View full question & answer→MCQ 881 Mark
If $S_r$ denotes the sum of the first $r$ terms of an A.P. Then, $S_{3 n}:\left(S_{2 n}-S_n\right)$ is
View full question & answer→MCQ 891 Mark
In an $AP , S_p=q, S_q=p$ and $S_r$ denotes the sum of first $r$ terms. Then, $S_{p+q}$ is equal to
- A
$0$
- ✓
$-(p+q)$
- C
$p+q$
- D
$p q$
AnswerCorrect option: B. $-(p+q)$
View full question & answer→MCQ 901 Mark
If $S_n$ 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
View full question & answer→MCQ 911 Mark
If in an A.P., $S_n=n^2 p$ and $S_m=m^2 p$, where $S_r$ denotes the sum of $r$ terms of the A.P., then $S_p$ is equal to
- A
$\frac{1}{2} p^3$
- B
$m n p$
- ✓
$p^3$
- D
$(m+n) p^2$
View full question & answer→MCQ 921 Mark
If $S_1$ is the sum of an arithmetic progression of ' $n$ ' odd number of terms and $S_2$ the sum of the terms of the series in odd places, then $\frac{S_1}{S_2}=$
- ✓
$\frac{2 n}{n+1}$
- B
$\frac{n}{n+1}$
- C
$\frac{n+1}{2 n}$
- D
$\frac{n+1}{n}$
AnswerCorrect option: A. $\frac{2 n}{n+1}$
View full question & answer→MCQ 931 Mark
If the first, second and last term of an A.P. are $a, b$ and $2 a$ respectively, its sum is
- A
$\frac{a b}{2(b-a)}$
- B
$\frac{a b}{b-a}$
- ✓
$\frac{3 a b}{2(b-a)}$
- D
AnswerCorrect option: C. $\frac{3 a b}{2(b-a)}$
View full question & answer→MCQ 941 Mark
If the sum of first $n$ even natural numbers is equal to $k$ times the sum of first $n$ odd natural numbers, then $k=$
- A
$\frac{1}{n}$
- B
$\frac{n-1}{n}$
- C
$\frac{n+1}{2 n}$
- ✓
$\frac{n+1}{n}$
AnswerCorrect option: D. $\frac{n+1}{n}$
View full question & answer→MCQ 951 Mark
The first and last term of an A.P. are $a$ and $l$ respectively. If $S$ is the sum of all the terms of the A.P. and the common difference is given by $\frac{l^2-a^2}{k-(l+a)}$, then $k=$
View full question & answer→MCQ 961 Mark
Let $S_n$ denote the sum of $n$ terms of an A.P. whose first term is $a$. If the common difference $d$ is given by $d=S_n-k S_{n-1}+S_{n-2}$, then $k=$
View full question & answer→MCQ 971 Mark
If the sum of $P$ terms of an A.P. is $q$ and the sum of $q$ terms is $p$, then the sum of $p+q$ terms will be
- A
$0$
- B
$p-q$
- C
$p+q$
- ✓
$-(p+q)$
AnswerCorrect option: D. $-(p+q)$
View full question & answer→MCQ 981 Mark
The sum of first 24 terms of the sequence whose $n$th terms is given by $a_n=3+\frac{2}{3} n$
View full question & answer→MCQ 991 Mark
If the $n^{\text {th }}$ term of an A.P. is $2 n+1$, then the sum of first $n$ terms of the A.P. is
- A
$n(n-2)$
- ✓
$n(n+2)$
- C
$n(n+1)$
- D
$n(n-1)$
AnswerCorrect option: B. $n(n+2)$
View full question & answer→MCQ 1001 Mark
The sum of $n$ terms of an A.P. is $3 n^2+5 n$, then 164 is its
- A
$24^{\text {th }}$ term
- ✓
$27^{\text {ih }}$ term
- C
$26^{\text {th }}$ term
- D
$25^{\text {th }}$ term
AnswerCorrect option: B. $27^{\text {ih }}$ term
View full question & answer→