MCQ 11 Mark
Mark the correct alternative in the following: If the sum of three consecutive terms of an increasing $A.P$. is $51$ and the product of the first and third of these terms is $273,$ then the third term is:
Answer Given,
$a_1+a_2+a_3=51 \ldots . . \text { (i) } $
and $ a_1 \times a_3=273 \ldots . .(\text { ii) } $
We know $, a_n=a+(n-1) d $
$ 1^{\text {st }} \text { term, } a_1=a+(1-1) d $
$ \Rightarrow a_1=a+0 $
$ \Rightarrow a_1=a $
$ 2^{\text {nd }} \text { term, } a_2=a+(2-1) d $
$ \Rightarrow a_2=a+d $
$ 3^{\text {rd }} \text { term, } a_3 a+(3-1) d $
$ \Rightarrow a_3=a+2 d \ldots . . \text { (iii) }$
Put the values in eq $(i)$
$ \Rightarrow a+a+d+a+2 d=51 $
$ \Rightarrow 3 a+3 d=51 $
$ \Rightarrow 3(a+d)=51 $
$ \Rightarrow a+d=\frac{51}{3} $
$ \Rightarrow a+d=17 $
$ \Rightarrow a=17-d$
Put the value of $a_1$ and $a_3$ in eq. $(ii)$
$\Rightarrow(a)(a+2 d)=273 $
$ \Rightarrow(17-d)(17-d+2 d)=273 $
$ \Rightarrow(17-d)(17+d)=273 $
$ \Rightarrow 17^2-d^2=273$
$ {[\because(a-b)(a+b)=a 2-b 2]} $
$ \Rightarrow 289-d^2=273 $
$ \Rightarrow 289-273=d^2 $
$ \Rightarrow d^2=16 $
$ \Rightarrow d=\sqrt{16}=4$
$d=4$ because $d$ never be negative in incueasing $A.P$.
To find $a_3$ put the value of $d$ and $a$ eq. $(iii)$
$ a_3=a+2 d $
$ \Rightarrow a_3=17-d+2 d $
$ \Rightarrow a_3=17-4+2(4) $
$ \Rightarrow a_3=13+8 $
$ \Rightarrow a_3=21$
Hence, carrect chice is $(C)$.
View full question & answer→MCQ 21 Mark
Mark the correct alternative in the following:
If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times, the least, then the numbers are:
AnswerHere, we are given that four numbers are in A.P., such that their sum is 50 and the greatest number is 4 times the smallest.
So, let us take the four terms as a - d, a, a + d, a + 2d.
Now, we are given that sum of these numbers is 50, so we get,
(a - d) + (a) + (a + d) + (a + 2d) = 50
a - d + a + a + d + a + 2d = 50
4a + 2d = 50
2a + d = 50 .....(i)
Also, the greatest number is 4 times the smallest, so we get,
a + 2d = 4(a - d)
a + 2d = 4a - 4d
4d + 2d = 4a - a
6d = 3a
$\text{d}=\frac{3}{6}\text{d}\ .....(\text{ii})$
Now, using (ii) in (i), we get,
$2\text{a}+\frac{3}{6}\text{a}=25$
$\frac{12\text{a}+3\text{a}}{6}=25$
$15\text{a}=150$
$\text{a}=\frac{150}{15}$
$\text{a}=10$
Now, using the value of a in (ii), We get,
$\text{d}=\frac{3}{6}(10)$
$\text{d}=\frac{10}{2}$
$\text{d}=5$
So, first term is given by,
a - d = 10 - 5
= 5
Second term is given by,
a = 10
Third term is given by,
a + d = 10 + 5
= 15
Fourth term is given by,
a + 2d = 10 + (2)(5)
= 10 + 10
= 20
Therefore, the four terms are 5, 10, 15, 20.
Hence, the correct opting is (A).
View full question & answer→MCQ 31 Mark
Mark the correct alternative in the following:
The common difference of the A.P. is $\frac{1}{2\text{q}},\frac{1-2\text{q}}{2\text{q}},\frac{1-4\text{q}}{2\text{q}}, .....$ is
AnswerLet a be the first term and d be the common difference.
The given A.P. is $\frac{1}{2\text{q}},\frac{1-2\text{q}}{2\text{q}},\frac{1-4\text{q}}{2\text{q}}, .....$
Common difference = d = Second term - First term
$=\frac{1-2\text{q}}{2\text{q}}-\frac{1}{2\text{q}}$
$=\frac{-2\text{q}}{2\text{q}}=-1$
Hence, the correct option is (A).
View full question & answer→MCQ 41 Mark
Mark the correct alternative in the following:
The next term of the A.P. $\sqrt{7},\sqrt{28},\sqrt{63},\ .....$
- A
$\sqrt{70}$
- B
$\sqrt{84}$
- C
$\sqrt{97}$
- ✓
$\sqrt{112}$
AnswerCorrect option: D. $\sqrt{112}$
Let a be the first term and d be the common difference.
The given A.P. is $\sqrt{7},\sqrt{28},\sqrt{63},\ .....$
$\text{i.e.},\sqrt{7},\sqrt{4\times7},\sqrt{9\times7}$
$\text{i.e.},\sqrt{7},2\sqrt{7},3\sqrt{7},\ .....$
Common difference = d = Second term - First term
$=2\sqrt{7}-\sqrt{7}$
$=\sqrt{7}$
$\therefore$ Next term of the A.P. $=3\sqrt{7}+\sqrt{7}$
$=4\sqrt{7}$
$=\sqrt{16\times7}$
$=\sqrt{112}$
Hence, the correct option is (D).
View full question & answer→MCQ 51 Mark
Mark the correct alternative in the following:
The common difference of the A.P. $\frac{1}{3},\frac{1-3\text{b}}{3},\frac{1-6\text{b}}{3}, ....$ is
- A
$\frac{1}{3}$
- B
$-\frac{1}{3}$
- ✓
$-\text{b}$
- D
$\text{b}$
AnswerCorrect option: C. $-\text{b}$
A.P. is $\frac{1}{3},\frac{1-3\text{b}}{3},\frac{1-6\text{b}}{3}, .....\ \text{is}$
Common difference $\text{d}=\frac{1-3\text{b}}{3}-\frac{1}{3}=\frac{1-6\text{b}}{3}-\frac{(1-3\text{b})}{3}$
$\frac{1-3\text{b}-1}{3}=\frac{1-6\text{b}-1+3\text{b}}{3}$
$-3\text{b}=-3\text{b}$
$-\text{b}=-\text{b}$
View full question & answer→MCQ 61 Mark
Mark the correct alternative in the following: If in an $A.P. S_n= n^2p$ and $S_m= m^2p$, where $S_r$ denotes the sum of $r$ terms of the $A.P.,$ then $S_p$ is equal to:
- A
$\frac{1}{2}\text{p}^2$
- B
$\ce{mnp}$
- ✓
$p^3$
- D
$(m + n)p^2$.
Answer$S_n=n^2 p, S_m=m^2 p$
$\therefore \mathrm{S}_{\mathrm{r}}=\mathrm{r}^2 \mathrm{p} \text { and } \mathrm{S}_{\mathrm{p}}=\mathrm{p}^2 \mathrm{q}=\mathrm{p}^3$
Hence $S_p= p^3$.
View full question & answer→MCQ 71 Mark
Mark the correct alternative in the following:
The number of terms of the $A.P. 3, 7, 11, 15, .....$ to be taken so that the sum is $406$ is:
AnswerGiven, $A.P. 3, 7, 11, 15, .....$
$S_n= 406$
First term, $a = 3$
and Difference, $d = 7 - 3 = 4$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 406=\frac{\text{n}}{2}[2(3)+(\text{n}-1)4]$
$\Rightarrow 406 \times 2=n[6+4 n-4]$
$\Rightarrow 812=n[4 n+2]$
$\Rightarrow 812=n \times 2[2 n+1]$
$\Rightarrow 406=n[2 n+1]$
$\Rightarrow 406=2 n^2+n$
$\Rightarrow 2 n^2+n-406=0$
$\Rightarrow 2 n^2+29 n-28 n-406=0$
$\Rightarrow n(2 n+29)-14(2 n+29)=0$
$\Rightarrow(n-14)(2 n+29)=0$
Here,
$n - 14 = 0$
$\Rightarrow n = 14$
and $2n + 29 = 0$
$2n = -29$
$\text{n}=\frac{-29}{2}$
We know term never ne negative
So, number of terms is $14$
Hence, correct choice is $(D).$
View full question & answer→MCQ 81 Mark
Mark the correct alternative in the following:
If the first, second and last term of an $A.P,$ are $a, b$ and $2a$ respectively, its sum is
- A
$\frac{\text{ab}}{2(\text{b}-\text{a})}$
- B
$\frac{\text{ab}}{(\text{b}-\text{a})}$
- ✓
$\frac{3\text{ab}}{2(\text{b}-\text{a})}$
- D
AnswerCorrect option: C. $\frac{3\text{ab}}{2(\text{b}-\text{a})}$
In the given problem, we are given first, second and last term of an $A.P.$ We need to find its sum.
So, here
First term $= a$
Second term $(a_2) = b$
Last term $(l) = 2a$
Now, using the formula $a_n= a + (n - 1)d$
$a_2= a + (2 - 1)d$
$b = a + d$
$d = b - a .....(i)$
Also,
$a_n= a + (n - 1)d$
$2a = a + nd - d$
$2a - a = nd - d$
$\frac{\text{a}+\text{d}}{\text{d}}=\text{n}\ .....\text{(ii)} $
Furthere as we know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
Substituting $(ii)$ in the above equation, we get
Using $(i),$ we get
$\text{S}_\text{n}=\frac{\text{a}+(\text{b}-\text{a})}{2(\text{b}-\text{a})}(3\text{a})$
$\text{S}_\text{n}=\frac{\text{b}}{2(\text{b}-\text{a})}(3\text{a})$
Thus,
$\text{S}_\text{n}=\frac{3\text{ab}}{2(\text{b}-\text{a})}$
Therefore, the correct option is $(C).$
View full question & answer→MCQ 91 Mark
Mark the correct alternative in the following:
It the sums of $n$ terms of two arithmetic progressions are in the ratio $\frac{3\text{n}+5}{5\text{n}-7}$, then their $n^{th}$ terms are in the ratio.
- A
$\frac{3\text{n}-1}{5\text{n}-1}$
- ✓
$\frac{3\text{n}+1}{5\text{n}+1}$
- C
$\frac{5\text{n}+1}{3\text{n}+1}$
- D
$\frac{5\text{n}-1}{3\text{n}-1}$
AnswerCorrect option: B. $\frac{3\text{n}+1}{5\text{n}+1}$
Given,
$\frac{\text{Sum of A.P}_1}{\text{Sum of A.P}_2}=\frac{\text{S}_\text{n}}{\text{S}^1_\text{n}}=\frac{3\text{n}+5}{5\text{n}+7}\ .....\text{(i)}$
We know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P_1\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P_2\ \text{S}'_\text{n}=\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}']$
Put the value in eq. $(i)$
$\Rightarrow\ \frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}}{\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{3\text{n}+5}{5\text{n}+7}=\frac{[2\text{a}+(\text{n}-1)\text{d}]}{[2\text{a}'+(\text{n}-1)\text{d}']}$
Now, put the $n = 2n - 1$
$\Rightarrow\ \frac{3(2\text{n}-1)+5}{5(2\text{n}-1)+7}=\frac{[2\text{a}+(2\text{n}-1-1)\text{d}]}{[2\text{a}'+(2\text{n}-1-1)\text{d}']}$
$\Rightarrow\ \frac{6\text{n}-3+5}{10\text{n-5+7}}=\frac{[2\text{a}+(2\text{n}-2)\text{d}]}{[2\text{a}'+(2\text{n}-2)\text{d}']}$
$\Rightarrow\ \frac{6\text{n}+2}{10\text{n}+2}=\frac{2\text{a}+2(\text{n}-1)\text{d}}{2\text{a}'+2(\text{n}-1)\text{d}'}$
$\Rightarrow\ \frac{2(3\text{n}+1)}{2(5\text{n}+1)}=\frac{2[\text{a}+(\text{n}-1)\text{d}]}{2[\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{3\text{n}+1}{5\text{n}+1}=\frac{[\text{a}+(\text{n}-1)\text{d}]}{[\text{a}'+(\text{n-1})\text{d}']}$
$\Rightarrow\ \frac{3\text{n}+1}{5\text{n}+1}=\frac{[\text{a}+(\text{n}-1)\text{d}]}{[\text{a}'+(\text{n}-1)\text{d}']}$
We know, $a_n=a+(n-1) d$
$\Rightarrow\ \frac{3\text{n}+1}{5\text{n}+1}=\frac{\text{a}_\text{n}}{\text{a}'_\text{n}}$
Hence, correct choice is $(B).$
View full question & answer→MCQ 101 Mark
Mark the correct alternative in the following: The common difference of an $A.P.,$ the sum of whose $n$ terms is $S_n$, is
AnswerCorrect option: A. $S_n - 2S_{n - 1} + S_{n - 2}$
The sum of $u$ term of an $A.P.=S_n$
There fore sum of $n-1$ term of $A.P. =S_{n-1}$
Similarly sum of $n-2$ term of $A.P.=S_{n-2}$
$a_n= S _{ n }- S _{ n -1}$ and $a _{ n -1}= S _{ n -1}- S _{ n -2}$
Common difference,
$d=a_n-a_{n-1}=S_n-S_{n-1}-\left[S_{n-1}-S_{n-2}\right]$
$=S_n-S_{n-1}-S_{n-1}+S_{n-2}$
$=S_n-2 S_{n-1}+S_{n-2}$
Corrent option is $(a).$
View full question & answer→MCQ 111 Mark
Mark the correct alternative in the following:
The first and last term of an $A.P.$ are a and $l$ respectively. If $S$ is the sum if all the terms of the $A.P.$ and the common difference is given by $\frac{\text{l}^2-\text{a}^2}{\text{k}-(\text{l}+\text{a})}$, then $k =$
AnswerGiven,
First term, $a$
Last term, $l$
Sum of $l$ terms, $S_l$
and Difference, $\text{d}=\frac{\text{l}^2-\text{a}^2}{\text{k}-(\text{l}+\text{a})}\ .....(\text{i})$
We know, $l = a + (n - 1)d$
put in eq. $(i) \text{d}=\frac{(\text{a}+(\text{n}-1)\text{d})^2-\text{a}^2}{\text{k}-[(\text{a}+(\text{n}-1)\text{d})+\text{a}]}$
$\Rightarrow\ \text{d}=\frac{\text{a}^2+((\text{n}-1)\text{d})^2+2\text{a}(\text{n}-1)\text{d}-\text{a}^2}{\text{k}-[\text{a}+\text{nd}-\text{d}+\text{a}]}$
$\text{d}=\frac{(\text{n}-1)^2\text{d}^2+2\text{a}(\text{n}-1)\text{d}}{\text{k}-[2\text{a}+(\text{n}-1)\text{d}]}$
$\Rightarrow\ \text{d}=\frac{\text{d}[(\text{n}-1)^2\text{d}+2\text{a}(\text{n}-1)]}{\text{k}-[2\text{a}+(\text{n}-1)\text{d}]}$
$\Rightarrow k - [2a+ (n - 1)d] = (n - 1)2 d + 2a (n - 1)$
$\Rightarrow k = (n - 1)2 d + 2a (n - 1) + [2a + (n - 1)d]$
$\Rightarrow k = (n - 1)[(n - 1)d + 2a] + [2a + (n - 1)d]$
$\Rightarrow k = 2a + (n - 1)d [n - 1 + 1]$
$\Rightarrow k = (2a + (n - 1)d)n$
We know, $\text{S}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Now divide by $2$ both side
$\Rightarrow\ \frac{\text{k}}{2}=(2\text{a}+(\text{n}-1)\text{d})\frac{\text{n}}{2}$
$\Rightarrow\ \frac{\text{k}}{2}=\text{S}$
$\Rightarrow\ \text{k}=2\text{S}$
Hence, correct chice is $(B).$
View full question & answer→MCQ 121 Mark
Mark the correct alternative in the following:If the first term of an $A.P.$ is a and $n^{th}$ term is $b,$ then its common difference is:
- A
$\frac{\text{b}-\text{a}}{\text{n}+1}$
- ✓
$\frac{\text{b}-\text{a}}{\text{n}-1}$
- C
$\frac{\text{b}-\text{a}}{\text{n}}$
- D
$\frac{\text{b}+\text{a}}{\text{n}-1}$
AnswerCorrect option: B. $\frac{\text{b}-\text{a}}{\text{n}-1}$
Here,
We are given the first term of the $A.P.$ as a and the $n ^{\text {th }}$ term $\left(a_n\right)$ as $b$.
So, let us take the common difference of the $A.P.$ as $d$.
Now, as we know,
$a_n = a + (n - 1)d$
On substituting the values given in the question, we get.
$b = a + (n - 1)d$
$(n - 1)d = b - a$
$\text{d}=\frac{\text{b}-\text{a}}{\text{n}-1}$
$\therefore\text{d}=\frac{\text{b}-\text{a}}{\text{n}-1}$
Hence the correct option is $(B).$
View full question & answer→MCQ 131 Mark
Mark the correct alternative in the following:
The sum of first 20 odd natural numbers is:
AnswerLet a be the first term and d be the common difference.
We know that, sum of first n terms $=\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{a}]$
The given series is 1 + 3 + 5 + .....
First term = a = 1
Common difference = d = 3 - 1 = 2
$\therefore\ \text{S}_{20}=\frac{20}{2}[2\times1+(20-1)2]$
= 10(2 + 19 × 2)
= 10(40)
= 400
Hence, the correct option is (C).
View full question & answer→MCQ 141 Mark
Mark the correct alternative in the following:
The common difference of the A.P. $\frac{1}{2\text{b}},\frac{1-6\text{b}}{2\text{b}},\frac{1-12\text{b}}{2\text{b}}, .....$ is
AnswerLet a be the first term and d be the common difference.
The given A.P. is $\frac{1}{2\text{b}},\frac{1-6\text{b}}{2\text{b}},\frac{1-12\text{b}}{2\text{b}},\ .....$
Common difference = d = Second - First term
$=\frac{1-6\text{b}}{2\text{b}}-\frac{1}{2\text{b}}$
$=\frac{-6\text{b}}{2\text{b}}=-3$
Hence, the correct option is (D).
View full question & answer→MCQ 151 Mark
Mark the correct alternative in the following:
If $7^{\text {th }}$ and $13^{\text {th }}$ terms of an $A.P$. be $34$ and $64$ respectively, then its $18^{\text {th }}$ term is:
AnswerGiven,
$a_7=34$
and $a_{13}=64$
We know, $a_n=a+(n-1) d$
$7^{\text {th }}$ term, $a_7=a+(7-1) d$
$\Rightarrow 34=a+6 d \ldots . .(i)$
$13^{\text {th }}$ term, $a_{13}=a+(13-1) d$
$\Rightarrow 64=a+12 d . . . . .(\text { ii) }$
By subracting eq. $(i)$ from $(ii)$ from eq. $(ii)$
$\Rightarrow 64-34=a+21 d-(a+6 d)$
$\Rightarrow 30=a+12 d-a-6 d$
$\Rightarrow 30=6 d$
$\Rightarrow d=\frac{30}{6}$
Put the value of $d$ in eq. $(i)$
$\Rightarrow 34=a+6(5)$
$\Rightarrow 34=a+30$
$\Rightarrow a=34-30=4$
Now we have to find $18^{\text {th }}$ term,
$\Rightarrow a_{18}=4+(18-1) 5$
$\Rightarrow a_{18}=4+17 \times 5$
$\Rightarrow a_{18}=4+85$
$\Rightarrow a_{18}=89$
Hence, correct choice is $(C).$
View full question & answer→MCQ 161 Mark
Mark the correct alternative in the following:
The first three terms of an A.P. respectively are 3y - 1, 3y + 5 and 5y + 1. Then, y equals:
AnswerSince, 3y - 1, 3y + 5 and 5y + 1 are first three terms of an A.P.
Then, Second term - First term = Third term - Second term = d (common difference)
⇒ 3y + 5 - (3y - 1) = 5y + 1 - (3y + 5)
⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5
⇒ 6 = 2y - 4
⇒ 2y = 6 + 4
⇒ 2y = 10
⇒ y = 5
Hence, the correct option is (C).
View full question & answer→MCQ 171 Mark
Mark the correct alternative in the following: The $n^{th}$ term of an $A.P.,$ the sum of whose $n $ terms is $S_n$, is
- A
$S_n + S_{n-1}$
- ✓
$S_n - S_{n-1}$
- C
$S_n + S_{n+1}$
- D
$S_n - S_{n+1}$
AnswerCorrect option: B. $S_n - S_{n-1}$
$S_n$ is the sum of first $n$ terms
Last term $n ^{\text {th }}$ term $= S _{ n }- S _{ n -1}$
View full question & answer→MCQ 181 Mark
Mark the correct alternative in the following:
If the sum of $n$ terms of an $A.P.$ is $3n^2 + 5n$ then which of its terms is $164?$
- A
$26^{th}$
- ✓
$27^{th}$
- C
$28^{th}$
- D
AnswerCorrect option: B. $27^{th}$
Here, the sum of first $n$ terms is given by the expression,
$S_n=3 n^2+5 n$
We need to find which term of the $A.P.$ is $164.$
Let us take $164$ as the $n ^{\text {th }}$ term
So we know that the $n ^{\text {th }}$ term of an $A.P.$ is given by,
$a_n=S_n-S_{n-1}$
So,
$164=S_n=S_{n-1}$
$164=3 n^2+5 n-\left[3(n-1)^2+5(n-1)\right]$
using the property,
$(a-b)^2+a^2+b^2-2 a b$
We get,
$164=3 n^2+5 n-\left[3\left(n^2+1-2 n\right)+5(n-1)\right]$
$164=3 n^2+5 n-\left[3 n^2+3-6 n+5 n-5\right]$
$164=3 n^2+5 n-\left(3 n^2+n-2\right)$
$164=3 n^2+5 n-3 n^2+n+2$
$164=6 n+2$
Further solving for $n$, we get
$6 n=164-2$
$n=\frac{162}{6}$
$n=27$
Therefore, $164$ is the $27^{\text {th }}$ term of the given $A.P.$
Hence the correct option is $(b).$
View full question & answer→MCQ 191 Mark
Mark the correct alternative in the following:
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 $k =$
AnswerIn the given problem, we are given $d = S_n - kS_{n-1} + S_{n-1}$
We need to find the value of $k$
So here,
First term $= a$
Common difference $= d$
Sum of $n$ terms $= S_n$
Now, as we know
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\ ....\text{(i)}$
Also, for $n-1$ terms
$\text{S}_{\text{n}-1}=\frac{\text{n}-1}{2}[2\text{a}+[(\text{n}-1)-1]\text{d}]$
$=\frac{\text{n}-1}{2}[2\text{a}+[(\text{n}-1)-1]\text{d}]\ .....\text{(ii)}$
Further, for $n-2$ terms,
$\text{S}_{\text{n}-1}=\frac{\text{n}-1}{2}[2\text{a}+[(\text{a}-2)-1]\text{d}]$
$=\frac{\text{n}-1}{2}[2\text{a}+[(\text{a}-3)\text{d}]\ .....(\text{iii})$
Now, we are given
$\text{d}=\text{S}_\text{n}-\text{kS}_{\text{n}-1}+\text{S}_{\text{n}-2}$
$\text{d}+\text{kS}_{\text{n}-1} =\text{S}_{\text{n}}+\text{S}_{\text{n}-2}$
$\text{k}=\frac{\text{S}_\text{n}+\text{S}_{\text{n}-1}-\text{d}}{\text{S}_{\text{n}-1}}$
Using $(i), (ii)$ and $(iii)$ in the given equation, we get
$\text{k}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]+\frac{\text{n}-2}{2}[2\text{a}+(\text{n}-3)\text{d}]-\text{d}}{\frac{\text{n-1}}{2}[2\text{a}+(\text{n}-2)\text{d}]}$
Taking $\frac{1}{2}$ common, we get,
$\text{k}=\frac{\text{n}[2\text{a}+(\text{n}-1)\text{d}]+(\text{n}-2)[2\text{a}+(\text{n}-3)\text{d}]-2\text{d}}{(\text{n}-1)[2\text{a}+(\text{n}-2)\text{d}]}$
$=\frac{2\text{an}+\text{n}^2\text{d}-\text{nd}+2\text{an}+\text{n}^2\text{d}-3\text{nd}-4\text{a}-2\text{nd}+6\text{d}-2\text{d}}{2\text{an}+\text{n}^2\text{d}-2\text{nd}-2\text{d}-\text{nd}+2\text{d}}$
$=\frac{2\text{n}^2\text{d}+4\text{an}-6\text{nd}-4\text{a}+4\text{d}}{\text{n}^2\text{d}+2\text{an}-3\text{nd}+-2\text{a}+2\text{d}}$
Taking $2$ common from the numerator, we get,
$\text{k}=\frac{2(\text{n}^2\text{d}+2\text{an}-3\text{nd}+-2\text{a}+2\text{d})}{\text{n}^2\text{d}+2\text{an}-3\text{nd}+-2\text{a}+2\text{d}}$
$=2$
Therefore, $k = 2$
Hence, the correct option is $(B).$
View full question & answer→MCQ 201 Mark
Mark the correct alternative in the following:The sum of first $n$ odd natural numbers is:
- A
$2n - 1$
- B
$2n + 1$
- ✓
$n^2$
- D
$n^2 - 1$
AnswerLet, odd numbers are,
$1, 3, 5, .....$
Here,
First term, $a = 1$
and Difference, $d = 3 - 1 = 2$
We know, sum of $n$ terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2(1)+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2[1+\text{n}-1]$
$\Rightarrow\ \text{S}_\text{n}=\text{n}^2$
Hence, correct choice is $(C).$
View full question & answer→MCQ 211 Mark
Mark the correct alternative in the following:
If the sum of $n$ terms of an $A.P.$ be $3n^2 + n$ and its common difference is $6$, then its first term is:
AnswerGiven,
$S_n=3 n^2+n$
and $d=6$
Putting $n =0$ and $1$
$S_0=3(0)^2+0=3 \times 0+0=0+0=0$
And $S_1=3(1)^2+1=3 \times 1+1=3+1=4$
We know, $a_n+S_n=S_{n-1}$
First term $a _1= S _1- S _0$
$\Rightarrow a_1=4-0$
$\Rightarrow a_1=4$
Hence, correnct choice is $(d).$
View full question & answer→MCQ 221 Mark
Mark the correct alternative in the following:
If the sum of $n$ terms of an $A.P.$ is $2 n^2+5 n$, then its $n^{\text {th }}$ term is:
- A
$4n - 3$
- B
$3n - 4$
- ✓
$4n + 3$
- D
$3n + 4$
AnswerCorrect option: C. $4n + 3$
Given,
$s_n=2 n^2-5 n$
We know, $a_n=S_n-S_{n-1}$
$S_n=2 n^2+5 n$
$S_{n-1}=2(n-1)^2+5(n-1)$
$\Rightarrow S_{n-1}=2\left(n^2+1-2 n\right)+5 n-5$
$[\because \left.(a-b)^2=a^2+b^2-2 a b\right]$
$\Rightarrow S_{n-1}=2 n^2+2-4 n+5 n-5$
$\Rightarrow S_{n-1}=2 n^2+n-3$
Now, $a_n=2 n^2+5 n-\left(2 n^2+n-3\right)$
$\Rightarrow a_n=2 n^2+5 n-2 n^2-n+3$
$\Rightarrow a_n=4 n+3$
Hence, correct choice is $(C).$
View full question & answer→MCQ 231 Mark
Mark the correct alternative in the following:
The $9^{\text {th }}$ term of an $A.P.$ is $449$ and $449^{\text {th }}$ term is $9$ . The term which is equal to zero is:
- A
$501^{\text {th }}$
- B
$502^{\text {th }}$
- C
$508^{\text {th }}$
- ✓
AnswerIn the given problem, let us take the first term as a and the common difference as $d.$
Here, we are given that,
$a_9=449 \ldots . . . . \text { (i) }$
$a_{449}=9 \ldots . . \text { (iii) }$
We need to find $n$
Also, we know,
$a_n=a+(n-1) d$
For the $9^{\text {th }}$ term $(n=9)$,
$a_9=a+(9-1) d$
$449=a+8 d \text { (Using i) }$
$a=449-8 d \text {.....(iii) }$
Similarly, for the $449^{\text {th }}$ term $(n=449)$,
$a 449=a+(449-1) d$
$9=a+448 d \text { (Using ii) }$
$a=9-448 d \text {.....(iv) }$
Subtracting $(3)$ from $(4),$ we get,
$a-a=(9-448 d)-(449-8 d)$
$0=9-448 d-449+8 d$
$0=-440-440 d$
$440 d=-440$
$d=-1$
Now, to find a, we substitute the value of $d$ in $(3)$
$a=449-8(-1)$
$a=449+8$
$a=457$
So, for the given $A.P\ d = -1$ and $a=457$
So, let us take the term equal to zero as the $n ^{\text {th }}$ term. So,
$a_n=457+(n-1)(-1)$
$0=457-n+1$
$n=458$
So, $n =458$
Therefore, the correct option is $(D).$
View full question & answer→MCQ 241 Mark
Mark the correct alternative in the following:
If $S_1$ is the sum of an arithnetic progression of $'n\ '$ odd number of terms and $S_2$ the sum of the terms of the series in odd places, then $\frac{\text{S}_1}{\text{S}_2}=$
- ✓
$\frac{2\text{n}}{\text{n}+1}$
- B
$\frac{\text{n}}{\text{n}+1}$
- C
$\frac{\text{n}+1}{2\text{n}}$
- D
$\frac{\text{n}+1}{\text{n}}$
AnswerCorrect option: A. $\frac{2\text{n}}{\text{n}+1}$
Given,
$S_1$ is the sum of an $A.P.$ of n odd number of terms,
$S_2$ is the sum of the terms if the series in odd plaves,
Let, $A.P.$ is $a_2, a_2, a_3, \ldots . . ., a_n$
Here,
$n$ is odd
d is difference
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}$
then, $\text{S}_1=\frac{\text{n}}{2}[2\text{a}_1+(\text{n}-1)\text{d}]$
Now, $S^2$ be the sum if the terms of the series in odd places.
Number of terms, $\frac{\text{n}+1}{2}$
Difference, $d = 2d$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_2=\frac{\frac{\text{n}+1}{2}}{2}\bigg[2\text{a}_1+\Big(\frac{\text{n}+1}{2}-1\Big)2\text{d}\bigg]$
$\Rightarrow\ \text{S}_2=\frac{\text{n}+1}{2}\Big[2\text{a}_1+\frac{\text{n}-1}{2}\times2\text{d}\Big]$
$\Rightarrow\ \text{S}_2=\frac{\text{n}+1}{4}[2\text{a}_1+(\text{n}-1)\text{d}]$
Now, we have to find $\frac{\text{S}_1}{\text{S}_2}$,
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}+1}{4}[2\text{a}_1+(\text{n}-1)\text{d}]}$
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{4\text{n}}{2(\text{n}+1)}$
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{2\text{n}}{(\text{n}+1)}$
Hence, correct chice is $(A).$
View full question & answer→MCQ 251 Mark
Mark the correct alternative in the following:In an $A P\ S_p=q, S_q=p$ 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
$pq.$
AnswerCorrect option: B. $-(p + q)$
Given,
$S_p = q$
and $S_q = p$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
then, $\text{S}_\text{p}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$\Rightarrow\ \text{q}=\frac{\text{P}}{2}[2\text{a}+\text{pd}-\text{d}]$
$\Rightarrow\ 2\text{q}=2\text{ap}+\text{p}^2\text{d}-\text{pd}\ .....\text{(i)}$
Again, $\text{Sq}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$\Rightarrow\ \text{p}=\frac{\text{q}}{2}[2\text{a}+\text{qd}-\text{d}]$
$\Rightarrow\ 2\text{p}=2\text{aq}+\text{q}^2\text{d}-\text{qd}\ .....\text{(ii)}$
By subtracting eq. $(i)$ from eq. $(ii)$
$\Rightarrow 2p - 2q = 2aq + q^2d - qd - (2ap + p^2d - pd)$
$\Rightarrow 2(p - q) = 2aq + q^2d - qd - 2ap - p^2d + pd$
$\Rightarrow 2(p - q) = 2aq - 2ap + q^2d - p^2d - qd + pd$
$\Rightarrow -2(q - p) = 2a(q - p) + d(q^2 - p^2) - d (q - p)$
$\Rightarrow -2(q - p) = 2a(q - p) + d(q + p)(q - p) -d(q - p)$
$\Rightarrow -2(q - p) = (q - p)(2a + d(q + p)-d)$
$\Rightarrow\ \frac{-2(\text{q}-\text{p})}{\text{q}-\text{p}}=2\text{a}+\text{dq}+\text{dp}-\text{d}$
$\Rightarrow -2 = 2a + (q + p - 1)d$
Now, we have to find sum of $p + q$
$\Rightarrow\ \text{S}_{\text{p}+\text{q}}=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$\Rightarrow\ \text{S}_{\text{p}+\text{q}}=\frac{\text{p}+\text{q}}{2}\times-2\ [\because\ \text{From eq. (iii)}]$
$\Rightarrow\ \text{S}_{\text{p}+\text{q}}=-(\text{p}+\text{q})$
Hence, correct choice is $(B).$
View full question & answer→MCQ 261 Mark
Mark the correct alternative in the following:
If $18^{\text {th }}$ and $11^{\text {th }}$ term of an $A.P.$ are in the ratio $3: 2$, then its $21^{\text {st }}$ and $5^{\text {th }}$ terms are in the ratio:
- A
$3 : 2$
- ✓
$3 : 1$
- C
$1 : 3$
- D
$2 : 3$
AnswerCorrect option: B. $3 : 1$
$18^{th}$ term : $11^{th}$ term $= 3 : 2$
$\Rightarrow\ \frac{\text{a}_{18}}{\text{a}_{11}}=\frac{3}{2}$
$\Rightarrow\ \frac{\text{a}+17\text{d}}{\text{a}+10\text{d}}=\frac{3}{2}$
$\Rightarrow\ 2\text{a}+34\text{d}=3\text{a}+30\text{d}$
$\Rightarrow\ 34\text{d}-30\text{d}=3\text{a}-2\text{a}$
$\Rightarrow\ \text{a}=4\text{d}$
Now $\frac{\text{a}_{21}}{\text{a}_5}=\frac{\text{a}+20\text{d}}{\text{a}+4\text{d}}=\frac{4\text{d}+20\text{d}}{4\text{d}+4\text{d}}$
$=\frac{24\text{d}}{8\text{d}}=\frac{3}{1}$
$\text{a}_{21}:\text{a}_5=3:1$
View full question & answer→MCQ 271 Mark
Mark the correct alternative in the following:
If the n$^{th}$ term of an $A.P.$ is $2n + 1,$ 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. $n(n + 2)$
Given,
$a_n=2 n+1=1$
$\text { Putting } n=1,2,3, \ldots . .$
$a_1=2(1)+1=2+1=3$
$a_2=2(2)+1=4+1=5$
and $a_3=2(3)=1=6+1=7$
Now, $A.P.$ is $3,5,7, \ldots .(2 n+1)$
Here,
First term, $a=3$
and Difference, $d=5-3=2$
We know, sum of $n$ terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}(3+2\text{n}+1)$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}(2\text{n}+4)$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2(\text{n}+2)$
$\Rightarrow\ \text{S}_\text{n}=\text{n}(\text{n}+2)$
Hence, correct choice is $(B).$
View full question & answer→MCQ 281 Mark
Mark the correct alternative in the following:
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:
AnswerHere, we are given an $A.P.$ whose sum of $n$ terms is $S_n$ and $S_{2n} = 3S_n$
We need to find $\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}$.
Here we use the following formula for the sum of $n$ terms of an $A.P.$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Where; $a =$ first term for the given $A.P.$
$d =$ common difference of the given $A.P.$
$n =$ number of terms
So, first we find $S_{3n}$
$\text{S}_{3\text{n}}=\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]$
$=\frac{3\text{n}}{2}[2\text{a}+3\text{nd}-\text{d}] ..... (\text{i})$
Similarly,
$\text{S}_{2\text{n}}=\frac{2\text{n}}{\text{2}}[2\text{a}+(2\text{n}-1)\text{d}]$
$=\frac{2\text{n}}{2}[2\text{a}+2\text{nd}-\text{d}]\ .....(\text{ii})$
Also,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2\text{a}+\text{nd}-\text{d}]\ .....(\text{iii})$
Now, $S_{2n} = 3S_n$
So, using $(ii)$ and $(iii),$ we get,
$\frac{2\text{n}}{2}(2\text{a}+2\text{nd}-\text{d})=3\Big[\frac{\text{n}}{2}(2\text{a}+\text{nd}-\text{d})\Big]$
$\frac{2\text{n}}{2}(2\text{a}+2\text{nd}-\text{d})=\frac{3\text{n}}{2}(2\text{a}+\text{nd}-\text{d})$
On further solving, we get,
$2(2\text{a}+2\text{nd}-\text{d})=3(2\text{a}+\text{nd}-\text{d})$
$4\text{a}+4\text{nd}-2\text{d}=6\text{a}+3\text{nd}-3\text{d}$
$2\text{a}=\text{nd}+\text{d}\ .....\text{iv}$
So,
$\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=\frac{\frac{3\text{n}}{2}[2\text{a}+3\text{nd}-\text{d}]}{\frac{\text{n}}{(2)}[2\text{a}+\text{nd}-\text{d}]}$
Taking $\frac{\text{n}}{2}$ common, we get,
$\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=\frac{3(2\text{a}+3\text{nd}-\text{d})}{(2\text{a}+\text{nd}-\text{d})}$
$=\frac{3(\text{nd}+\text{d}+3\text{nd}-\text{d})}{(\text{nd}+\text{d}+\text{nd}-\text{d})}$
$=\frac{3(4\text{nd})}{2\text{nd}}$
$=6$
$\therefore\ \frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=6$
Hence, the correct option is $(b).$
View full question & answer→MCQ 291 Mark
Mark the correct alternative in the following:The first and last terms of an $A.P.$ are $1$ and $11$. If the sum of its terms is $36,$ then the number of terms will be:
AnswerGiven,
First term, $a = 1$
Last term, $a_n=11 = 1$
and Sum of $11$ term, $S _{11}=36$
We know, Sum of $n$ terms,
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\Rightarrow\ 36=\frac{\text{n}}{2}(1+11)$
$\Rightarrow\ 36\times2=\text{n}\times12$
$\Rightarrow\ \text{n}=\frac{36\times2}{12}$
$\Rightarrow\ \text{n}=6$
Hence, correct choice is $(b).$
View full question & answer→MCQ 301 Mark
Mark the correct alternative in the following:
If $\frac{1}{\text{x}+2},\frac{1}{\text{x}+3},\frac{1}{\text{x}+5}$ are in A.P. then, x =
AnswerGiven,
$\frac{1}{\text{x}+2},\frac{1}{\text{x}+3},\frac{1}{\text{x}+5}$ are in A.P.
Then difference between terms is equal
$\Rightarrow\ \frac{1}{\text{x}+3}-\frac{1}{\text{x}+2}=\frac{1}{\text{x}+5}-\frac{1}{\text{x}+3}$
$\Rightarrow\ \frac{(\text{x}+2)-(\text{x}+3)}{(\text{x}+3)(\text{x}+2)}=\frac{(\text{x}+3)-(\text{x}+5)}{(\text{x}+5)(\text{x}+3)}$
$\Rightarrow\ \frac{\text{x}+2-\text{x}-3}{\text{x}+2}=\frac{\text{x}+3-\text{x}-5}{\text{x}+5}$
$\Rightarrow\ \frac{-1}{\text{x}+2}=\frac{-2}{\text{x}+5}$
$\Rightarrow\ -1(\text{x}+5)=-2(\text{x}+2)$
$\Rightarrow\ -\text{x}-5=-2\text{x}-4$
$\Rightarrow\ -\text{x}+2\text{x}=-4+5$
$\Rightarrow\ \text{x}=1$
Hence, carrect chice is (C).
View full question & answer→MCQ 311 Mark
Mark the correct alternative in the following:
The sum of n terms of an $A.P.$ is $3n^2 + 5n$, then $164$ is its:
- A
$24^{th}$ term
- ✓
$27^{th}$ term
- C
$26^{th}$ term
- D
$25^{th}$ term
AnswerCorrect option: B. $27^{th}$ term
Here, the sum of first $n$ terms is given by the expression,
$S_n=3 n^2+5 n$
We need to find which term of the $A.P.$ is $164.$
Let us take $164$ as the $n ^{\text {th }}$ term.
So we know that the $n ^{\text {th }}$ term of an $A.P.$ is given by,
$A_n=S_n-S_{n-1}$
So,
$164=S_n-S_{n-1}$
$164=3 n^2+5 n-\left[3(n-1)^2+5(n-1)\right]$
Using the property,
$(a-b)^2=a^2+b^2-2 a b$
We get,
$164=3 n^2+5 n-\left[3\left(n^2+1-2 n\right)+5(n-1)\right]$
$164=3 n^2+5 n-\left[3 n^2+3-6 n+5 n-5\right]$
$164=3 n^2+5 n-\left(3 n^2-n-2\right)$
$164=3 n^2+5 n-3 n^2+n+2$
$164=6 n+2$
Further solving for $n$, we get
$6 n=164-2$
$n=\frac{162}{6}$
$n=27$
Therefore, $164$ is the $27^{\text {th }}$ term of the given $A.P.$
Hence the correct option is $(B).$
View full question & answer→MCQ 321 Mark
Mark the correct alternative in the following:
The sum of n terms of two $A.P.'s$ are in the ratio $5n + 9 : 9n + 6$. Then, the ratio of their $18^{th}$ term is
- A
$\frac{179}{321}$
- B
$\frac{178}{321}$
- C
$\frac{175}{321}$
- ✓
$\frac{184}{321}$
AnswerCorrect option: D. $\frac{184}{321}$
Given,
$\frac{\text{Sum of A.P.}_1}{\text{Sum of A.P.}_2}=\frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{5\text{n}+9}{9\text{n}+6}\ .....{\text{(i)}}$
We know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P_1\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P_2\ \text{S}'_\text{n}=\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}]$
Put the value in Eq. $(i)$
$\Rightarrow\ \frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{5\text{n}+9}{9\text{n}+6}=\frac{[2\text{a}+(\text{n}-1)\text{d}]}{[2\text{a}'+(\text{n}-1)\text{d}']}$
Now, put the $n = 2n - 1$
$\Rightarrow\ \frac{5(2\text{n}-1)+9}{9(2\text{n}-1)+6}=\frac{[2\text{a}+(2\text{n}-1-1)\text{d}]}{[2\text{a}'+(2\text{n}-1-1)\text{d}']}$
$\Rightarrow\ \frac{10\text{n}-5+9}{18\text{n}-9+6}=\frac{[2\text{a}+(2\text{n}-2)\text{d}]}{[2\text{a}'+(2\text{n}-2)\text{d}']}$
$\Rightarrow\ \frac{10\text{n}+4}{18\text{n}-3}=\frac{2\text{a}+2(\text{n}-1)\text{d}}{2\text{a}'+2(\text{n}-1)\text{d}'}$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{2[\text{a}+(\text{n}-1)\text{d}]}{2[\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{[\text{a}+(\text{n}-1)\text{d}]}{[\text{a}'+(\text{n}-1)\text{d}']}$
We kmow, $a_n = a + (n - 1)d$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{\text{a}_\text{n}}{\text{a}'_\text{n}}$
Now put $n = 18$
$\Rightarrow\ \frac{2[5(18)+2]}{3[6(18)-1]}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{2[90+2]}{3[108-1]}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{2\times92}{3\times107}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{184}{321}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
Hence, the correct option is $(D)$
View full question & answer→MCQ 331 Mark
Mark the correct alternative in the following:
If k, 2k - 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is
AnswerSince, k, 2k - 1 and 2k + 1 are three consecutive terms of an A.P.
Then, Second term - First term = Third term - Second term = d(common difference)
⇒ 2k -1 - k = 2k + 1 - (2k - 1)
⇒ k - 1 = 2k + 1 - 2k + 1
⇒ k - 1 = 2
⇒ k = 2 + 1
⇒ k = 3
Hence, the correct option is (B).
View full question & answer→MCQ 341 Mark
Mark the correct alternative in the following:
$\text{If }\frac{5+9+13+, ....\text{ to n terms}}{7+9+11+, .....\text{to }(\text{n}+1)\text{ term}}=\frac{17}{16},\text{then n }=$
AnswerSum of $5 + 9 + 13 + .....$ to $n$ term
$=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Here, $a = 5, d = 9 - 5 = 4$
$\therefore\ \text{Sum}=\frac{\text{n}}{2}[2\times5+(\text{n}-1)\times4]$
$=\frac{\text{n}}{2}[10+4\text{n}-4]$
$=\frac{\text{n}}{2}[6+4\text{n}=\text{n}(3+2\text{n})]$
and sum of $7 + 9 + 11 + .....$ to $(n + 1) $terms
$=\frac{\text{n}+1}{2}[2\times7+(\text{n}+1-1)2]$
$=\frac{\text{n}+1}{2}[14+2\text{n}]=(\text{n}+1)(7+\text{n})$
$\therefore\ \frac{5+9+13+\ .....\text{to n terms}}{7+9+11+\ .....\text{ to }(\text{n}+1)\text{ terms}}=\frac{17}{16}$
$\Rightarrow\ \frac{\text{n}(3+2\text{n})}{(\text{n}+1)(7+\text{n})}=\frac{17}{16}$
$\Rightarrow 16n(3 + 2n) = 17(n + 1)(7 + n)$
$\Rightarrow 48n + 32n^2 = 17(n^2 + 8n + 7)$
$\Rightarrow 48n + 32n^2 = 17n^2 + 136n + 119$
$\Rightarrow 48n + 32n^2 - 17n^2 - 136n - 119 = 0$
$\Rightarrow 15n^2 - 88n - 119 = 0$
$\Rightarrow 15n^2 - 105n + 17n - 119 = 0$
$\begin{Bmatrix}\because\ 15\times(-119)=1785 \\ -1785 = 17\times(105) \\ -88 = 17 - 105 \end{Bmatrix}$
$\Rightarrow 15n(n - 7) + 17(n - 7) = 0$
$\Rightarrow (n - 7)(15n + 17) = 0$
Either, $n - 7 = 0,$ then $n = 7$ or $15n + 17 = 0,$ then $\text{n}=\frac{-17}{15}$ which is not possible being fraction.
$\therefore\ \text{n}=7$
View full question & answer→MCQ 351 Mark
Mark the correct alternative in the following:If $18, a, b, -3$ are in $A.P.,$ the $a + b =$
AnswerHere, we are given four terms which are in $A.P.,$
First term $( a _1 ) =$
Second term $\left( a _2\right)=$
Third term $\left( a _3\right)=$
Fourth term $\left( a _4\right)=$
So, in an $A.P.$ the difference of two adjacent terms is always constant. So, we get,
$d=a_2-a_1$
$d=a-18 \ldots . . \text { (i) }$
Also,
$d=a_4-a_3$
$d=-3-b \ldots . . .(\text { ii) }$
Now, on equating $(1)$ and $(2),$ we get,
$a-18=-3-b$
$a+b=18-3$
$a+b=15$
Therefore, $a+b=15$
Hence the correct option is $(D).$
View full question & answer→MCQ 361 Mark
Mark the correct alternative in the following:
If Sn denote the sum of n terms of an A.P. with first term a and common difference d such that $\frac{\text{S}_\text{x}}{\text{Sk}_\text{x}}$ is independent of x, then
AnswerSn is the sum of first n terms a is the first term add d is the common difference
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}$
$\frac{\text{S}_\text{x}}{\text{S}_\text{kx}}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}}{\frac{\text{kx}}{2}[2\text{a}+(\text{kx}-1)\text{d]}}$
$\because\ \frac{\text{S}_\text{x}}{\text{S}_\text{kx}}$ is independent of x
$\therefore\ \frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}}{\frac{\text{kx}}{2}[2\text{a}+(\text{kx}-1)\text{d]}}$ is independent of x
$\therefore\ \frac{\frac{\text{n}}{2}[2\text{a}+\text{xd}-\text{d]}}{\frac{\text{kx}}{2}[2\text{a}+\text{kdx}-\text{d]}}$
$\Rightarrow\ \frac{2\text{a}-\text{d}}{\text{k}(2\text{a}-\text{d})}$ is in dependent of x if $2\text{a}-\text{d}\neq0$
If 2a - d = 0, then d = 2a.
View full question & answer→MCQ 371 Mark
Mark the correct alternative in the following:
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. $-(p + q)$
In the given problem, we are given $S_p = q$ and $S_q = p$
We need to find $S_{p+q}$
Now, as we know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
So
$\text{S}_\text{p}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$\text{q}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$2\text{q}=2\text{ap}+\text{p}(\text{q}-1)\text{d}\ .....(\text{i})$
Similarly,
$\text{S}_\text{q}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$\text{p}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$2\text{p}=2\text{aq}+\text{q}(\text{q}-1)\text{d}\ .....\text{(ii)}$
Subtracting $(ii)$ from $(i),$ we get
$2 q-2 p=2 a p+[p(p-1) d]-2 a q-[q(q-1) d]$
$2 q-2 p=2 a(p-q)+[p(p-1)-1(q-1)] d$
$-2(p-q)=2 a(p-q)+\left[\left(p^2-q^2\right)-(p-q)\right]$
$-2=2 a+(p+q-1) d \ldots . .(\text { iii })$
Now,
$\text{S}_{\text{p}+\text{q}}=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$\text{S}_{\text{p}+\text{q}}=\frac{(\text{p}+\text{d})}{2}(-2)$
$\text{S}_{\text{p}+\text{q}}= -(\text{p}+\text{d})$
Thus, $S_{p+q} = -(p + q)$
Hence, the correct option is $(d).$
View full question & answer→MCQ 381 Mark
Mark the correct alternative in the following:
If the first term of an $A.P.$ is $2$ and common difference is $4$, then the sum of its $40$ terms is:
- ✓
$3200$
- B
$1600$
- C
$200$
- D
$2800$
AnswerCorrect option: A. $3200$
In the given problem, we need to find the sum of $40$ terms of an arithmetic progression, where we are given the first term and the common difference.
So, here we use the following formula for the sum of $n$ terms of an $A.P.,$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Where; $a=$ first term for the given $A.P.$
$d =$ common difference of the given $A.P.$
$n =$ number of terms
Given,
First term $(a) = 2$
Common difference $( d )=4$
Number of terms $( n ) =40$
So, using the formula we get,
$S_{40}=\frac{40}{2}[2(2)+(40-1)(4)]$
$=(20)[4+(39)(4)]$
$=(20)[4+156]$
$=(20)(160)$
$=3200$
Therefore, the sum of first $40$ terms for the given $A.P.$ is $S_{40}=3200$.
So, the correct option is $(a).$
View full question & answer→MCQ 391 Mark
Mark the correct alternative in the following:
Sum of n term of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ .....\text{ is}$
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
The series is given
$\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ ....$
$\Rightarrow\ \sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+\ .....$
Here $\text{a}=\sqrt{2}\text{ and d} = 2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2\sqrt{2}+(\text{n}-1)\sqrt{2}]$
$=\frac{\text{n}}{2}[2\sqrt{2}+\sqrt{2}\text{n}-\sqrt{2}]$
$=\frac{\text{n}}{2}(\sqrt{2}\text{n}+\sqrt{2})$
$=\frac{\text{n}\sqrt{2}}{2}(\text{n}+1)=\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
View full question & answer→MCQ 401 Mark
Mark the correct alternative in the following:Two $A.P.'s$ have the same common difference. The first term of one of these is $8$ and that of the other is $3$. The difference between their $30_{th}$ term is:
AnswerIn two $A.P.'s$ common difference is same
Let A and a are two $A.P.'s$
First term of $A$ is $8$ and first term of a is $3$
$A_{30}-a_{30}=8+(30-1) d-3-(30-1) d$
$= 5 + 29d - 29d$
$= 5$
View full question & answer→MCQ 411 Mark
Mark the correct alternative in the following:
If $S_r$ denotes the sum of the first $r$ terms of an $A.P.$ Then, $S_{3n} : (S_{2n} - S_n)$ is:.
Answer$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}], \text{S}_{2\text{n}}=\frac{2\text{n}}{2}[2\text{a}+(2\text{n}-1)\text{d}]$
and $\text{S}_{3\text{n}}=\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]$
Now $\text{S}_{2\text{n}}-\text{S}_\text{n}=\frac{2\text{n}}{2}[2\text{a}+(2\text{n}-1)\text{d}]-\frac{\text{n}}{2}$
$[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[4\text{a}+(4\text{n}-2)\text{d}]-]2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[4\text{a}-2\text{a}+(4\text{n}-2-\text{n}+1)\text{d}]=\frac{\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]$
$=\frac{1}{3}(\text{S}_{3\text{n}})$
$\therefore\ \text{S}_{3\text{n}}:(\text{S}_{2\text{n}}-\text{S}_\text{n})=3:1\text{ or }\frac{3}{1}=3$
View full question & answer→MCQ 421 Mark
Mark the correct alternative in the following:
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
$\frac{1}{\text{n}}$
- B
$\frac{\text{n}-1}{\text{n}}$
- C
$\frac{\text{n}+1}{2\text{n}}$
- ✓
$\frac{\text{n}+1}{\text{n}}$
AnswerCorrect option: D. $\frac{\text{n}+1}{\text{n}}$
Sun of $n$ even natural number $=n(n+1)$
and sum of $n$ odd natural numbers $=n^2$
Stherefore $n ( n +1)=k n ^2$
$\Rightarrow\ \text{k}=\frac{\text{n}(\text{n}+1)}{\text{n}^2}=\frac{\text{n}+1}{\text{n}}$
View full question & answer→MCQ 431 Mark
The sum of first five multiples of 3 is
MCQ 441 Mark
If the $2^{\text {nd }}$ term of an A.P. is 13 and $5^{\text {th }}$ term is 25 , what is its $7^{\text {th }}$ term?
View full question & answer→MCQ 451 Mark
Which term of the A.P. $21,42,63,84, \ldots$ is 210 ?
- A
$9^{\text {th }}$
- ✓
$10^{\text {th }}$
- C
$11^{1 h}$
- D
$12^{\text {th }}$
AnswerCorrect option: B. $10^{\text {th }}$
View full question & answer→MCQ 461 Mark
The $4^{\text {th }}$ term from the end of the $A P:-11,-8,-5, \ldots, 49$ is
View full question & answer→MCQ 471 Mark
If the first term of an A.P. is -5 and the common difference is 2 , then the sum of first 6 terms is
MCQ 481 Mark
The sum of first 16 terms of the A.P.: $10,6,2, \ldots$, is
View full question & answer→MCQ 491 Mark
Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3 . The difference between their $30^{\text {th }}$ terms is
View full question & answer→MCQ 501 Mark
In an A.P., if the first term $a=7, n^{\text {th }}$ term $a_n=84$ and the sum of first $n$ terms $S_n=\frac{2093}{2}$. Then, $n$ is equal to
Answer(C)23
We have, $a_1=7, a_n=84$ and $S_n=\frac{2093}{2}$
$
\begin{array}{ll}
\therefore & S_n=\frac{n}{2}\left(a_1+a_n\right) \\
\Rightarrow & \frac{2093}{2}=\frac{n}{2}(7+84) \Rightarrow 91 n=2093 \Rightarrow n=23
\end{array}
$
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