Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
Which term of the series $ \sqrt{2}, \dfrac{2}{3}, \dfrac{2 \sqrt{2}}{9},… $ is $ \dfrac{16}{2187} $?
- $ 10^{th} $ term
- $ 8^{th} $ term
- $ 9^{th} $ term
- $ 11^{th} $ term
Correct

Incorrect

Hint

$Tn=\dfrac{16}{2187}$
$\implies ar^{n-1}=\dfrac{16}{2187}$
$\implies (\sqrt{2})\left(\frac{\sqrt{2}}{3}\right)^{n-1}=\frac{16}{2187}=\frac{8\sqrt{2}}{2187}=\left(\frac{\sqrt{2}}{3}\right)^7$
$\implies n-1=7$
$\implies n=8$
Question 2 of 25

2. Question
1 point(s)
Let $ S_k $ be sum of an infinite $ \text{GP} $ whose first term is $ k $ and common ratio is $ \dfrac{k}{(k+1)}(k >0) $. Then the value of $ \sum_{k=1}^{\infty}\frac{(-1)^k}{S_k} $ is equal to_______
- $ \ln 4 $
- $ \ln 2-1 $
- $ 1-\ln 2 $
- $ 1-\ln 4 $
Correct

Incorrect

Hint

$S_k=\dfrac{k}{1-k}=k(k+1)$
$\sum_{k=1}^{\infty}\dfrac{(-1)^k}{S_k} = \sum_{k=1}^{\infty}\dfrac{(-1)^k}{k(k+1)}=\sum_{k=1}^{\infty}\left[ \dfrac{1}{k}-\dfrac{1}{k+1} \right]$
$=\sum_{k=1}^{\infty}\left[ -\dfrac{(-1)^{k+1}}{k}+\dfrac{(-1)^{k+1}}{k}\right]=-\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{k}-\sum_{k=2}^{\infty}\dfrac{(-1)^{k+1}}{k}$
$=1-\sum_{k=2}^{\infty}\dfrac{(-1)^{k+1}}{k}-\sum_{k=2}^{\infty}\dfrac{(-1)^{k+1}}{k}$
$=1-2\sum_{k=2}^{\infty}\dfrac{(-1)^{k+1}}{k}$
$=1-2\ln 2=1-\ln 4$
Question 3 of 25

3. Question
1 point(s)
The first term of an $ \text{AP} $ is $ a $ and the sum of the first $ p $ terms is zero, then the sum of its next $ q $ terms is ________
- $ \dfrac{p+q}{p-1} $
- $ \dfrac{-a(p+q)q}{p-1} $
- $ \dfrac{-a(p+q)q}{p-1} $
- None of these
Correct

Incorrect

Hint

Let common difference of $\text{AP}$ is $d$ .
$S_p=0\implies \dfrac{p}{2}\left[2a+(p-1)d\right]=0$
$\implies 2a+(p-1)d=0\implies d=\dfrac{-2a}{p-1}$
Sum of next $q$ terms: $S_{p+q}-Sp$
$=S_{p+q}-0=\left(\dfrac{p+q}{2}\right)\left[2a+(p+q-1)d\right]$
$=\left(\dfrac{p+q}{2}\right)\left[2a+(p-1)d+qd\right]$
$=\left(\dfrac{p+q}{2}\right)\left[2a+(p-1)\cdot\dfrac{-2a}{p-1}+\dfrac{q(-2a)}{p-1}\right]$
$=\left(\dfrac{p+q}{2}\right)\left[2a+(-2a)-\dfrac{2aq}{p-1}\right]$
$=\dfrac{p+q}{2}\left[\dfrac{-2aq}{p-1}\right]=\dfrac{-1(p+q)q}{p-1}$
Question 4 of 25

4. Question
1 point(s)
A man saved $ ₹66000 $ in 20 yrs. In such succeeding years after the first year, he saved $ ₹200 $ more than what he saved in the previous year. How much did he save in the first year?
- $ 1200 $
- $ 1300 $
- $ 1400 $
- $ 1500 $
Correct

Incorrect

Hint

First term $=a, c \cdot =200, n=20$ yr.
$S_{20} =\dfrac{20}{2}\left[2a+(20-1)d\right]$
$\implies 66000=10\left[2a+19d\right]$
$\implies 66000=20a+190d$
$\implies 66000=20a+190\times 200$
$\implies 20a=66000-38000$
$\implies 20a=28000$
$\implies a=\dfrac{28000}{20}=1400$
Hence, he saved $₹1400$ in the first year.
Question 5 of 25

5. Question
1 point(s)
A man accepts a position with an initial salary of $ ₹5200 $ per month. It is understood that he will receive an automatic increase of $ ₹320 $ in the very next month and each month thereafter. What is his salary for the tenth month?
- $ 8080 $
- $ 8060 $
- $ 8090 $
- $ 8030 $
Correct

Incorrect

Hint

First term $=5200$ , Common difference $=320$
Salary for tenth month, for $n=10$
$a_{10}=a+(n-1)d$
$\begin{aligned}\implies a_{10}&=5200+(10-1)\times 320\\&=5200+9\times 320\\&=5200+2880\\\therefore a_{10}&=8080 \end{aligned}$
Question 6 of 25

6. Question
1 point(s)
A Carpenter was hired to build $ 192 $ window frames. The first day he made five frames and each day, threafter he made two more frames than he made the day before. How many days did it take him to finish the job?
- $ 11 $
- $ 12 $
- $ 13 $
- $ 15 $
Correct

Incorrect

Hint

Here $a=5$ and $d=2$
Then $S_n=192$
$\begin{aligned} \implies S_n&=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\\&=\dfrac{n}{2}\left[ 2\times 5+(n-1)2 \right]=\dfrac{n}{2}\left[ 10+2n-2 \right] \end{aligned}$
$\implies 192=\dfrac{n}{2}\left[ 8+2n \right]=4n+n^2$
$n^2+4n-192=0\implies (n-12)(n+16)=0$
$\implies n=12,-16$
$\therefore n=12$
Question 7 of 25

7. Question
1 point(s)
Find the sum of the series $ (3^3-2^3)+(5^3-4^3)+(7^3-6^3)+\ … $ to $ n $ terms _______
- $ 4n^3+9n^2+6n $
- $ 4n^3-9n^2-6n $
- $ 4n^3-9n^2+6n $
- $ 4n^3+9n^2-6n $
Correct

Incorrect

Hint

Given, $(3^3-2^3)+(5^3-4^3)+(7^3-6^3)+...$
$=(3^3+5^3+7^3+...)-(2^3+4^3+6^3+...)$
$\begin{aligned} Tn&=(2n+1)^3-(2n)^3\\&=\left[ 4n^2+1+4n+4n^2+2n+4n^2 \right]\\ &=\left[ 12n^2+6n \right]+1 \end{aligned}$
$\begin{aligned} Sn&=\Sigma\ T_n=\Sigma(12n^2+6n) \\&=12\Sigma\ n^2+6\ \Sigma\ n+\Sigma\ n \\&=12\cdot\frac{n(n+1)(2n+1)}{6}+\frac{6n(n+1)}{2}+n \\&=2n(n+1)(2n+1)+3n(n+1)+n \\&=2n(n+1)(2n+1)+3n^2+3n+n \\&=4n^3+2n^2+4n^2+2n+3n^2+3n+n \\&=4n^3+9n^2+6n \end{aligned}$
Question 8 of 25

8. Question
1 point(s)
Find the $ r^{th} $ term of an $ \text{AP} $, where sum of whose first $ n $ terms is $ 2n+3n^2 $______
- $ 6r+1 $
- $ 6r-1 $
- $ 3r+1 $
- $ 3r-1 $
Correct

Incorrect

Hint

$Sn=2n+3n^2$
$\begin{aligned} Tn&=Sn-S_{n-1}\\&= \left( 2n+3n^2 \right)-\left[ 2(n-1)+3(n-1)^2 \right] \\&=\left( 2n+3n^2 \right)-\left[ 2n-2+3\left( n^2+1-2n \right) \right] \\&=\left( 2n+3n^2 \right)-\left( 2n-2+3n^2+3-6n \right) \\&=2n+3n^2-2n+2-3n^2-3+6n \\&=6n-1 \end{aligned}$
$\therefore r^{th}$ term $T_r=6r-1$
Question 9 of 25

9. Question
1 point(s)
If $ A $ is the $ \text{A.M} $ and $ G_1,\ G_2 $ be two $ \text{GM} $ between any two numbers, then find $ \dfrac{G_1^2}{G_2}+\dfrac{G_2^2}{G_1} $.
- $ A $
- $ 3A $
- $ A^2 $
- $ 2A $
Correct

Incorrect

Hint

$A=\dfrac{a+b}{2}\implies 2A=a+b$
$G_1,\ G_2$ be $\text{GM}$ between $a$ and $b$ , then $a,\ G_1,\ G_2,\b$ are in $\text{GP}$
Let $r$ be the common ratio.
Then, $b=ar^{4-1}=ar^3\implies \dfrac{b}{a}=r^3\implies r=\left( \dfrac{b}{a} \right)^{1/3}$
$G_1=ar=a\left( \frac{b}{a} \right)^{\frac{1}{3}}\text{ and }\ G_2=ar^2=a\left( \dfrac{b}{a} \right)^{\frac{2}{3}}$
$\therefore \dfrac{G_1^2}{G_2}+\dfrac{G_2^2}{G1}=\dfrac{\left[ a\left( \dfrac{b}{a} \right)^{\frac{1}{3}} \right]^2}{a\left( \dfrac{b}{a} \right)^{\frac{2}{3}}}+\dfrac{\left[ a\left( \dfrac{b}{a} \right)^{\frac{2}{3}} \right]^2}{a\left( \dfrac{b}{a} \right)^{\frac{1}{3}}}$
$=\dfrac{a^2\left( \dfrac{b}{a} \right)^{\frac{2}{3}}}{a\left( \dfrac{b}{a} \right)^{\frac{2}{3}}}+\dfrac{a^2\left( \dfrac{b}{a} \right)^{\frac{4}{3}}}{a\left( \dfrac{b}{a} \right)^{\frac{1}{3}}}=a+a\left( \dfrac{b}{a} \right)$
$=a+b=2A$
Question 10 of 25

10. Question
1 point(s)
If the sum of $ n $ terms of an $ \text{AP} $ is given by $ S_n=3n+2n^2, $ then the common difference of the AP is_______
- $ 3 $
- $ 2 $
- $ 6 $
- $ 4 $
Correct

Incorrect

Hint

$Sn=3n+2n^2$
First term of the $\text{AP}$
$\therefore T_1=3\times 1+2(1)^2=3+2=5$
and $\begin{aligned} T_2&=S_2-S_1 \\&=\left[ 3\times 2+2\times(2)^2 \right] - \left[ 3 \times 1+2 \times (1)^2 \right] \\&=14-5=9 \end{aligned}$
$\therefore$ Common difference $(d)=T_2-T_1=9-5=4$
Question 11 of 25

11. Question
1 point(s)
If $ 9 $ times the $ 9^{th} $ term of an $\text{ AP} $ is equal to $ 13 $ times the $ 13^{th} $ term, then the $ 22^{nd} $ term of the $ \text{AP} $ is_______
- $ 0 $
- $ 22 $
- $ 198 $
- $ 220 $
Correct

Incorrect

Hint

$9\times T_9=13\times T_13$
$\implies 9(a+8d)=13(a+12d)$
$\implies 9a+72d=13a+156d$
$\implies 9a-13a=156d-72d$
$\implies -4a=84d$
$\implies a=-21d$
$\implies a+21d=0$
$T_{22}=a+21d=0$
$\therefore T_{22}=0$
Question 12 of 25

12. Question
1 point(s)
If $ x,\ 2y, $ and $ 3z $ are in $ \text{AP} $ where the distinct numbers $ x,\ y, $ and $ z $ are in $ \text{GP} $, then the common ratio of the $ \text{GP} $ is__________
- $ 3 $
- $ \dfrac{1}{3} $
- $ 2 $
- $ \dfrac{1}{2} $
Correct

Incorrect

Hint

$x,\ 2y,\ 3z$ are in $\text{AP}\implies 2y=\dfrac{x+32}{2}$
$\implies y=\dfrac{x+3z}{4}\implies 4y=x+32$ —(1)
$x,\ y,\z$ are in $\text{GP}$
$\implies \dfrac{y}{x}=\dfrac{z}{y}=\lambda\implies y=x\lambda\ \text{ and }\ z=\lambda y=\lambda^2x$
From (1)
$4\left( x\lambda \right)=x+3\left( \lambda^2x \right)$
$\implies 4x\lambda=x+3\lambda^2x$
$\implies 4\lambda=1+3\lambda^2$
$\implies 3\lambda^2-4\lambda+1=0$
$\implies\left( 3\lambda=1 \right)\left( \lambda-1 \right)=0$
$\implies\lambda=\dfrac{1}{3},\ \lambda=1$
Question 13 of 25

13. Question
1 point(s)
If in an $ \text{AP},\ S_n=qn^2 $ and $ S_m=qm^2, $ where $ Sr $ donates the sum of $ r $ terms of the $ \text{AP} $ then $ Sq $ equals to________
- $ \dfrac{q^3}{2} $
- $ mnq $
- $ q^3 $
- $ (m+n)q^2 $
Correct

Incorrect

Hint

$Sn=qn^2\ \text{ and }\ Sm=qm^2$
$\therefore S_1=q,\ S_2=4q,\ S_3=9q\ \text{ and }\ S_4=16q$
Now, $T_1=q$
$T_2=S_2-S_1=4q-q=3q$
$T_3=S_3-S_2=9q-4q=5q$
$T_4=S_4-S_3=16q-9q=7q$
So, the series is $q,\ 3q,\ 5q,\ 7q,\ ...$
Here $a=q\ \text{ and }\ d=3q-q=2q$
$Sq=\dfrac{q}{2}\left[ 2\times q+\left( q-1 \right)2q \right]$
$=\dfrac{q}{2}\left[ 2q+2q^2-2q \right]$
$=\dfrac{q}{2}\times2q^2=q^3$
Question 14 of 25

14. Question
1 point(s)
The minimum value of $ 4^x+4^{1-x}, x\in R $ is_______
- $ 2 $
- $ 4 $
- $ 1 $
- $ 0 $
Correct

Incorrect

Hint

$AM\geq GM$
$\implies\dfrac{4^x+4^{1-x}}{2}\geq \sqrt{4^x\cdot4^{1-x}}$
$\implies 4^x+4^{1-x}\ge 2\sqrt{4}$
$\implies 4^x+4^{1-x}\geq 2\times2$
$\implies 4^x+4^{1-x}\geq 4$
$\therefore$ minimum value of $4^x+4^{1-x}$ is $4$
Question 15 of 25

15. Question
1 point(s)
If $ Tn $ donates the $ n^{th} $ term of the series $ 2+3+6+11+18+… \ , $ then $ T_{50} $ is _______
- $ 49^2-1 $
- $ 49^2 $
- $ 50^2+1 $
- $ 49^2+2 $
Correct

Incorrect

Hint

$Sn=2+3+6+11+18+...+T_{50}$
$\text{ and }\ Sn=0+2+3+6+...+T_{49}+T_{50}$
$Sn-Sn=\left( 2-0 \right)+\left( 3-2 \right)+\left( 6-3 \right)+...$
$\implies 0=2+1+3+...-T_{50}$
$\begin{aligned} \implies T_{50}&=2+1+3+5+... \\&=2+\left[ 1+3+5+...\text{Up to}\ 49\ \text{terms}\right] \\&=2+\frac{49}{2}\left[ 2\times1+48\times2 \right] \\&=2+\frac{49}{2}\left[ 2+96 \right] \\&=2+\left( 49+49\times48 \right) \\&=2+49\times49 \\&=2+(49)^2 \end{aligned}$
Question 16 of 25

16. Question
1 point(s)
If $ a,\ b, $ and $ c $ are in $ \text{GP} $ then the value of $ \dfrac{a-b}{b-c} $ is equal to______
- $ \dfrac{a}{b} $
- $ \dfrac{c}{b} $
- $ \dfrac{-b}{a} $
- $ -\dfrac{b}{c} $
Correct

Incorrect

Hint

$\dfrac{b}{a}=\dfrac{c}{b}=r$ (Since $a,\ b,\$ ) and $\ c$ are in $\text{GP}$ )
$b=ar,\ c=br$
$\dfrac{a-b}{b-c}=\dfrac{a-ar}{ar-br}=\dfrac{a(1-r)}{r(a-b)}=\dfrac{a(1-r)}{r(a-ar)}=\dfrac{a(1-r)}{ar(1-r)}=\dfrac{1}{r}$
$\therefore \dfrac{a-b}{b-c}=\dfrac{1}{r}=\dfrac{a}{b}\ \text{ or }\ \dfrac{b}{c}$
$\therefore \dfrac{a-b}{b-c}=\dfrac{a}{b}$
Question 17 of 25

17. Question
1 point(s)
The series $ 4,\ 1,\ \dfrac{1}{4},\ \dfrac{1}{16} $ are in ______
- $ AP $
- $ GP $
- $ HP $
- $ AGP $
Correct

Incorrect

Hint

$T_1=4,\ T_2=1,\ T_3=\dfrac{1}{4},\ T_4=\dfrac{1}{16}$
$\implies \dfrac{T_2}{T_1}=\dfrac{1}{4}$
$\implies \dfrac{T_3}{T_2}=\dfrac{1}{4}$
$\implies \dfrac{T_4}{T_3}=\dfrac{1\16}{1\4}=\dfrac{1}{4}$
$\therefore$ It is an $GP$
Question 18 of 25

18. Question
1 point(s)
The series $ 13,\ 8,\ 3,\ -2,\ -7 $ are in_____
- $ AP $
- $ GP $
- $ HP $
- $ AGP $
Correct

Incorrect

Hint

Here $T_1=13,\ T_2=8,\ T_3=3,\ T_4=-2,\ T_5=-7$
$\therefore T_2-T_1=8-13=-5$
$T_3-T_2=3-8=-5$
$T_4-T_3=-2-3=-5$
$T_5-T_4=-7-(-2)=-5$
$\therefore$ It is an $AP$
Question 19 of 25

19. Question
1 point(s)
If $ a^2,\ b^2,\ c^2 $ are in $ \text{AP} $ which of the following is also an $\text{ AP} $?
- $ \sin A,\sin B,\sin C $
- $ \tan A,\tan B,\tan C $
- $ \cot A,\cot B,\cot C $
- None of these
Correct

Incorrect

Hint

$\sin^2 B-\sin^2 A=\sin^2 C-\sin^2 B$

$\implies \sin (B-A) \sin (B+A)=\sin (C-B)\cdot \sin (C+B)$
$\implies \sin C \left[ \sin B \cdot \cos A-\cos B \cdot \sin A \right]$
$=\sin A \left[ \sin C \cdot \cos B-\cos C \cdot \sin B \right]$
$\implies 2 \cot B =\cot A+\cot C$
[Taking mathjax] $a=\sin A, b=\sin B, c=\sin C$ ]
$\implies \cot A, \cot B,\cot C$ are an AP.
Question 20 of 25

20. Question
1 point(s)
If the $ AM $ of $ a $ and $ b $ is $ \dfrac{a^n+b^n}{a^{n-1}+b^{n-1}} $, then the value of $ n $ is _______
- $ -1 $
- $ 0 $
- $ 1 $
- None of these
Correct

Incorrect

Hint

$\dfrac{a+b}{2}=\dfrac{a^n+b^n}{a^{n-1}+b^{n-1}}$
$\implies a^n\left( \dfrac{a-b}{a} \right)=b^n\left( \dfrac{a-b}{b} \right)$
$\implies \dfrac{a^n}{b^n}=\dfrac{a}{b}=\left( \dfrac{a}{b} \right)^1$
$\implies \left( \dfrac{a}{b} \right)^n=\left( \dfrac{a}{b} \right)^1$
$\implies n=1$
Question 21 of 25

21. Question
1 point(s)
$ \sum_{r=1}^{\infty} \dfrac{1+a+a^2+…+a^{r-1}}{r!} $ is equal to_______
- $ \dfrac{e^a-e}{a-1} $
- $ e^a-e $
- $ \dfrac{e^a}{a-1} $
- $ \dfrac{e^a-e}{e-1} $
Correct

Incorrect

Hint

$\sum_{r=1}^{\infty } \dfrac{\left[ 1+a+a^2+...+a^{r-1} \right]}{r!}=\sum_{r=1}^{\infty }\dfrac{1\left( a^r-1 \right)}{\left( a-1 \right)r!}$
$=\dfrac{1}{a-1}\left[ \sum_{r=1}^{\infty }\dfrac{a^r}{r!}-\sum_{r=1}^{\infty }\dfrac{1}{r!} \right]$
$=\dfrac{1}{a-1}\left[ \dfrac{a}{1!}+\frac{a^2}{2!}+...-\left( \dfrac{1}{1!}+\dfrac{1}{2!}+\frac{1}{3!}+...\right) \right]$
$=\dfrac{1}{a-1}\left[ e^a-1-\left( e-1 \right) \right]$
$=\dfrac{e^a-e}{a-1}$
Question 22 of 25

22. Question
1 point(s)
If $ \left\{ a_n \right\} $ is in $ \text{GP{ $ such that $ \dfrac{a_4}{a_6}=\dfrac{1}{4} $ and $ a_2+a_5=216 $, then $ a_1 $ is equal to______
- $ 12 $ or $ \dfrac{108}{7}$
- $ 10 $
- $ 7 $ or $ \dfrac{57}{7}$
- None of these
Correct

Incorrect

Hint

$\dfrac{T_4}{T_6}=\dfrac{1}{4}\implies \dfrac{ar^3}{ar^5}=\dfrac{1}{4}$
$\implies r^2=4\implies r=\pm 2$
$T_2+T_5=216$
$\implies ar+ar^4=216$
when $r=2, a=12$
$r=-2,\ a=\dfrac{108}{7}$
$\therefore a=12\ \text{ or }\dfrac{108}{7}$
Question 23 of 25

23. Question
1 point(s)
The value of $ x $ which satisfies $ 1+\cos x+\cos^2x+…=64 $ in $ \left[-\pi,\pi\right] $ is_______
- $ \pm \dfrac{\pi}{2},\pm \dfrac{\pi}{3} $
- $ \pm \dfrac{\pi}{3} $
- $ \pm \dfrac{\pi}{2},\pm\dfrac{\pi}{6} $
- None of these
Correct

Incorrect

Hint

$8^{1+\cos x+\cos^2x+...}=8^2$
$\implies 8^{\dfrac{1}{1-Cos\ x}}=8^2$
$\implies \dfrac{1}{1-\cos x}=2$
$\implies \cos x=\dfrac{1}{2}\implies x=\pm \dfrac{\pi}{3}$
Question 24 of 25

24. Question
1 point(s)
If $ S $ is the sum, $ P $ is the product and $ R $ is the sum of the reciprocals of $ n $ terms of a $ GP $, then $ P^2 $ is equal to ______
- $ \left( \dfrac{S}{R} \right)^n$
- $ \dfrac{S}{R} $
- $ \left( \dfrac{R}{S} \right)^n $
- $ \dfrac{R}{S} $
Correct

Incorrect

Hint

$S=a+ar+ar^2+...+ar^{n-1}=\dfrac{a(1-r^n)}{1-r}$
$\begin{aligned} P&=a\cdot ar \cdot ar^2...ar^{n-1}=a^nr^{1+2+...+(n-1)}\\&=a^nr^{\frac{n(n-1)}{2}} \end{aligned}$
$\begin{aligned} R&=\dfrac{1}{a}+\dfrac{1}{ar}+\dfrac{1}{ar^2}+...+\dfrac{1}{ar^{n-1}}=\dfrac{\dfrac{1}{a}\left( \dfrac{1}{r^n} \right)-1}{\dfrac{1}{r}-1}\\&=\dfrac{\dfrac{1}{a}(1-r^n)}{r^{n-1}(1-r)}\\\therefore P^2&=\left\{ a^n r^n\dfrac{n(n-1)}{2}=a^{2n}r^{n(n-1)} \right\} \\&=\left| \dfrac{\dfrac{a(1-r^n)}{1-r}}{\dfrac{\dfrac{1}{a(1-r^n)}}{r^{n-1}(1-r)}}\right|=\left( \dfrac{S}{R} \right)^n \end{aligned}$
Question 25 of 25

25. Question
1 point(s)
If $ a_1,\ a_2,\ …\ a_{50} $ are in $\text{ GP} $, then $ \dfrac{a_1-1_3+a_5-…+a_{49}}{a_2-a_4+a_6-…+a_{50}} $ is equal to _______
- $ 0 $
- $ 1 $
- $ \dfrac{a_1}{a_2} $
- $ \dfrac{a_{25}}{a_{24}} $
Correct

Incorrect

Hint

$a_2=r\ a_1,\ a_3=r^2a_1,\ ...$ (Say)
$\therefore \dfrac{a_1-a_3+a_5-...+a_{49}}{a_2-a_4+a_6-...+a_{50}}$
$=\dfrac{a_1-r^2a_1+r^4a_1-...+r^{48}a_1}{a_2r-r^3a_1+r^4a_1-...+r^{48}a_1}$
$=\dfrac{a_1\left( 1-r^2+r^4-...+r^{48} \right)}{a_1r\left( 1-r^2+r^4-...+r^{48} \right)}=\dfrac{1}{r}=\dfrac{a_1}{a_2}$

Mathematics Class XI

Practice Questions 2-Sequence ans Series-Class XI

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