Sequence-Series – JEE Main Previous Year Question with Solutions

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Q. The sum to infinity of the series $1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\frac{14}{3^{4}}+\ldots .$ is :-

(1) 4          (2) 6          (3) 2           (4) 3


Sol. (4)

Q. A person is to count 4500 currency notes. Let a denote the number of notes he counts in the $\mathrm{n}^{\mathrm{th}}$ minute. If $\mathrm{a}_{1}=\mathrm{a}_{2}=\ldots=\mathrm{a}_{10}=150$ and $\mathrm{a}_{10} \mathrm{a}_{11}, \ldots$ are in an AP with common difference – $2,$ then the time taken by him to count all notes is :-

(1) 24 minutes        (2) 34 minutes          (3) 125 minutes         (4) 135 minutes


Sol. (2)

Q. 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 the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after :-

(1) 20 months        (2) 21 months            (3) 18 months        (4) 19 months


Sol. (2)

Saving after first 3 month $=600$

$600+\left\{\frac{240+280+\ldots .}{\text { let } n \text { month }}\right\}=11040$

$| 240+280+\ldots . .$ n terms $|=10440$

n/2 $[480+\ldots . . . \text { n terms }]=10440$

n/2 $[480+(n-1) 40]=10440$

n $\{440+40 n\}=20880$

$\mathrm{n}^{2}+11 \mathrm{n}-522=0$

$\mathrm{n}=18,-29 \quad(-29 \text { rejected })$

Total months $=\mathrm{n}+3$

18 $+3=21$ Months

Q. Let $a_{n}$ be the $n^{\text {th }}$ term of an A.P. If $\sum_{r=1}^{100} a_{2 r}=\alpha$ and $\sum_{r=1}^{100} a_{2 r-1}=\beta,$ then the common difference of the A.P. is :

(1) $\frac{\alpha-\beta}{200}$

(2) $\alpha-\beta$

(3) $\frac{\alpha-\beta}{100}$

(4) $\beta-\alpha$


Sol. (3)

Q. Statement-1: The sum of the series $1+(1+2+4)+(4+6+9)+(9+12+16)$

$\quad+\ldots \ldots+(361+380+400)$ is $8000 .$

Statement-2: $\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\mathrm{k}^{3}-(\mathrm{k}-1)^{3}\right)=\mathrm{n}^{3},$ for any natural number $\mathrm{n}$

(1) Statement-1 is true, Statement-2 is false.

(2) Statement- 1 is false, Statement- 2 is true.

(3) Statement- 1 is true, Statement-2 is true ; Statement-2 is a correct explanation for


(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for

Statement- – $1 .$


Sol. (4)


$\left(1^{3}-0^{3}\right)+\left(2^{3}-1^{3}\right)+\left(3^{3}-2^{3}\right)+\ldots+\left(20^{3}-19^{3}\right)=20^{3}=8000$ Statement-l is true.


$\sum_{k=1}^{n} k^{3}-(k-1)^{3}=\left(1^{3}-0^{3}\right)+\left(2^{3}-1^{3}\right)+\left(3^{3}-2^{3}\right)+\ldots . n^{3}+(n-1)^{3}=n^{3}$

Statement-2 is true and Statement-2 is a correct explanation of Statement- $1 .$

Q. If 100 times the 100th term of an A.P. with non-zero common difference equals the

50 times its 50th term, then the 150th term of this A.P. is :

(1) zero             (2) –150           (3) 150 times its               50th term (4) 150


Sol. (1)

Q. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……, is :

(1) $\frac{7}{81}\left(179-10^{-20}\right)$

(2) $\frac{7}{9}\left(99-10^{-20}\right)$

(3) $\frac{7}{81}\left(179+10^{-20}\right)$

(4) $\frac{7}{9}\left(99-10^{-20}\right)$


Sol. (3)

$S=\frac{7}{10}+\frac{77}{100}+\frac{777}{1000}+\ldots \ldots \quad S=\frac{7}{9}\left\{\frac{10-1}{10}+\frac{100-1}{100}+\frac{1000-1}{1000}+\ldots\right\}$

$=\frac{7}{9}\left\{20-\frac{1}{10}\left(\frac{1-10^{-20}}{9 / 10}\right)\right\}=\frac{7}{9}\left\{20-\frac{1}{9}\left(1-10^{-20}\right)\right\}=\frac{7}{81}\left(179+10^{-20}\right)$

Q. Let $\alpha$ and $\beta$ be the roots of equation $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0, \mathrm{p} \neq 0 .$ If $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in A.P. and $\frac{1}{\alpha}+\frac{1}{\beta}=4,$ then the value of $|\alpha-\beta|$ is:

(1) $\frac{\sqrt{61}}{9}$

(2) $\frac{2 \sqrt{17}}{9}$

(3) $\frac{\sqrt{34}}{9}$

(4) $\frac{2 \sqrt{13}}{9}$


Sol. (4)

$\frac{\alpha+\beta}{\alpha \beta}=4=-\frac{q}{r} \Rightarrow q=-4 r$

$\because p, q \& r$ are in A.P $2 q=p+r$

$\Rightarrow \quad-8 r=p+r \Rightarrow p=-9 r$

$|\alpha-\beta|=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}=\sqrt{\frac{16 r^{2}+36 r^{2}}{p^{2}}}=\sqrt{\frac{52 r^{2}}{p^{2}}}=\sqrt{\frac{52}{81}}=\frac{2 \sqrt{13}}{9}$

Q. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is :

(1) $\sqrt{2}+\sqrt{3}$

(2) $3+\sqrt{2}$

(3) $2-\sqrt{3}$

(4) $2+\sqrt{3}$


Sol. (4)

Let a, ar, ar $^{2}$ are in G.P

$\therefore \quad$ a, 2 ar, $\operatorname{ar}^{2}$ are in AP

$\Rightarrow \quad 4$ ar $=a+a r^{2}$

$\Rightarrow \quad r^{2}-4 r+1=0$

$\Rightarrow r=2+\sqrt{3}, 2-\sqrt{3}$

since GP is an increasing G.P

$\Rightarrow r=2+\sqrt{3}$

Q. If $(10)^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+\ldots \ldots+10(11)^{9}=\mathrm{k}(10)^{9},$ then $\mathrm{k}$ is equal to :

(1) $\frac{121}{10}$

(2) $\frac{441}{100}$

(3) 100

(4) 110


Sol. (3)


$\frac{11}{10} \mathrm{S}=(11)(10)^{8}+2(11)^{2}(10)^{7}+\ldots 11^{10}$

$-\frac{\mathrm{S}}{10}=10^{9}+\left(11.10^{8}+11^{2} .10^{7}+\ldots .+11^{9}\right)-11^{10}$

$\begin{aligned}-\frac{\mathrm{S}}{10} &=10^{9}+11.108^{8} \frac{\left(1-\left(\frac{11}{10}\right)^{9}\right)}{\left(1-\frac{11}{10}\right)}-11^{10} \\ &=10^{9}+10^{8} \cdot 11 \frac{\left(10^{9}-11^{9}\right)}{10^{9}(-1)} \cdot 10-11^{10} \\ &=10^{9}+11\left(11^{9}-11^{9}\right) \cdot 10-11^{10} \\ &=10^{9}+11\left(11^{9}-10^{9}\right)-11^{10} \\ \mathrm{S} &=10^{11}=\mathrm{K} 10^{9} \\ \Rightarrow \mathrm{K} &=100 \end{aligned}$

Q. If $\mathrm{m}$ is the A.M. of two distinct real numbers 1 and $\mathrm{n}(1, \mathrm{n}>1)$ and $\mathrm{G}_{1}, \mathrm{G}_{2}$ and $\mathrm{G}_{3}$ are three geometric means between 1 and $\mathrm{n}$, then $\mathrm{G}_{1}^{4}+2 \mathrm{G}_{2}^{4}+\mathrm{G}_{3}^{4}$ equals –

(1) $4 \operatorname{lmn}^{2}$

(2) $41^{2} \mathrm{m}^{2} \mathrm{n}^{2}$

(3) $4 \mathrm{l}^{2} \mathrm{mn}$

(4) $4 \ln ^{2} n$


Sol. (4)

$\ell, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{n}$ in G.P.

Let $\mathrm{r}$ be the common ratio $\Rightarrow \mathrm{r}^{4}=\frac{\mathrm{n}}{\ell}$

Here $\mathrm{G}_{1}^{4}+2 \mathrm{G}_{2}^{4}+\mathrm{G}_{3}^{4}=(\ell \mathrm{r})^{4}+2\left(\ell \mathrm{r}^{2}\right)^{4}+\left(\ell \mathrm{r}^{3}\right)^{4}=\mathrm{n} \ell\left[\ell^{2}+2 \ell \mathrm{n}+\mathrm{n}^{2}\right]=\mathrm{n} \ell(\ell+\mathrm{n})^{2}=\mathrm{n} \ell 4 \mathrm{m}^{2}$

$=4 \mathrm{m}^{2} \mathrm{n} \ell \quad(\because 2 \mathrm{m}=\mathrm{n}+\ell)$

Q. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this

( 1)$\frac{7}{4}$

(2) $\frac{8}{5}$

(3) $\frac{4}{3}$

(4) 1


Sol. (3)

Let ‘a’ be the first term and d be the common difference

$2^{\text {nd }} \operatorname{term}=a+d, 5^{\text {th }} \operatorname{term}=a+4 d$,

9th term $=4+8 d$

$\therefore$ Common ratio $=\frac{a+4 d}{a+d}=\frac{a+8 d}{a+4 d}=\frac{4 d}{3 d}=\frac{4}{3}$

Q. If the sum of the first ten terms of the series $\left(1 \frac{3}{5}\right)^{2}+\left(2 \frac{2}{5}\right)^{2}+\left(3 \frac{1}{5}\right)^{2}+4^{2}+\left(4 \frac{4}{5}\right)^{2}+\ldots,$ is $\frac{16}{5} \mathrm{m}$, then $\mathrm{m}$ is equal to :-

(1) 99         (2) 102          (3) 101          (4) 100


Sol. (3)

Given series is

$\mathrm{S}=\frac{8^{2}}{5^{2}}+\frac{12^{2}}{5^{2}}+\frac{16^{2}}{5^{2}}+\ldots 10 \mathrm{terms}$

$=\frac{4^{2}}{5^{2}}\left(2^{2}+3^{2}+4^{2}+\ldots .10 \mathrm{terms}\right)$

$=\frac{16}{25}\left(\frac{11.12 .23}{6}-1\right)=\frac{16}{25} \times 505$

$\therefore \mathrm{m}=101$

Q. If, for a positive integer n, the quadratic equation,

$\mathrm{x}(\mathrm{x}+1)+(\mathrm{x}+1)(\mathrm{x}+2)+\ldots \ldots+(\mathrm{x}+\overline{\mathrm{n}-1})(\mathrm{x}+\mathrm{n})$= 10n has two consecutive integral solutions, then n is equal to :

(1) 11       (2) 12        (3) 9         (4) 10


Sol. (1)

We have

$\sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{x}+\mathrm{r}-1)(\mathrm{x}+\mathrm{r})=10 \mathrm{n}$

$\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{x}^{2}+(2 \mathrm{r}-1) \mathrm{x}+\left(\mathrm{r}^{2}-\mathrm{r}\right)\right)=10 \mathrm{n}$

$\therefore$ On solving, we get

Q. For any three positive real numbers a, b and $c, 9\left(25 a^{2}+b^{2}\right)+25\left(c^{2}-3 a c\right)=15 b(3 a+c)$ Then :

(1) a, b and c are in G.P.

(2) b, c and a are in G.P.

(3) b, c and a are in A.P.

(4) a, b and c are in A.P.


Sol. (3)

$(15 a)^{2}+(3 b)^{2}+(5 c)^{2}-(15 a)(5 c)-(15 a)(3 b)-(3 b)(5 c)=0$

$\frac{1}{2}\left[(15 a-3 b)^{2}+(3 b-5 c)^{2}+(5 c-15 a)^{2}\right]=0$

it is possible when $15 a=3 b=5 c$

$\therefore b=\frac{5 c}{3}, a=\frac{c}{3}$

$a+b=2 c$

$\Rightarrow b, c, a$ in A.P.

Q. Let $a, b, c \in R .$ If $f(x)=a x^{2}+b x+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+$ $\mathrm{xy}, \forall \mathrm{x}, \mathrm{y} \in \mathrm{R},$ then $\sum_{\mathrm{n}=1}^{10} \mathrm{f}(\mathrm{n})$ is equal to :

(1) 255        (2) 330          (3) 165          (4) 190


Sol. (2)

Q. Let $a_{1}, a_{2}, a_{3}, \ldots ., a_{49}$ be in A.P. such that $\sum_{\mathrm{k}=0}^{12} \mathrm{a}_{4 \mathrm{k}+1}$ $=416$ and $a_{9}+a_{43}=66 .$ If $a_{1}^{2}+a_{2}^{2}+\ldots \ldots+a_{17}^{2}$ = 140m, then m is equal to-

(1) 68 (2) 34 (3) 33 (4) 66


Sol. (2)

Q. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series $1^{2}+2 \cdot 2^{2}+3^{2}+2 \cdot 4^{2}+5^{2}+2 \cdot 6^{2}+\ldots \ldots \ldots \quad$ If $\mathrm{B}-2 \mathrm{A}=100 \lambda,$ then $\lambda$ is equal to :

(1) 248 (2) 464 (3) 496 (4) 232


Sol. (1)


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