JEE Advanced Maths – Arithmetic Progression PYQ with Solutions

Q 1 :

Let $b_{i} > 1$ for $i = 1, 2, \dots, 101 .$ Suppose $\log_{e} b_{1}, \log_{e} b_{2}, \dots, \log_{e} b_{101}$ are in Arithmetic Progression (A.P.) with common difference $\log_{e} 2 .$ Suppose $a_{1}, a_{2}, \dots, a_{101}$ are in A.P. such that $a_{1} = b_{1}$ and $a_{51} = b_{51} .$ If $t = b_{1} + b_{2} + \dots + b_{51}$ and $s = a_{1} + a_{2} + \dots + a_{53}$ , then [2016]

$s > t and a_{101} > b_{101}$
$s > t and a_{101} < b_{101}$
$s < t and a_{101} > b_{101}$
$s < t and a_{101} < b_{101}$

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(2)

$\log_{e} b_{1}, \log_{e} b_{2}, \dots, \log_{e} b_{101} are in A.P.$

$\Rightarrow b_{1}, b_{2}, \dots, b_{101} are in G.P.$

$Also a_{1}, a_{2}, \dots, a_{101} are in A.P., where a_{1} = b_{1}$ are $a_{51} = b_{51} .$

$∴$ $b_{2}, b_{3}, \dots, b_{50} are G.M.'s and a_{2}, a_{3}, \dots, a_{50} are A.M.'s between b_{1} and b_{51} .$

$∵$ $G.M. < A.M. \Rightarrow b_{2} < a_{2}, b_{3} < a_{3}, \dots, b_{50} < a_{50}$

$∵$ $b_{1} + b_{2} + \dots + b_{51} < a_{1} + a_{2} + \dots + a_{51}$

$\Rightarrow t < s$

$Also a_{1}, a_{51}, a_{101} are in A.P. and b_{1}, b_{51}, b_{101} are in G.P.$

$∵$ $a_{1} = b_{1} and a_{51} = b_{51}$ $∴$ $b_{101} > a_{101}$

Q 2 :

In the sum of first $n$ terms of an A.P. is $c n^{2},$ then the sum of squares of these $n$ terms is [2009]

$\frac{n (4 n^{2} - 1) c^{2}}{6}$
$\frac{n (4 n^{2} + 1) c^{2}}{3}$
$\frac{n (4 n^{2} - 1) c^{2}}{3}$
$\frac{n (4 n^{2} + 1) c^{2}}{6}$

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(3)

$Given: For an A.P., S_{n} = c n^{2}$

$Then T_{n} = S_{n} - S_{n - 1} = c n^{2} - c {(n - 1)}^{2}$

$= (2 n - 1) c$

$∴$ $Sum of squares of n terms of this A.P.$

$= \sum T_{n}^{2} = \sum {(2 n - 1)}^{2} c^{2}$

$= c^{2} [4 \sum n^{2} - 4 \sum n + n]$

$= c^{2} [\frac{4 n (n + 1) (2 n + 1)}{6} - \frac{4 n (n + 1)}{2} + n]$

$= c^{2} n [\frac{2 (2 n^{2} + 3 n + 1) - 6 (n + 1) + 3}{3}]$

$= c^{2} n [\frac{4 n^{2} - 1}{3}]$

$= \frac{n (4 n^{2} - 1) c^{2}}{3}$

Q 3 :

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots,$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots,$ then $n$ equals [2001]

10
12
11
13

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(3)

$Given: 2 + 5 + 8 + \dots 2 n terms = 57 + 59 + 61 + \dots n terms$

$\Rightarrow \frac{2 n}{2} [4 + (2 n - 1) \cdot 3] = \frac{n}{2} [114 + (n - 1) \cdot 2]$

$\Rightarrow 6 n + 1 = n + 56$

$\Rightarrow 5 n = 55 \Rightarrow n = 11$

Q 4 :

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is [2015]

(9)

$\frac{S_{7}}{S_{11}} = \frac{6}{11} \Rightarrow \frac{\frac{7}{2} [2 a + 6 d]}{\frac{11}{2} [2 a + 10 d]} = \frac{6}{11} \Rightarrow a = 9 d$

$a_{7} = a + 6 d = 15 d$

$Given 130 < a_{7} < 140$

$\Rightarrow 130 < 15 d < 140 \Rightarrow d = 9$

[Since $d$ is a natural number because all terms are natural numbers.]

Q 5 :

A pack contains $n$ cards numbered from 1 to $n$ . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is $k$ , then $k - 20 =$ [2013]

(5)

$Let k, k + 1 be removed from pack.$

$∴$ $(1 + 2 + 3 + \dots + n) - (k + k + 1) = 1224$

$\frac{n (n + 1)}{2} - 2 k - 1 = 1224 \Rightarrow \frac{n (n + 1)}{2} - 2 k = 1225$

$\Rightarrow k = \frac{n (n + 1) - 2450}{4}$

$for n = 50, k = 25$

$∴$ $k - 20 = 5$

Q 6 :

Let $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100 .$ For any integer $n$ with $1 \leq n \leq 20,$ let $m = 5 n$ . If $\frac{S_{m}}{S_{n}}$ does not depend on $n,$ then $a_{2}$ is [2011]

(9)

$\frac{S_{m}}{S_{n}} = \frac{S_{5 n}}{S_{n}} = \frac{\frac{5 n}{2} [2 \times 3 + (5 n - 1) d]}{\frac{n}{2} [6 + (n - 1) d]}$ $[∵ m = 5 n]$

$= \frac{5 [(6 - d) + 5 n d]}{(6 - d) + n d},$ which will be independent of $n$ if $d = 6$ or $d = 0$

For a proper A.P., we take $d = 6$

$∴$ $a_{2} = a_{1} + d = 3 + 6 = 9$

Q 7 :

Let $S_{k}, k = 1, 2, \dots, 100,$ denote the sum of the infinite geometric series whose first term is $\frac{k - 1}{k!}$ and the common ratio is $\frac{1}{k}$ . Then the value of $\frac{100^{2}}{100!} + \sum_{k = 1}^{100}$ $| (k^{2} - 3 k + 1) S_{k} |$ is [2010]

(3)

$We know that S_{\infty} = \frac{a}{1 - r}$

$∴$ $S_{k} =$ ${\begin{matrix} \frac{\frac{k - 1}{k!}}{1 - \frac{1}{k}}, & k \neq 1 \\ 0, & k = 1 \\ \frac{1}{(k - 1)!}, & k \geq 2 \end{matrix}$

$Now \sum_{k = 1}^{100}$ $| (k^{2} - 3 k + 1) S_{k} |$ $= \sum_{k = 2}^{100}$ $| (k^{2} - 3 k + 1) |$ $\frac{1}{(k - 1)!}$

$=$ $| - 1 |$ $+ \sum_{k = 3}^{100} \frac{(k^{2} - 1) + 1 - 3 (k - 1) - 2}{(k - 1)!}$

$since k^{2} - 3 k + 1 > 0$ $\forall k \geq 3$

$= 1 + \sum_{k = 3}^{100} (\frac{1}{(k - 3)!} - \frac{1}{(k - 1)!})$

$= 1 + (1 - \frac{1}{2!}) + (\frac{1}{1!} - \frac{1}{3!}) + (\frac{1}{2!} - \frac{1}{4!}) + \dots + (\frac{1}{96!} - \frac{1}{98!}) + (\frac{1}{97!} - \frac{1}{99!})$

$= 3 - \frac{1}{98!} - \frac{1}{99!} = 3 - \frac{9900}{100!} - \frac{100}{100!} = 3 - \frac{10000}{100!} = 3 - \frac{{(100)}^{2}}{100!}$

$\Rightarrow \frac{100^{2}}{100!} + \sum_{k = 1}^{100}$ $| (k^{2} - 3 k + 1) S_{k} |$ $= 3$

Q 8 :

Let $a_{1}, a_{2}, a_{3}, \dots, a_{11}$ be real numbers satisfying $a_{1} = 15, 27 - 2 a_{2} > 0$ and $a_{k} = 2 a_{k - 1} - a_{k - 2}$ for $k = 3, 4, \dots, 11 .$ If $\frac{a_{1}^{2} + a_{2}^{2} + \dots + a_{11}^{2}}{11} = 90,$ then the value of $\frac{a_{1} + a_{2} + \dots + a_{11}}{11}$ is equal to [2010]

(0)

$Given: a_{k} = 2 a_{k - 1} - a_{k - 2}$

$\Rightarrow \frac{a_{k - 2} + a_{k}}{2} = a_{k - 1}, 3 \leq k \leq 11$

$\Rightarrow a_{1}, a_{2}, a_{3}, \dots, a_{11} are in A.P.$

$If a is the first term and d the common difference, then$

$a_{1}^{2} + a_{2}^{2} + \dots + a_{11}^{2} = 990$

$\Rightarrow 11 a^{2} + d^{2} (1^{2} + 2^{2} + \dots + 10^{2}) + 2 a d (1 + 2 + \dots + 10) = 990$

$\Rightarrow 11 a^{2} + \frac{10 \times 11 \times 21}{6} d^{2} + 2 a d \times \frac{10 \times 11}{2} = 990$

$\Rightarrow a^{2} + 35 d^{2} + 10 a d = 90$

$Since a = a_{1} = 15$

$∴$ $35 d^{2} + 150 d + 135 = 0 \Rightarrow 7 d^{2} + 30 d + 27 = 0$

$\Rightarrow (d + 3) (7 d + 9) = 0 \Rightarrow d = - 3 or - \frac{9}{7}$

$Then a_{2} = 15 - 3 = 12 or 15 - \frac{9}{7} = \frac{96}{7} > \frac{27}{2}$

$∴$ $d \neq - \frac{9}{7}$

$Hence \frac{a_{1} + a_{2} + \dots + a_{11}}{11} = \frac{\frac{11}{2} [2 \times 15 + 10 (- 3)]}{11} = 0$

Q 9 :

Let $ℝ$ denote the set of all real numbers. Let $f : ℝ \to ℝ$ be a function such that $f (x) > 0$ for all $x \in ℝ$ , and $f (x + y) = f (x) f (y)$ for all $x, y \in ℝ$ . Let the real numbers $a_{1}, a_{2}, \dots, a_{50}$ be in an arithmetic progression. If $f (a_{31}) = 64 f (a_{25})$ , and $\sum_{i = 1}^{50} f (a_{i}) = 3 (2^{25} + 1),$ then the value of $\sum_{i = 6}^{30} f (a_{i})$ is [2025]

(96)

Given $f (x + y) = f (x) \cdot f (y)$

$\Rightarrow f (x) = k^{x} (f (x) > 0, \forall x \in R)$

$∵$ $f (a_{31}) = 64 f (a_{25})$

$\Rightarrow k^{(a + 30 d)} = 64 \cdot k^{(a + 24 d)} \Rightarrow k^{6 d} = 64 \Rightarrow k^{d} = 2$

$\sum_{i = 1}^{50} f (a_{i}) = f (a_{1}) + f (a_{2}) + \dots + f (a_{50}) = k^{a} + k^{a + d} + \dots + k^{a + 49 d} = \frac{k^{a} (k^{50 d} - 1)}{k^{d} - 1}$ $(∵ k^{d} = 2)$

$= k^{a} (2^{50} - 1) = 3 (2^{25} + 1) (Given)$

$\Rightarrow k^{a} = \frac{3}{2^{25} - 1}$

$∴$ $\sum_{i = 6}^{30} f (a_{i}) = k^{a + 5 d} + k^{a + 6 d} + \dots + k^{a + 29 d}$

$= k^{a + 5 d} (\frac{k^{25 d} - 1}{k^{d} - 1}) = k^{a} . {(k^{d})}^{5} (2^{25} - 1)$

$= \frac{3}{2^{25} - 1} \cdot 2^{5} (2^{25} - 1) = 96$

Q 10 :

Let $l_{1}, l_{2}, \dots, l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_{1}$ , and let $w_{1}, w_{2}, \dots, w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_{2}$ , where $d_{1} d_{2} = 10$ . For each $i = 1, 2, \dots, 100$ , let $R_{i}$ be a rectangle with length $l_{i}$ , width $w_{i}$ and area $A_{i}$ . If $A_{51} - A_{50} = 1000$ , then the value of $A_{100} - A_{90}$ is _______. [2022]

(18900)

For A.P. $l_{1}, l_{2}, \dots, l_{100}$

$Let T_{1} = a and common difference = d_{1}$ $and similarly now for A.P. w_{1}, w_{2}, \dots, w_{100}$

$let T_{1} = b and common difference = d_{2}$

$A_{51} - A_{50} = l_{51} w_{51} - l_{50} w_{50}$

$= (a + 50 d_{1}) (b + 50 d_{2}) - (a + 49 d_{1}) (b + 49 d_{2})$

$= 50 b d_{1} + 50 a d_{2} + 2500 d_{1} d_{2} - 49 a d_{2} - 49 b d_{1} - 2401 d_{1} d_{2}$

$= b d_{1} + a d_{2} + 99 d_{1} d_{2} = 1000$

$\Rightarrow b d_{1} + a d_{2} = 1000 - 990 = 10 ...(i) (As d_{1} d_{2} = 10)$

$∴$ $A_{100} - A_{90} = l_{100} w_{100} - l_{90} w_{90}$

$= (a + 99 d_{1}) (b + 99 d_{2}) - (a + 89 d_{1}) (b + 89 d_{2})$

$= 99 b d_{1} + 99 a d_{2} + 99^{2} d_{1} d_{2} - 89 b d_{1} - 89 a d_{2} - 89^{2} d_{1} d_{2}$

$= 10 (b d_{1} + a d_{2}) + 1880 d_{1} d_{2}$

$\Rightarrow 10 (10) + 1880 (10) = 18900$

Q 11 :

Let $AP (a; d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d > 0$ . If $AP (1; 3) AP (2; 5) AP (3; 7) = AP (a; d)$ , then $a + d$ equals ______. [2019]

(157)

$AP (1, 3) : 1, 4, 7, 10, 13, \dots$

$AP (2, 5) : 2, 7, 12, 17, 22, \dots$

$AP (3, 7) : 3, 10, 17, 24, 31, \dots$

$For AP (1, 3) \cap AP (2, 5) \cap AP (3, 7)$ first term will be the minimum common value of a term.

$∴$ we need to find that minimum number which.

when divided by 7 leaves remainder $3 \to (7 m + 3)$

and when divided by 5 leaves remainder $2 \to (5 p + 2)$

and when divided by 3 leaves remainder $1 \to (3 q + 1)$

By hit and trial 52 is such number $(7 \times 7 + 3)$

$∴$ first term $' a'$ of intersection AP = 52

Also common difference $' d'$ of intersection AP

$= LCM (7, 5, 3) = 105$

$∴$ $a + d = 52 + 105 = 157$

Q 12 :

Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1, 6, 11, \dots,$ and Y be the set consisting of the first 2018 terms of the arithmetic progression $9, 16, 23, \dots$ . Then, the number of elements in the set $X \cup Y$ is _______. [2018]

(3748)

The given sequences upto 2018 terms are

$1, 6, 11, 16, \dots, 10086 and 9, 16, 23, \dots, 14128$

The common terms are

$16, 51, 86, \dots upto n terms, where T_{n} \leq 10086$

$\Rightarrow 16 + (n - 1) 35 \leq 10086$

$\Rightarrow 35 n - 19 \leq 10086$

$\Rightarrow n \leq \frac{10105}{35} = 288.7$ $∴$ $n = 288$

$∴$ $n (X \cup Y) = n (X) + n (Y) - n (X \cap Y)$

$= 2018 + 2018 - 288 = 3748$

Q 13 :

The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side? [2018]

(6)

Let the sides be $a - d, a, a + d$ where $d$ is positive. Using Pythagoras theorem,

${(a + d)}^{2} = {(a - d)}^{2} + a^{2} \Rightarrow a = 4 d$

$∴$ $Sides are 3 d, 4 d, 5 d$

$Area = 24 \Rightarrow \frac{1}{2} \times 3 d \times 4 d = 24 \Rightarrow d^{2} = 4 \Rightarrow d = 2$

$∴$ $Smallest side = 3 d = 6 .$

Q 14 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression with $a_{1} = 7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, \dots$ be such that $T_{1} = 3$ and $T_{n + 1} - T_{n} = a_{n}$ for $n \geq 1$ . Then, which of the following is/are TRUE? [2022]

$T_{20} = 1604$
$\sum_{k = 1}^{20} T_{k} = 10510$
$T_{30} = 3454$
$\sum_{k = 1}^{30} T_{k} = 35610$

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Select one or more options

(2, 3)

Given, $a_{1} = 7, d = 8$

$Hence, a_{n} = 7 + (n - 1) 8 and T_{1} = 3$

$Also T_{n + 1} = T_{n} + a_{n}$

$T_{n} = T_{n - 1} + a_{n - 1}$

$⋮$

$T_{2} = T_{1} + a_{1}$

$∴$ $T_{n + 1} = (T_{n - 1} + a_{n - 1}) + a_{n}$

$= T_{n - 2} + a_{n - 2} + a_{n - 1} + a_{n}$

$⋮$

$\Rightarrow T_{n + 1} = T_{1} + a_{1} + a_{2} + \dots + a_{n}$

$\Rightarrow T_{n + 1} = T_{1} + \frac{n}{2} [2 (7) + (n - 1) 8]$

$\Rightarrow T_{n + 1} = 3 + n (4 n + 3) ...(i)$

$Hence, for n = 19; T_{20} = 3 + (19) (79) = 1504$

$For n = 29; T_{30} = 3 + (29) (119) = 3454 \to (c)$

$\sum_{k = 1}^{20} T_{k} = 3 + \sum_{k = 2}^{20} T_{k} = 3 + \sum_{k = 1}^{19} (3 + 4 n^{2} + 3 n)$

$= 3 + 3 (19) + \frac{3 (19) (20)}{2} + \frac{4 (19) (20) (39)}{6}$

$= 3 + 10507 = 10510 \to (b)$

$And similarly$ $\sum_{k = 1}^{30} T_{k} = 3 + \sum_{k = 1}^{29} (4 n^{2} + 3 n + 3) = 35615$

Q 15 :

Let $S_{n} = \sum_{k = 1}^{4 n} {(- 1)}^{\frac{k (k + 1)}{2}} k^{2} .$ Then $S_{n}$ can take value(s) [2013]

1056
1088
1120
1332

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4

Select one or more options

(1, 4)

$S_{n} = - 1^{2} - 2^{2} + 3^{2} + 4^{2} - 5^{2} - 6^{2} + \dots$

$= (3^{2} + 7^{2} + 11^{2} + \dots) + (4^{2} + 8^{2} + 12^{2} + \dots) - (1^{2} + 5^{2} + 9^{2} + \dots) - (2^{2} + 6^{2} + 10^{2} + \dots)$

$= \sum_{r = 1}^{n} {(4 r - 1)}^{2} + \sum_{r = 1}^{n} {(4 r)}^{2} - \sum_{r = 1}^{n} {(4 r - 3)}^{2} - \sum_{r = 1}^{n} {(4 r - 2)}^{2}$

$= [\sum_{r = 1}^{n} {(4 r - 1)}^{2} - {(4 r - 3)}^{2}] + 4 [\sum_{r = 1}^{n} {(2 r)}^{2} - {(2 r - 1)}^{2}]$

$= 8 \sum_{r = 1}^{n} (2 r - 1) + 4 \sum_{r = 1}^{n} (4 r - 1)$

$= 8 [2 \frac{n (n + 1)}{2} - n] + 4 [4 \frac{n (n + 1)}{2} - n]$

$= 8 n^{2} + 8 n^{2} + 4 n = 16 n^{2} + 4 n$

$For n = 8, 16 n^{2} + 4 n = 1056$

$and for n = 9, 16 n^{2} + 4 n = 1332$

Q 16 :

Let $V_{r}$ denote the sum of first $r$ terms of an arithmetic progression (A.P.) whose first term is $r$ and the common difference is $(2 r - 1)$ . Let $T_{r} = V_{r + 1} - V_{r} - 2$ and $Q_{r} = T_{r + 1} - T_{r}$ for $r = 1, 2, \dots$ . [2007]

Q. The sum $V_{1} + V_{2} + \dots + V_{n}$ is

$\frac{1}{12} n (n + 1) (3 n^{2} - n + 1)$
$\frac{1}{12} n (n + 1) (3 n^{2} + n + 2)$
$\frac{1}{2} n (2 n^{2} - n + 1)$
$\frac{1}{3} (2 n^{3} - 2 n + 3)$

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(2)

$V_{1} + V_{2} + \dots + V_{n} = \sum_{r = 1}^{n} V_{r}$ $= \sum_{r = 1}^{n} (r^{3} - \frac{r^{2}}{2} + \frac{r}{2})$

$= \sum n^{3} - \frac{\sum n^{2}}{2} + \frac{\sum n}{2}$

$= \frac{n^{2} {(n + 1)}^{2}}{4} - \frac{n (n + 1) (2 n + 1)}{12} + \frac{n (n + 1)}{4}$

$= \frac{n (n + 1)}{4} [n (n + 1) - \frac{2 n + 1}{3} + 1]$

$= \frac{n (n + 1) (3 n^{2} + n + 2)}{12}$

Q 17 :

Let $V_{r}$ denote the sum of first $r$ terms of an arithmetic progression (A.P.) whose first term is $r$ and the common difference is $(2 r - 1)$ . Let $T_{r} = V_{r + 1} - V_{r} - 2$ and $Q_{r} = T_{r + 1} - T_{r}$ for $r = 1, 2, \dots$ . [2007]

Q. $T_{r}$ is always

an odd number
an even number
a prime number
a composite number

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(4)

$T_{r} = V_{r + 1} - V_{r} - 2$

$= [{(r + 1)}^{3} - \frac{{(r + 1)}^{2}}{2} + \frac{r + 1}{2}] - [r^{3} - \frac{r^{2}}{2} + \frac{r}{2}] - 2$

$= 3 r^{2} + 2 r + 1 = (r + 1) (3 r - 1)$

For each $r, T_{r}$ has two different factors other than 1 and itself.

$∴$ $T_{r}$ is always a composite number.

Q 18 :

Let $V_{r}$ denote the sum of first $r$ terms of an arithmetic progression (A.P.) whose first term is $r$ and the common difference is $(2 r - 1)$ . Let $T_{r} = V_{r + 1} - V_{r} - 2$ and $Q_{r} = T_{r + 1} - T_{r}$ for $r = 1, 2, \dots$ . [2007]

Q. Which one of the following is a correct statement?

$Q_{1}, Q_{2}, Q_{3}, \dots$ are in A.P. with common difference 5
$Q_{1}, Q_{2}, Q_{3}, \dots$ are in A.P. with common difference 6
$Q_{1}, Q_{2}, Q_{3}, \dots$ are in A.P. with common difference 11
$Q_{1} = Q_{2} = Q_{3} = \dots$

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(2)

$Q_{r + 1} - Q_{r} = T_{r + 2} - T_{r + 1} - (T_{r + 1} - T_{r})$

$= T_{r + 2} - 2 T_{r + 1} + T_{r}$

$= (r + 3) (3 r + 5) - 2 (r + 2) (3 r + 2) + (r + 1) (3 r - 1)$

$∵$ $Q_{r + 1} - Q_{r} = 6 (r + 1) + 5 - 6 r - 5 = 6 (constant)$

$∴$ $Q_{1}, Q_{2}, Q_{3}, \dots are in A.P. with common difference 6 .$

Topic Question Set

Q 2 :

In the sum of first $n$ terms of an A.P. is $c n^{2},$ then the sum of squares of these $n$ terms is [2009]

Q 3 :

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots,$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots,$ then $n$ equals [2001]

Q 4 :

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is [2015]

Q 5 :

A pack contains $n$ cards numbered from 1 to $n$ . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is $k$ , then $k - 20 =$ [2013]

Q 6 :

Let $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100 .$ For any integer $n$ with $1 \leq n \leq 20,$ let $m = 5 n$ . If $\frac{S_{m}}{S_{n}}$ does not depend on $n,$ then $a_{2}$ is [2011]

Q 7 :

Let $S_{k}, k = 1, 2, \dots, 100,$ denote the sum of the infinite geometric series whose first term is $\frac{k - 1}{k!}$ and the common ratio is $\frac{1}{k}$ . Then the value of $\frac{100^{2}}{100!} + \sum_{k = 1}^{100}$ $| (k^{2} - 3 k + 1) S_{k} |$ is [2010]

Q 11 :

Let $AP (a; d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d > 0$ . If $AP (1; 3) AP (2; 5) AP (3; 7) = AP (a; d)$ , then $a + d$ equals ______. [2019]

Q 12 :

Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1, 6, 11, \dots,$ and Y be the set consisting of the first 2018 terms of the arithmetic progression $9, 16, 23, \dots$ . Then, the number of elements in the set $X \cup Y$ is _______. [2018]

Q 13 :

The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side? [2018]

Q 14 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression with $a_{1} = 7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, \dots$ be such that $T_{1} = 3$ and $T_{n + 1} - T_{n} = a_{n}$ for $n \geq 1$ . Then, which of the following is/are TRUE? [2022]

Q 15 :

Let $S_{n} = \sum_{k = 1}^{4 n} {(- 1)}^{\frac{k (k + 1)}{2}} k^{2} .$ Then $S_{n}$ can take value(s) [2013]

Q 17 :

Let $V_{r}$ denote the sum of first $r$ terms of an arithmetic progression (A.P.) whose first term is $r$ and the common difference is $(2 r - 1)$ . Let $T_{r} = V_{r + 1} - V_{r} - 2$ and $Q_{r} = T_{r + 1} - T_{r}$ for $r = 1, 2, \dots$ . [2007]

Q. $T_{r}$ is always

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Topic Question Set

Q 1 : Let bi>1 for i=1,2,…,101. Suppose logeb1,logeb2,…,logeb101 are in Arithmetic Progression (A.P.) with common difference loge2. Suppose a1,a2,…,a101 are in A.P. such that a1=b1 and a51=b51. If t=b1+b2+⋯+b51 and s=a1+a2+⋯+a53, then [2016]

Q 2 : In the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is [2009]

Q 3 : If the sum of the first 2n terms of the A.P. 2,5,8,…, is equal to the sum of the first n terms of the A.P. 57,59,61,…, then n equals [2001]

Q 4 : Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is [2015]

Q 5 : A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k-20= [2013]

Q 6 : Let a1,a2,a3,…,a100 be an arithmetic progression with a1=3 and Sp=∑i=1pai, 1≤p≤100. For any integer n with 1≤n≤20, let m=5n. If SmSn does not depend on n, then a2 is [2011]

Q 7 : Let Sk, k=1,2,…,100, denote the sum of the infinite geometric series whose first term is k-1k! and the common ratio is 1k. Then the value of 1002100!+∑k=1100|(k2-3k+1)Sk| is [2010]

Q 8 : Let a1,a2,a3,…,a11 be real numbers satisfying a1=15, 27-2a2>0 and ak=2ak-1-ak-2 for k=3,4,…,11. If a12+a22+⋯+a11211=90, then the value of a1+a2+⋯+a1111 is equal to [2010]

Q 11 : Let AP(a;d) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d>0. If AP(1;3)AP(2;5)AP(3;7)=AP(a;d), then a+d equals ______. [2019]

Q 12 : Let X be the set consisting of the first 2018 terms of the arithmetic progression 1,6,11,…, and Y be the set consisting of the first 2018 terms of the arithmetic progression 9,16,23,…. Then, the number of elements in the set X∪Y is _______. [2018]

Q 13 : The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side? [2018]

Q 14 : Let a1,a2,a3,… be an arithmetic progression with a1=7 and common difference 8. Let T1,T2,T3,… be such that T1=3 and Tn+1-Tn=an for n≥1. Then, which of the following is/are TRUE? [2022]

Q 15 : Let Sn=∑k=14n(-1)k(k+1)2 k2. Then Sn can take value(s) [2013]

Q 16 : Let Vr denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r-1). Let Tr=Vr+1-Vr-2 and Qr=Tr+1-Tr for r=1,2,…. [2007] Q. The sum V1+V2+⋯+Vn is

Q 17 : Let Vr denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r-1). Let Tr=Vr+1-Vr-2 and Qr=Tr+1-Tr for r=1,2,…. [2007] Q. Tr is always

Q 18 : Let Vr denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r-1). Let Tr=Vr+1-Vr-2 and Qr=Tr+1-Tr for r=1,2,…. [2007] Q. Which one of the following is a correct statement?

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Q 2 :

In the sum of first $n$ terms of an A.P. is $c n^{2},$ then the sum of squares of these $n$ terms is [2009]

Q 3 :

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots,$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots,$ then $n$ equals [2001]

Q 4 :

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is [2015]

Q 5 :

A pack contains $n$ cards numbered from 1 to $n$ . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is $k$ , then $k - 20 =$ [2013]

Q 6 :

Let $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100 .$ For any integer $n$ with $1 \leq n \leq 20,$ let $m = 5 n$ . If $\frac{S_{m}}{S_{n}}$ does not depend on $n,$ then $a_{2}$ is [2011]

Q 7 :

Let $S_{k}, k = 1, 2, \dots, 100,$ denote the sum of the infinite geometric series whose first term is $\frac{k - 1}{k!}$ and the common ratio is $\frac{1}{k}$ . Then the value of $\frac{100^{2}}{100!} + \sum_{k = 1}^{100}$ $| (k^{2} - 3 k + 1) S_{k} |$ is [2010]

Q 11 :

Let $AP (a; d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d > 0$ . If $AP (1; 3) AP (2; 5) AP (3; 7) = AP (a; d)$ , then $a + d$ equals ______. [2019]

Q 12 :

Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1, 6, 11, \dots,$ and Y be the set consisting of the first 2018 terms of the arithmetic progression $9, 16, 23, \dots$ . Then, the number of elements in the set $X \cup Y$ is _______. [2018]

Q 13 :

The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side? [2018]

Q 14 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression with $a_{1} = 7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, \dots$ be such that $T_{1} = 3$ and $T_{n + 1} - T_{n} = a_{n}$ for $n \geq 1$ . Then, which of the following is/are TRUE? [2022]

Q 15 :

Let $S_{n} = \sum_{k = 1}^{4 n} {(- 1)}^{\frac{k (k + 1)}{2}} k^{2} .$ Then $S_{n}$ can take value(s) [2013]

Q 17 :

Let $V_{r}$ denote the sum of first $r$ terms of an arithmetic progression (A.P.) whose first term is $r$ and the common difference is $(2 r - 1)$ . Let $T_{r} = V_{r + 1} - V_{r} - 2$ and $Q_{r} = T_{r + 1} - T_{r}$ for $r = 1, 2, \dots$ . [2007]

Q. $T_{r}$ is always