Sequences and Series jee mains questions | Sequences and Series Previous Year Question

Q 1 :

The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + ... is [2023]

3450
3420
3520
3250

1

2

3

4

(3)

$Let S = 5 + 11 + 19 + 29 + \dots + T_{n}$

$\Rightarrow S = 5 + 11 + 19 + \dots + T_{n - 1} + T_{n}$

$Subtracting above two equations, we get:$

$0 = 5 + {6 + 8 + 10 + 12 + \dots + (n - 1) terms} - T_{n}$

$\Rightarrow T_{n} = 5 + 2 (3 + 4 + 5 + \dots upto (n - 1) terms)$

$= 5 + 2 (1 + 2 + 3 + \dots upto (n + 1) terms - 1 - 2)$

$\Rightarrow T_{n} = 5 + 2 \cdot \frac{(n + 1) (n + 2)}{2} - 6 = n^{2} + 3 n + 2 - 1$

$\Rightarrow T_{n} = n^{2} + 3 n + 1$ $∴$ $S_{n} = \sum n^{2} + 3 \sum n + \sum 1$

$\Rightarrow \frac{n (n + 1) (2 n + 1)}{6} + \frac{3 (n + 1) n}{2} + n$

$S_{20} = \frac{20 \times 21 \times 41}{6} + \frac{3 \times 20 \times 21}{2} + 20 = 3520$

Q 2 :

If gcd (m, n) = 1 and $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . + {(2021)}^{2} - {(2022)}^{2} + {(2023)}^{2} = 1012$ $m^{2} n$ , then $m^{2} - n^{2}$ is equal to [2023]

200
180
220
240

1

2

3

4

(4)

$We have, (1^{2} - 2^{2}) + (3^{2} - 4^{2}) + \dots + ({(2021)}^{2} - {(2022)}^{2}) + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow \underset{It is an A.P.}{\underset{⏟}{{- 3 - 7 - 11 - \dots - 4043}}} + {(2023)}^{2} = 1012 m^{2} n$

$Now, - 4043 = - 3 + (n - 1) (- 4) \Rightarrow n = 1011$

$So, - (3 + 7 + \dots + 4043) + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow - \frac{1011}{2} [3 + 4043] + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow - 2023 \times 1011 + {(2023)}^{2} = 1012 m^{2} n$

$\Rightarrow 2023 (2023 - 1011) = 1012 m^{2} n$

$\Rightarrow 2023 \times 1012 = 1012 m^{2} n \Rightarrow m^{2} n = 2023 = 7 \times 17^{2}$

$\Rightarrow m = 17$ and $n = 7$

$∴$ $m^{2} - n^{2} = 17^{2} - 7^{2} = 289 - 49 = 240$

Q 3 :

If $S_{n} = 4 + 11 + 21 + 34 + 50 + . . .$ to $n$ terms, then $\frac{1}{60} (S_{29} - S_{9})$ is equal to [2023]

223
220
226
227

1

2

3

4

(1)

$Given, S_{n} = 4 + 11 + 21 + 34 + \dots up to n terms \dots (i)$

$Also, S_{n} = 4 + 11 + 21 + \dots + \dots up to n terms \dots (i i)$

$Subtracting (ii) from (i), we get:$

$0 = [4 + 7 + 10 + 13 + \dots + \dots] - a_{n}$

$∴$ $a_{n} = \frac{n}{2} [2 (4) + (n - 1) (3)] = \frac{n}{2} [3 n + 5] = \frac{3 n^{2} + 5 n}{2}$

$Now, S_{n} = \sum a_{n} = \frac{3}{2} \sum n^{2} + \frac{5}{2} \sum n$

$= \frac{3}{2} \frac{n \cdot (n + 1) (2 n + 1)}{6} + \frac{5}{2} \frac{n (n + 1)}{2} = \frac{n (n + 1)}{4} [(2 n + 1) + 5]$

$= \frac{n (n + 1) (2 n + 6)}{4} = \frac{n (n + 1) (n + 3)}{2}$

$Now, \frac{1}{60} [S_{29} - S_{9}] = \frac{1}{60} [\frac{(29) (30) (32)}{2} - \frac{(9) (10) (12)}{2}]$

$= \frac{1}{60} [13920 - 540] = \frac{13380}{60} = 223$

Q 4 :

Let $S_{1}, S_{2}, S_{3}, . . . . . . . . . . . . S_{10}$ respectively be the sum to 12 terms of 10 A.P. whose first terms are 1, 2, 3, ....... 10 and the common difference are 1, 3, 5, .........., 19 respectively. Then $\sum_{i = 1}^{10} S_{i}$ is equal to [2023]

7220
7380
7260
7360

1

2

3

4

(3)

$s_{1} = 1 + 2 + 3 + \dots + 12 \dots (i)$

$s_{2} = 2 + 5 + 8 + \dots up to 12 terms \dots (i i)$

$s_{3} = 3 + 8 + 13 + \dots up to 12 terms \dots (i i i)$

$∵$ $s_{10} = 10 + 29 + 48 + \dots up to 12 terms \dots (i v)$

$From (i), s_{1} = \frac{12 (13)}{2} = 78$

$From (ii), s_{2} = \frac{12}{2} [2 (2) + 11 \times 3] = 6 [4 + 33] = 222$

$From (iii), s_{3} = \frac{12}{2} [2 (3) + 11 \times 5] = 6 [6 + 55] = 366$

$From (iv), s_{10} = \frac{12}{2} [2 (10) + 11 \times 19] = 6 [20 + 209] = 1374$

$Thus, s_{k} = 6 [2 k + 11 (2 k - 1)] = 6 (2 k + 22 k - 11) = 144 k - 66$

$∴$ $\sum_{k = 1}^{10} s_{k} = 144 \sum_{k = 1}^{10} k - 66 \sum_{k = 1}^{10} 1 = 144 (\frac{10 \times 11}{2}) - 66 (10)$

$= 7920 - 660 = 7260$

Q 5 :

If $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + . . . + a_{25}$ is equal to [2023]

52/147
50/141
51/144
49/138

1

2

3

4

(2)

$a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15} = \frac{- 2}{(2 n - 3) (2 n - 5)}$

$a_{n} = \frac{1}{2 n - 3} - \frac{1}{2 n - 5}$

$For n = 1, a_{1} = - 1 + \frac{1}{3}; n = 2, a_{2} = 1 + 1; n = 3, a_{3} = \frac{1}{3} - 1$

$n = 4, a_{4} = \frac{1}{5} - \frac{1}{3}; n = 5, a_{5} = \frac{1}{7} - \frac{1}{5}$

$n = 25, a_{25} = \frac{1}{47} - \frac{1}{45}$

$Now, a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + \dots + a_{25}$

$= (- 1 + \frac{1}{3}) + (1 + 1) + (\frac{1}{3} - 1) + (\frac{1}{5} - \frac{1}{3}) + (\frac{1}{7} - \frac{1}{5}) + \dots + (\frac{1}{47} - \frac{1}{45})$

$= - 1 + \frac{1}{3} + 2 - 1 + \frac{1}{47} = \frac{50}{141}$

Q 6 :

Let $a_{1}, a_{2}, a_{3}, . . . .$ be an A.P. If $a_{7} = 3,$ the product $a_{1} a_{4}$ is minimum and the sum of its first $n$ terms is zero, then $n! - 4 a_{n (n + 2)}$ is equal to [2023]

381/4
9
33/4
24

1

2

3

4

(4)

$Given, a_{7} = 3$

$\Rightarrow a + 6 d = 3 \dots (1)$

$Let Z = a_{1} a_{4} = a (a + 3 d) = (a + 6 d - 6 d) (a + 6 d - 3 d)$

$= (3 - 6 d) (3 - 3 d) = 18 d^{2} - 27 d + 9$

$Differentiating with respect to d,$

$\Rightarrow 36 d - 27 = 0 \Rightarrow d = \frac{3}{4}$ , $From (1), a = \frac{- 3}{2} (Z is minimum)$

$Now, S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [- 3 + (n - 1) \frac{3}{4}] = 0$

$n (3 n - 15) = 0 \Rightarrow n = 5$

$Now, n! - 4 a_{n (n + 2)} = 5! - 4 a_{35} = 120 - 4 (a_{35})$

$= 120 - 4 (a + (35 - 1) d) = 120 - 4 (\frac{- 3}{2} + 34 (\frac{3}{4}))$

$= 120 - 4 (\frac{- 6 + 102}{4}) = 120 - 96 = 24$

Q 7 :

The sum of all those terms, of the arithmetic progression 3, 8, 13,..., 373, which are not divisible by 3, is equal to __________ . [2023]

(9525)

$The given A.P. is 3, 8, 13, \dots, 373$

$T_{n} = a + (n - 1) d$

$373 = 3 + (n - 1) 5$ $\Rightarrow n = 75$

$S_{n} = \frac{75}{2} [3 + 373] = 14100$

$Numbers which are divisible by 3 are 3, 18, \dots, 363 .$

$T'_{n} = 3 + (n' - 1) 15$

$\Rightarrow 363 = 3 + 15 n' - 15 \Rightarrow 15 n' = 375 \Rightarrow n' = 25$

$S'_{n} = \frac{25}{2} [3 + 363] = 4575$

$∴$ $Required sum = 14100 - 4575 = 9525$

Q 8 :

Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed? [2023]

(1260)

There are only two ways to write digits $a, b, c$ be in A.P.

$\begin{matrix} I^{st} & a b c \\ = & {}^{2}C_{1} \\ {II}^{nd} & c b a \end{matrix}$

-------------------------------------

Here, 9 places to put $a b c$ or $b c a$ but there will be only 7 possible ways A.P. to choose three consecutive numbers are i.e.

$123, 234, 345, 456, 567, 678, 789 = {}^{7}C_{1}$

Now, we have left only 6 places where $a, b, c$ are three such that three consecutive digits are in A.P.

$\Rightarrow \frac{6!}{2! 2! 2!}$

Required number $= {}^{2}C_{1} \cdot {}^{7}C_{1} \cdot \frac{6!}{2! 2! 2!}$

$= \frac{2 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2 \times 2 \times 2} = 1260$

Q 9 :

Let $a_{1} = 8, a_{2}, a_{3}, . . ., a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is _________ . [2023]

(754)

Sum of first four terms $= \frac{4}{2} [16 + 3 d]$

$\Rightarrow 50 = 32 + 6 d \Rightarrow d = 3$

Now, sum of last four terms = 170

$\Rightarrow \frac{4}{2} [2 (8 + (n - 4) d) + 3 d] = 170$ $[∵ last four terms are a_{n - 3}, a_{n - 2}, a_{n - 1}, a_{n}]$

$\Rightarrow 2 [2 (8 + (n - 4) \cdot 3) + 9] = 170$

$\Rightarrow 16 + 6 (n - 4) + 9 = 85$

$\Rightarrow 6 (n - 4) = 60 \Rightarrow n - 4 = 10 \Rightarrow n = 14$

$∴$ $Middle terms are 7th term and 8th term.$

$∴$ $T_{7} = 8 + 6 d = 8 + 18 = 26, T_{8} = 8 + 7 d = 8 + 21 = 29$

Now, $T_{7} \times T_{8} = 26 \times 29 = 754$

Q 10 :

The sum of the common terms of the following three arithmetic progressions. 3, 7, 11, 15, .., 399,

2, 5, 8, 11, ..., 359 and 2, 7, 12, 17, ..., 197, is equal to ________ . [2023]

(321)

$3, 7, 11, 15, \dots, 399$ are in A.P.

$So, common difference d_{1} = 7 - 3 = 4$

$2, 5, 8, 11, \dots, 359 are in A.P.$

$Then, d_{2} = 5 - 2 = 3; 2, 7, 12, 17, \dots, 197 are in A.P.$

$d_{3} = 7 - 2 = 5$

$L.C.M. of d_{1}, d_{2}, d_{3} = L.C.M. of (4, 3, 5) = 60$

$On expanding all A.P.s, common terms are 47, 107, 167$

$∴$ $Sum = 47 + 107 + 167 = 321$

Q 11 :

Let $A_{1}, A_{2}, A_{3}$ be the three A.P. with the same common difference $d$ and having their first terms as A, A + 1, A + 2, respectively. Let $a, b, c$ be the $7^{t h}, 9^{t h}, 17^{t h}$ terms of $A_{1}, A_{2}, A_{3},$ respectively such that $| \begin{matrix} a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1 \end{matrix} |$ $+ 70 = 0 .$ If $a = 29,$ then the sum of first 20 terms of an A.P. whose first term is $c - a - b$ and common difference is $\frac{d}{12}$ , is equal to ________ . [2023]

(495)

Given $| \begin{matrix} a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1 \end{matrix} |$ $+ 70 = 0$

$\Rightarrow$ $| \begin{matrix} A + 6 d & 7 & 1 \\ 2 (A + 1 + 8 d) & 17 & 1 \\ A + 2 + 16 d & 17 & 1 \end{matrix} |$ $+ 70 = 0$

$\Rightarrow (A + 6 d) (0) - 7 [(2 A + 2 + 16 d) - (A - 2 - 16 d)] + 1 [17 (2 A + 2 + 16 d - A - 2 - 16 d)] + 70 = 0$

$\Rightarrow - 7 (A) + 1 [17 (A)] + 70 = 0$

$\Rightarrow 10 A + 70 = 0 \Rightarrow A = - 7$

$\Rightarrow A + 6 d = 29 \Rightarrow - 7 + 6 d = 29 \Rightarrow 6 d = 36 \Rightarrow d = 6$

$c = A + 2 + 16 d = - 7 + 2 + 16 (6) = 91$

$b = A + 1 + 8 d = - 7 + 1 + 8 (6) = 42$

$\Rightarrow c - a - b = 91 - 29 - 42 = 20$

$S_{20} = \frac{20}{2} [2 \times 20 + 19 \times \frac{6}{12}] = 495$

Q 12 :

The $8^{t h}$ common term of the series

$S_{1} = 3 + 7 + 11 + 15 + 19 + . . .,$

$S_{2} = 1 + 6 + 11 + 16 + 21 + . . . . .$

is _________ . [2023]

(151)

$S_{1} = 3 + 7 + 11 + 15 + 19 + \dots$

$a_{1} = 3, d_{1} = 4$

$S_{2} = 1 + 6 + 11 + 16 + 21 + \dots$

$a_{2} = 1, d_{2} = 5$

First common term, $a$ = 11

$Common difference, d = LCM (d_{1}, d_{2}) = LCM (4, 5) = 20$

$8^{t h}$ common term of the series

$a_{8} = a + 7 d = 11 + 7 \times 20 = 151$

Q 13 :

Let $a_{1}, a_{2}, . . . ., a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18,$ then $12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + . . . + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$ is equal to _______ . [2023]

(8)

We have, $a_{1}, a_{2}, \dots, a_{n}$ are in A.P.

$a_{5} = 2 a_{7}$

$\Rightarrow a + 4 d = 2 (a + 6 d) \Rightarrow a + 8 d = 0 ...(1)$

$and a_{11} = 18 \Rightarrow a + 10 d = 18 ...(2)$

$On solving (1) and (2), we get 2 d = 18 \Rightarrow d = 9 \Rightarrow a = - 72$

$a_{10} = a + 9 d = - 72 + 81 = 9$

$a_{18} = a + 17 d = - 72 + 153 = 81$

$Now, 12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \dots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}}}{{(\sqrt{a_{10}})}^{2} - {(\sqrt{a_{11}})}^{2}} + \frac{\sqrt{a_{11}} - \sqrt{a_{12}}}{{(\sqrt{a_{11}})}^{2} - {(\sqrt{a_{12}})}^{2}} + \dots + \frac{\sqrt{a_{17}} - \sqrt{a_{18}}}{{(\sqrt{a_{17}})}^{2} - {(\sqrt{a_{18}})}^{2}})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}}}{a + 9 d - a - 10 d} + \frac{\sqrt{a_{11}} - \sqrt{a_{12}}}{a + 10 d - a - 11 d} + \dots + \frac{\sqrt{a_{17}} - \sqrt{a_{18}}}{a + 16 d - a - 17 d})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}} + \sqrt{a_{11}} - \sqrt{a_{12}} + \dots + \sqrt{a_{17}} - \sqrt{a_{18}}}{- d})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{18}}}{- 9}) = \frac{12 \times (\sqrt{9} - \sqrt{81})}{- 9} = \frac{12 \times (3 - 9)}{- 9} = 8$

Q 14 :

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to _________ . [2023]

(710)

$Number divisible by 3,1002,1005,…,2799$

$T_{n} = 1002 + 3 (n - 1) = 2799 \Rightarrow n = 600$

$Number divisible by 11 are 164$

$Number divisible by 33 are 54$

$Total number = 600 + 164 - 54 = 710$

Q 15 :

For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]

4
10
8
16

1

2

3

4

(2)

We have, $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are in A.P.

$\Rightarrow 2 \frac{K}{2} = (4^{1 + x} + 4^{1 - x}) + (16^{x} + 16^{- x})$

$\Rightarrow K = 4 \cdot 4^{x} + \frac{4}{4^{x}} + 4^{2 x} + \frac{1}{4^{2 x}}$

$= 4 (4^{x} + \frac{1}{4^{x}}) + (4^{2 x} + \frac{1}{4^{2 x}})$ $\geq 4 \cdot 2 + 2$ $[∵ A . M . \geq G . M .]$

$= 10 \Rightarrow K \geq 10$

So, least value of $K$ is 10

Q 16 :

A software company sets up $m$ number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $m$ is equal to: [2024]

160
180
150
125

1

2

3

4

(3)

Let the work done by each computer $= k$

$∴$ Total work $= 17$ $m k$

Work done on 1 day by $m$ computers $= m k$

Work done on day 2 by $m - 4$ computers $= (m - 4) k$

Work done on day 3 by $m - 8$ computers $= (m - 8) k$

Work done on day 25 $= (m - (24 \times 4)) k$

$= (m - 96) k$

$∴$ $m k + (m - 4) k + \dots + (m - 96) k = 17 m k$

$\Rightarrow 25 m - (4 + 8 + \dots + 96) = 17 m$

$\Rightarrow 8 m = \frac{24}{2} (4 + 96) \Rightarrow m = 150$

Q 17 :

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is 15 : 7, then $S_{15} - S_{5}$ is equal to: [2024]

800
890
790
690

1

2

3

4

(3)

We have, $S_{10} = 390$

$\Rightarrow 5 (2 a + 9 d) = 390 \Rightarrow 2 a + 9 d = 78$ ...(i)

Also, $\frac{a_{10}}{a_{5}} = \frac{15}{7} \Rightarrow \frac{a + 9 d}{a + 4 d} = \frac{15}{7}$

$\Rightarrow 7 a + 63 d = 15 a + 60 d \Rightarrow 8 a = 3 d$

$\Rightarrow a = \frac{3}{8} d$ ...(ii)

From (i) & (ii), we get $d = 8$ and $a = 3$

Now, $S_{15} - S_{5} = \frac{15}{2} [6 + 112] - \frac{5}{2} [6 + 32] = 790$

Q 18 :

The number of common terms in the progressions
4, 9, 14, 19, ........…, up to 25th term and 3, 6, 9, 12, ....…, up to 37th term is: [2024]

8
7
5
9

1

2

3

4

(2)

4, 9, 14, 19, ... are in A.P. with $d_{1} = 5, n_{1} = 25$

3, 6, 9, 12, ... are in A.P. with $d_{2} = 3, n_{2} = 37$

L.C.M. $(d_{1}, d_{2}) = 15$

Common terms = 9, 24, 39, 54, 69, 84, 99

Q 19 :

The $20^{th}$ term from the end of the progression $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is: [2024]

$- 118$
$- 100$
$- 110$
$- 115$

1

2

3

4

(4)

Given, $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is in A.P. with first term $(a) = - 129 \frac{1}{4}$

Common difference $(d) = 20 - 19 \frac{1}{4} = \frac{80 - 77}{4} = \frac{3}{4}$

$20^{th}$ term from the end $= a_{20} = a + (20 - 1) d$

$= - 129 \frac{1}{4} + 19 \times \frac{3}{4} = \frac{- 517}{4} + \frac{57}{4} = \frac{- 460}{4} = - 115$

$∴$ $a_{20} = - 115$

Q 20 :

In an A.P., the sixth term $a_{6} = 2 .$ If the product $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P. is equal to [2024]

$\frac{8}{5}$
$\frac{5}{8}$
$\frac{3}{2}$
$\frac{2}{3}$

1

2

3

4

(1)

We have, $a_{6} = 2 \Rightarrow a + 5 d = 2 \Rightarrow d = \frac{2 - a}{5}$

Let $a_{1} a_{4} a_{5} = λ$

$\Rightarrow λ = a (a + 3 d) (a + 4 d)$ $= a (a^{2} + 7 a d + 12 d^{2})$ $= a^{3} + 7 a^{2} d + 12 a d^{2}$

$= a^{3} + 7 a^{2} (\frac{2 - a}{5}) + 12 a {(\frac{2 - a}{5})}^{2}$ $[∵ d = \frac{2 - a}{5}]$

$= a^{3} + \frac{14}{5} a^{2} - \frac{7}{5} a^{3} + \frac{12 a}{25} (4 - 4 a + a^{2})$

$= a^{3} + \frac{14}{5} a^{2} - \frac{7}{5} a^{3} + \frac{48}{25} a - \frac{48}{25} a^{2} + \frac{12}{25} a^{3}$

$= \frac{2}{25} a^{3} + \frac{22}{25} a^{2} + \frac{48}{25} a = \frac{2}{25} (a^{3} + 11 a^{2} + 24 a)$

$∴$ $\frac{d λ}{d a} = \frac{2}{25} (3 a^{2} + 22 a + 24)$

$λ$ to be greatest

$∴$ $\frac{d λ}{d a} = 0 \Rightarrow 3 a^{2} + 22 a + 24 = 0 \Rightarrow a = - 6, - \frac{4}{3}$

$\frac{d^{2} λ}{d a^{2}} = \frac{2}{25} (6 a + 22), For a = - 6, \frac{d^{2} λ}{d a^{2}} < 0$

So, $λ$ is maximum

For $a = \frac{- 4}{3}, \frac{d^{2} λ}{d a^{2}} > 0, λ$ is minimum $∴$ $d = \frac{2 - (- 6)}{5} = \frac{8}{5}$

Q 21 :

If $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P. and $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are also in an A.P., then $a : b : c$ is equal to [2024]

16 : 4 : 1
9 : 6 : 4
6 : 3 : 2
25 : 10 : 4

1

2

3

4

(2)

We have, $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P.

$\Rightarrow 2 \log_{e} b = \log_{e} a + \log_{e} c \Rightarrow b^{2} = a c$ ...(i)

Also, $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are in an A.P.

$\Rightarrow 2 (\log_{e} 2 b - \log_{e} 3 c) = \log_{e} a - \log_{e} 2 b + \log_{e} 3 c - \log_{e} a$

$\Rightarrow \log_{e} {(\frac{2 b}{3 c})}^{2} = \log_{e} (\frac{3 c}{2 b}) \Rightarrow \frac{4 b^{2}}{9 c^{2}} = \frac{3 c}{2 b}$

$\Rightarrow \frac{4 b^{2}}{9 c^{2}} = \frac{3 c}{2 b}$ ...(ii)

From (i) and (ii), we have

$a = \frac{9 c^{2}}{4 c} \Rightarrow a = \frac{9}{4} c$ $∴$ $a : b : c = \frac{9}{4} : \frac{3}{2} : 1 = 9 : 6 : 4$

Q 22 :

Let $S_{n}$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145,$ then $S_{15} - S_{5}$ is [2024]

390
405
410
395

1

2

3

4

(4)

Given, $S_{20} = 790$

$\Rightarrow \frac{20}{2} (2 a + 19 d) = 790 \Rightarrow 2 a + 19 d = 79$ ...(i)

Given, $S_{10} = 145$

$\frac{10}{2} (2 a + 9 d) = 145 \Rightarrow 2 a + 9 d = 29$ ...(ii)

From (i) and (ii), we get $10 d = 50 \Rightarrow d = 5$

From (i), we get $a = \frac{79 - 95}{2} = - 8$

So, $S_{15} - S_{5} = \frac{15}{2} [- 16 + 70] - \frac{5}{2} [- 16 + 20] = 395$

Q 23 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be in an arithmetic progression of positive terms.

Let $A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2} .$

If $A_{3} = - 153, A_{5} = - 435$ and $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 66,$ then $a_{17} - A_{7}$ is equal to _______ . [2024]

(910)

$a_{1}, a_{2}, a_{3}, \dots$ is an A.P.

Let $d$ be the common difference

$∴$ $a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{2 k} - a_{2 k - 1} = d$

Now, $a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}$

$= (a_{1} - a_{2}) (a_{1} + a_{2}) + \dots + (a_{2 k - 1} - a_{2 k}) (a_{2 k - 1} + a_{2 k})$

$= - d [a_{1} + a_{2} + \dots + a_{2 k - 1} + a_{2 k}]$

$= - d \cdot [\frac{2 k}{2} [a_{1} + a_{2 k}]] = - d k (a_{1} + a_{2 k})$

$∴$ $A_{k} = - d k (a_{1} + a_{2 k})$

So, $A_{3} = - 3 d (a_{1} + a_{6}) = - 153$

$\Rightarrow - 3 d (a_{1} + a_{1} + 5 d) = - 153$

$\Rightarrow - 3 d (2 a_{1} + 5 d) = - 153 \Rightarrow 2 a_{1} + 5 d = \frac{51}{d}$ ...(i)

Similarly $A_{5} = - 435$

$\Rightarrow - 5 d (2 a_{1} + 9 d) = - 435 \Rightarrow 2 a_{1} + 9 d = \frac{87}{d}$ ...(ii)

Solving (i) and (ii) we get $4 d = \frac{36}{d} \Rightarrow d^{2} = 9 \Rightarrow d = 3$ [ $∵$ Given A.P. is a progression of positive terms]

$\Rightarrow a_{1} = 1$

Now, $a_{17} = a_{1} + 16 d = 1 + 48 = 49$

and $A_{7} = - 3 \times 7 (a_{1} + a_{14})$

$= - 21 (1 + 1 + 13 \times 3) = - 21 (41) = - 861$

$∴$ $a_{17} - A_{7} = 49 + 861 = 910$

Q 24 :

An arithmetic progression is written in the following way

2

5 8

11 14 17

20 23 26 29

--------------------------------------------------------------------

---------------------------------------------------------------------------

The sum of all the terms of the $10^{th}$ row is ______. [2024]

(1505)

First term of the $10^{th}$ row is $10^{th}$ term of sequence 2, 5, 11, 20, 32, ...

$S = 2 + 5 + 11 + 20 + 32 + \dots + T_{10}$

$S =$ $2 + 5 + 11 + 20 + 32 + \dots + T_{9} + T_{10}$

On subtracting, we get

$0 = 2 + \underset{9 terms}{\underset{⏟}{3 + 6 + 9 + 12 + \dots} - T_{10}}$

$\Rightarrow T_{10} = 2 + \underset{9 terms}{\underset{⏟}{3 + 6 + 9 + 12 + \dots}}$

$= 2 + \frac{9}{2} [6 + 8 \times 3] = 2 + 135 = 137$

Terms in $10^{th}$ row with common difference 3 are 137, 140, 143, .....

$S_{10} = \frac{10}{2} [2 \times 137 + 9 \times 3] = 5 \times 301 = 1505$

Q 25 :

Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

(6699)

A.P. : 3, 7, 11, 15, ....., 403

$d_{1} = 4$

A.P. : 2, 5, 8, 11, ....., 404

$d_{2} = 3$

L.C.M. $(d_{1}, d_{2}) = 12$

Common terms = 11, 23, 35 ... 395

$a = 11, d = 12, a_{n} = 395$

$a_{n} = a + (n - 1) d \Rightarrow 395 = 11 + (n - 1) (12)$

$\Rightarrow \frac{384}{12} = (n - 1) \Rightarrow n = 33$

$S = \frac{33}{2} (11 + 395) = \frac{33}{2} (406) = 6699$

Q 26 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then n is : [2025]

17
18
10
11

1

2

3

4

(4)

Let First term, $a_{1} = a$ , common difference = d

Given, $\sum_{k = 1}^{12} a_{2 k - 1} = \frac{- 72}{5} a, a_{1} \neq 0$

$\Rightarrow a_{1} + a_{3} + a_{5} + . . . . + a_{23} = - \frac{72}{5} a$

$\Rightarrow a + (a + 2 d) + (a + 4 d) + . . . . + (a + 22 d) = \frac{- 72}{5} a$

Now, $\sum_{k = 1}^{12} a_{2 k - 1}$ is an AP with common difference as 2d.

$∴ \frac{12}{2} [2 a + 11 \times 2 d] = - \frac{72}{5} a$

$\Rightarrow 12 a + 132 d = - \frac{72}{5} a$

$\Rightarrow 132 a + 132 \times 5 d = 0$

$\Rightarrow a = - 5 d$

Also, given $\sum_{k = 1}^{n} a_{k} = 0 \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = 0$

$\Rightarrow n (- 10 d + n d - d) = 0 \Rightarrow n d (n - 11) = 0 \Rightarrow n = 11$ .

Q 27 :

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is : [2025]

10
8
4
6

1

2

3

4

(2)

Let n be the number of terms of an A.P., $a$ be the first term and $d$ be the common difference.

Let n = 2m, so terms are a, a + d, a + 2d, a + (2m – 1)d

Odd terms = a, a + 2d, a + 4d, ...., a + (2m – 2)d

Even terms = a + d, a + 3d, a + 5d, .... a + (2m – 1)d

Now, $S_{even} - S_{odd}$ = (a + d – a) + (a + 3d – a – 2d) + .... + (a + (2m – 1)d – a – (2m – 2)d)

$\Rightarrow 30 - 24 = m d \Rightarrow m d = 6$ ... (i)

Now, $a + (2 m - 1) d - a = \frac{21}{2} \Rightarrow 2 m d - d = \frac{21}{2}$

$\Rightarrow 12 - d = \frac{21}{2} \Rightarrow d = 12 - \frac{21}{2} = \frac{3}{2}$ [From (i)]

Since, $m d = 6 \Rightarrow m \times \frac{3}{2} = 6 \Rightarrow m = 4$

Now, $n = 2 m = 2 \times 4 = 8$

So, number of terms = 8.

Q 28 :

The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

7240
8124
7130
6982

1

2

3

4

(1)

Let $S_{n}$ = 1 + 3 + 11 + 25 + 45 + 71 + ... + $T_{n}$

Here, series of differences i.e., (3 – 1), (11 – 3), (25 – 11) ..... i.e., 2, 8, 14, .... is in A.P.

If the second order differences of a square are in A.P. then general term is given by

$T_{n} = a n^{2} + b n + c$

Put n = 1, 2, 3, we get

$T_{1}$ = 1 = a + b + c ... (i)

$T_{2}$ = 3 = 4a + 2b + c ... (ii)

$T_{3}$ = 11 = 9a + 3b + c ... (iii)

On solving equation (i), (ii) and (iii), we get the general term of given series as

$T_{n} = 3 n^{2} - 7 n + 5$

Hence, $\sum_{n = 1}^{20} (3 n^{2} - 7 n + 5)$

$= 3 (\frac{20 \cdot 21 \cdot 41}{6}) - 7 (\frac{20 \cdot 21}{2}) + 5 (20) = 7240$ .

Q 29 :

Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n (A \cup B)$ is [2025]

4027
3761
4003
3814

1

2

3

4

(2)

Given A = {1, 6, 11, 16, ...}, B = {9, 16, 23, 30, ...} are two A.P. and n(A) = 2025, n(B) = 2025

$2025^{th} term of set A = 1 + 2024 \times 5 = 10121$

$2025^{th} term of set B = 9 + 2024 \times 7 = 14177$

$\Rightarrow$ A = {1, 6, 11, 16, ..., 10121} and B = {9, 16, 23, ...,14177}

Now $A \cap B$ = 16, 51, 86, .... which is an A.P. with a = 16, d = 35

then number of terms in $(A \cap B)$ = 16 + (n – 1) 35 < 10121

$\Rightarrow$ n – 1 < 288.7

$\Rightarrow$ n < 289.7

$\Rightarrow$ n = 289 [ $∵$ n is natural number]

$\Rightarrow n (A \cup B) = n (A) + n (B) - n (A \cap B)$

= 2025 + 2025 – 289 = 3761.

Q 30 :

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP's in A and B respectively such that D = d + 3, d > 0. If $\frac{p + q}{p - q} = \frac{19}{5}$ , then p – q is equal to [2025]

450
630
540
600

1

2

3

4

(3)

Let A = (a – d, a, a + d) and B = (b – D, b, b + D)

For set A, Sum = 36, product = p

For set B, Sum = 36, product = q

$\Rightarrow a = 12, p = 12 (144 - d^{2})$

$\Rightarrow b = 12, q = 12 (144 - D^{2})$

$= 12 (144 - {(d + 3)}^{2}) = 12 (144 - d^{2} - 6 d - 9)$

Now, $\frac{p + q}{p - q} = \frac{19}{5} \Rightarrow \frac{p}{q} = \frac{12}{7}$

$\Rightarrow \frac{144 - d^{2}}{144 - d^{2} - 6 d - 9} = \frac{12}{7}$

$\Rightarrow 5 d^{2} + 72 d - 612 = 0$

$\Rightarrow d = \frac{- 72 \pm \sqrt{{(132)}^{2}}}{10}$

$\Rightarrow d = 6 and D = 9$ [ $∵ d > 0$ ]

$∴ p - q = 12 (D^{2} - d^{2}) = 12 (9^{2} - 6^{2}) = 540$ .

Topic Question Set

Q 1 :

The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + ... is [2023]

Q 2 :

If gcd (m, n) = 1 and $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . + {(2021)}^{2} - {(2022)}^{2} + {(2023)}^{2} = 1012$ $m^{2} n$ , then $m^{2} - n^{2}$ is equal to [2023]

Q 3 :

If $S_{n} = 4 + 11 + 21 + 34 + 50 + . . .$ to $n$ terms, then $\frac{1}{60} (S_{29} - S_{9})$ is equal to [2023]

Q 4 :

Let $S_{1}, S_{2}, S_{3}, . . . . . . . . . . . . S_{10}$ respectively be the sum to 12 terms of 10 A.P. whose first terms are 1, 2, 3, ....... 10 and the common difference are 1, 3, 5, .........., 19 respectively. Then $\sum_{i = 1}^{10} S_{i}$ is equal to [2023]

Q 5 :

If $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + . . . + a_{25}$ is equal to [2023]

Q 6 :

Let $a_{1}, a_{2}, a_{3}, . . . .$ be an A.P. If $a_{7} = 3,$ the product $a_{1} a_{4}$ is minimum and the sum of its first $n$ terms is zero, then $n! - 4 a_{n (n + 2)}$ is equal to [2023]

Q 7 :

The sum of all those terms, of the arithmetic progression 3, 8, 13,..., 373, which are not divisible by 3, is equal to __________ . [2023]

Q 8 :

Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed? [2023]

Q 9 :

Let $a_{1} = 8, a_{2}, a_{3}, . . ., a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is _________ . [2023]

Q 10 :

The sum of the common terms of the following three arithmetic progressions. 3, 7, 11, 15, .., 399,

2, 5, 8, 11, ..., 359 and 2, 7, 12, 17, ..., 197, is equal to ________ . [2023]

Q 12 :

The $8^{t h}$ common term of the series

$S_{1} = 3 + 7 + 11 + 15 + 19 + . . .,$

$S_{2} = 1 + 6 + 11 + 16 + 21 + . . . . .$

is _________ . [2023]

Q 13 :

Let $a_{1}, a_{2}, . . . ., a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18,$ then $12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + . . . + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$ is equal to _______ . [2023]

Q 14 :

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to _________ . [2023]

Q 15 :

For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]

Q 17 :

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is 15 : 7, then $S_{15} - S_{5}$ is equal to: [2024]

Q 18 :

The number of common terms in the progressions
4, 9, 14, 19, ........…, up to 25th term and 3, 6, 9, 12, ....…, up to 37th term is: [2024]

Q 19 :

The $20^{th}$ term from the end of the progression $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is: [2024]

Q 20 :

In an A.P., the sixth term $a_{6} = 2 .$ If the product $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P. is equal to [2024]

Q 21 :

If $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P. and $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are also in an A.P., then $a : b : c$ is equal to [2024]

Q 22 :

Let $S_{n}$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145,$ then $S_{15} - S_{5}$ is [2024]

Q 25 :

Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

Q 26 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then n is : [2025]

Q 27 :

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is : [2025]

Q 28 :

The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

Q 29 :

Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n (A \cup B)$ is [2025]

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Topic Question Set

Q 1 : The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + ... is [2023]

Q 2 : If gcd (m, n) = 1 and 12-22+32-42+...+(2021)2-(2022)2+(2023)2=1012 m2n, then m2-n2 is equal to [2023]

Q 3 : If Sn=4+11+21+34+50+... to n terms, then 160(S29-S9) is equal to [2023]

Q 4 : Let S1,S2,S3, ............ S10 respectively be the sum to 12 terms of 10 A.P. whose first terms are 1, 2, 3, ....... 10 and the common difference are 1, 3, 5, .........., 19 respectively. Then ∑i=110Si is equal to [2023]

Q 5 : If an=-24n2-16n+15, then a1+a2+...+a25 is equal to [2023]

Q 6 : Let a1,a2,a3,.... be an A.P. If a7=3, the product a1a4 is minimum and the sum of its first n terms is zero, then n!-4an(n+2) is equal to [2023]

Q 7 : The sum of all those terms, of the arithmetic progression 3, 8, 13,..., 373, which are not divisible by 3, is equal to __________ . [2023]

Q 8 : Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed? [2023]

Q 9 : Let a1=8,a2,a3,...,an be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is _________ . [2023]

Q 10 : The sum of the common terms of the following three arithmetic progressions. 3, 7, 11, 15, .., 399, 2, 5, 8, 11, ..., 359 and 2, 7, 12, 17, ..., 197, is equal to ________ . [2023]

Q 12 : The 8th common term of the series S1=3+7+11+15+19+..., S2=1+6+11+16+21+.... . is _________ . [2023]

Q 13 : Let a1,a2,....,an be in A.P. If a5=2a7 and a11=18, then 12(1a10+a11+1a11+a12+...+1a17+a18) is equal to _______ . [2023]

Q 14 : Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to _________ . [2023]

Q 15 : For x≥0, the least value of K, for which 41+x+41-x, K2, 16x+16-x are three consecutive terms of an A.P., is equal to: [2024]

Q 17 : Let Sn denote the sum of the first n terms of an arithmetic progression. If S10=390 and the ratio of the tenth and the fifth terms is 15 : 7, then S15-S5 is equal to: [2024]

Q 18 : The number of common terms in the progressions 4, 9, 14, 19, ........…, up to 25th term and 3, 6, 9, 12, ....…, up to 37th term is: [2024]

Q 19 : The 20th term from the end of the progression 20,1914,1812,1734,…,-12914 is: [2024]

Q 20 : In an A.P., the sixth term a6=2. If the product a1a4a5 is the greatest, then the common difference of the A.P. is equal to [2024]

Q 21 : If logea,logeb,logec are in an A.P. and logea-loge2b,loge2b-loge3c,loge3c-logea are also in an A.P., then a:b:c is equal to [2024]

Q 22 : Let Sn denote the sum of first n terms of an arithmetic progression. If S20=790 and S10=145, then S15-S5 is [2024]

Q 23 : Let a1,a2,a3,… be in an arithmetic progression of positive terms. Let Ak=a12-a22+a32-a42+…+a2k-12-a2k2. If A3=-153, A5=-435 and a12+a22+a32=66, then a17-A7 is equal to _______ . [2024]

Q 24 : An arithmetic progression is written in the following way 2 5 8 11 14 17 20 23 26 29 -------------------------------------------------------------------- --------------------------------------------------------------------------- The sum of all the terms of the 10th row is ______. [2024]

Q 25 : Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

Q 26 : Let a1,a2,a3, ..... be in an A.P. such that ∑k=112a2k–1=–725a1,a1≠0. If ∑k=1nak=0, then n is : [2025]

Q 27 : The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by 212. Then the number of terms which are integers in the A.P. is : [2025]

Q 28 : The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

Q 29 : Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B) is [2025]

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Q 1 :

The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + ... is [2023]

Q 2 :

If gcd (m, n) = 1 and $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . + {(2021)}^{2} - {(2022)}^{2} + {(2023)}^{2} = 1012$ $m^{2} n$ , then $m^{2} - n^{2}$ is equal to [2023]

Q 3 :

If $S_{n} = 4 + 11 + 21 + 34 + 50 + . . .$ to $n$ terms, then $\frac{1}{60} (S_{29} - S_{9})$ is equal to [2023]

Q 4 :

Let $S_{1}, S_{2}, S_{3}, . . . . . . . . . . . . S_{10}$ respectively be the sum to 12 terms of 10 A.P. whose first terms are 1, 2, 3, ....... 10 and the common difference are 1, 3, 5, .........., 19 respectively. Then $\sum_{i = 1}^{10} S_{i}$ is equal to [2023]

Q 5 :

If $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + . . . + a_{25}$ is equal to [2023]

Q 6 :

Let $a_{1}, a_{2}, a_{3}, . . . .$ be an A.P. If $a_{7} = 3,$ the product $a_{1} a_{4}$ is minimum and the sum of its first $n$ terms is zero, then $n! - 4 a_{n (n + 2)}$ is equal to [2023]

Q 7 :

The sum of all those terms, of the arithmetic progression 3, 8, 13,..., 373, which are not divisible by 3, is equal to __________ . [2023]

Q 8 :

Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed? [2023]

Q 9 :

Let $a_{1} = 8, a_{2}, a_{3}, . . ., a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is _________ . [2023]

Q 10 :

The sum of the common terms of the following three arithmetic progressions. 3, 7, 11, 15, .., 399,

2, 5, 8, 11, ..., 359 and 2, 7, 12, 17, ..., 197, is equal to ________ . [2023]

Q 12 :

The $8^{t h}$ common term of the series

$S_{1} = 3 + 7 + 11 + 15 + 19 + . . .,$

$S_{2} = 1 + 6 + 11 + 16 + 21 + . . . . .$

is _________ . [2023]

Q 13 :

Let $a_{1}, a_{2}, . . . ., a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18,$ then $12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + . . . + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$ is equal to _______ . [2023]

Q 14 :

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to _________ . [2023]

Q 15 :

For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]

Q 17 :

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is 15 : 7, then $S_{15} - S_{5}$ is equal to: [2024]

Q 18 :

The number of common terms in the progressions
4, 9, 14, 19, ........…, up to 25th term and 3, 6, 9, 12, ....…, up to 37th term is: [2024]

Q 19 :

The $20^{th}$ term from the end of the progression $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is: [2024]

Q 20 :

In an A.P., the sixth term $a_{6} = 2 .$ If the product $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P. is equal to [2024]

Q 21 :

If $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P. and $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are also in an A.P., then $a : b : c$ is equal to [2024]

Q 22 :

Let $S_{n}$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145,$ then $S_{15} - S_{5}$ is [2024]

Q 25 :

Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

Q 26 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then n is : [2025]

Q 27 :

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is : [2025]

Q 28 :

The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

Q 29 :

Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n (A \cup B)$ is [2025]