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]

0%

(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]

0%

(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]

0%

(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]

0%

(321)

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

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

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

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

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

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

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

$∴$ $Sum = 47 + 107 + 167 = 321$

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]

0%

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]

0%

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]

0%

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]

0%

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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]

0%

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]

0%

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]

0%

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]

0%

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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]