Sequence and series JEE Main 2025 PYQ with Solution

Question :

If the First term of an AP is 3 and the Sum of its First Four times is equal to One – fifth of the sum of the next four terms then the sum of the first 20 terms is equal to

Options :

(A) -1200

(B) – 1080

(C) – 1020

(D) – 120

Sequence and series JEE Main 2025 PYQ with Solution

Solution :

Given

a = 3

S₄ = $\frac{1}{5}$ [ S₈ – S₄ ]

5S₄ = S₈ – S₄

6S₄ = S₈

6 [ $\frac{4}{2}$ [ 2a + 3d ] = $\frac{8}{2}$ [ 2a + 7d ]

6 [ 2( 6 + 3d ) ] = 4 [ 6 + 7d ]

12 [ 6 + 3d ] = 4 [ 6 + 7d ]

3 [ 6 + 3d ] = 6 + 7d

18 + 9d = 6 + 7d

2d = – 12

d = – 6

S₂₀ = $\frac{20}{2}$ [ 2(3) + 19 x -6 ]

S₂₀ = 10 [ 6 + ( -114)

S₂₀ = 10 [ – 108 ]

S₂₀ = – 1080

Correct answer : Option (B)

Practice Questions :

1. Find the indicated terms in each of the sequence in given below whose n^th terms are

i) a_n = 4n – 3 ; find a₁₇, a₂₄

a₁₇ = 4( 17) – 3 = 68 – 3 = 65

a₂₄ = 4(24) – 3 = 96 – 3 = 93

ii) a_n = $\frac{n^2}{2^n}$, find a₇

a₇ = $\frac{7^2}{2^7}$ = $\frac{49}{128}$

iii) a_n = (-1)^{n – 1} . n³ ; a₉

a₉ = (-1)^{9 – 1}. 9³ = (-1)⁸. 729 = 729

iv) a_n = $\frac{n(n – 2)}{n +3}$ ; a₂₀

a₂₀ = $\frac{ 20(20 – 2)}{20 + 3}$ = $\frac{20(18)}{23}$ = $\frac{360}{23}$

2. Write the first five terms of each of the sequence given below and obtain the corresponding series :

i) a₁ = 3, a_n = 3a_{n – 1} + 2 for all n > 1

Solution :

Given a₁ = 3 , a_n = 3a_{n – 1} + 2

a₂ = 3a_{2 – 1} + 2 = 3( a₁) + 2 = 3(3) + 2 = 9 + 2 = 11

a₃ = 3a_{3 – 1} + 2 = 3a₂ + 2 = 3(11) + 2 = 33 + 2 = 35

a₄ = 3a_{4 – 1} + 2 = 3a₃ + 2 = 3(35) + 2 = 105 + 2 = 107

a₅ = 3a_{5 – 1} + 2 = 3a₄ + 2 = 3( 107) + 2 = 321 + 2 = 323

• The first five terms are 3, 11, 35, 107, 323
• The corresponding series = 3 + 11 + 35 + 107 + 323+……………..

ii) a₁ = a₂ = 2, a_n = a_{n – 1} – 1, n > 2

Solution :

Given

a₁ = 2, a₂ = 2, a_n = a_{n – 1} – 1

a₃ = a_{3 – 1} – 1 = a₂ -1 = 2 – 1 = 1

a₄ = a_{4 – 1} – 1 = a₃ – 1 = 1 – 1 = 0

a₅ = a_{5 – 1} – 1 = a₄ – 1 = 0 – 1 = -1

• The first five terms are 2, 2, 1 , 0, -1
• The corresponding series are 2 + 2 + 1 + 0 + (-1) + ………………………

3. The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_{n – 1} + a_{n – 2}, n >2 Find $\frac{a_n+1}{a_n}$, for n = 1, 2, 3, 4, 5…………

Solution :

Given

a₁ = 1, a₂ = 1 a_n = a_{n – 1} + a_{n – 2},

a₃ = a_{3 – 1} + a_{3 – 2} = a₂ + a₁ = 1 + 1 = 2

a₄ = a_{4 – 1} + a_{4 – 2} = a₃ + a₂ = 2 + 1 = 3

a₅ = a_{5 – 1} + a_{5 – 2} = a₄ + a₃ = 3 +2 = 5

a₆ = a_{6 -1} + a_{6 -2} = a₅ + a₄ = 5 + 3 = 8

For n = 1 $\rightarrow$ $\frac{a_n+1}{a_n}$ = $\frac{a_2}{a_1}$ = $\frac{1}{1}$

For n = 2 $\rightarrow$ $\frac{a_2+1}{a_2}$ = $\frac{a_3}{a_2}$ = $\frac{2}{1}$ = 2

For n = 3 $\rightarrow$ $\frac{a_3+1}{a_3}$ = $\frac{a_4}{a_3}$ = $\frac{3}{2}$

For n = 4 $\rightarrow$ $\frac{a_4+1}{a_4}$ = $\frac{a_5}{a_4}$ = $\frac{5}{3}$

For n = 5 $\rightarrow$ $\frac{a_5+1}{a_5}$ = $\frac{a_6}{a_5}$ = $\frac{8}{5}$

The Fibonacci sequence = 1, 2, $\frac{3}{2}$, $\frac{5}{8}$, $\frac{8}{5}$

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