Sequence and series JEE Main 2025 shift – 2 with solution

Question :

Let a_n be the n^th term of an AP If S_n = a₁ + a₂ + a₃ + ……………..+ a_n = 700, a₆ = 7 and S₇ = 7 then a_n is equal to

Options :

(a) 56

(b) 65

(c) 64

(d) 70

Sequence and series JEE Main 2025 shift – 2 with solution

Answer :

Given

a₆ = 7

a + 5d = 7 ———-1

S₇ = 7

$\frac{7}{2}$(2a + 6d ) = 7

a + 3d = 1 ————–2

Solving 1 and 2 equation

a + 5d – a – 3d = 7 – 1

2d = 6

d = 3

substitude d = 3 in 1st equation

a + 5d = 7

a + 5(3) = 7

a = 7 – 15

a = -8

To find ‘n’

S_n = a₁ + a₂ + a₃ + ……………..+ a_n = 700

S_n = $\frac{n}{2}$( 2a + (n – 1)d ) = 700

$\frac{n}{2}$( – 16 + (n – 1)30 = 700

-16n + 3n² – 3n = 1400

3n² – 19n – 1400 = 0

3n² – 75n + 56n – 1400 = 0

3n (n – 25) +56 ( n – 25 ) = 0

(n – 25 ) ( 3n + 56 ) = 0

n = 25

Now To find a_n

substitude a = -8, d = 3, n = 25

a_n = a + (n – 1)d = -8 + (24)3 = 64

Correct answer : Option (c)

Formula Used :

a_n = a + (n – 1)d

S_n = $\frac{n}{2}$( 2a + (n – 1)d