Question

Sum of n term of AP is 2n^2 find 5th term of the series

Answer 1

Hey there !!

Given :-

→ $S_n$ = 2n² .

To find :-

→ 5th term ( a₅ ) .

Solution :-

∵ S₁ = 2× 1² .

∴ S₁ = 2.

∵ S₂ = 2 × 2² .

⇒ S₂ = 2 × 4 .

∵ S₂ = 8 .

Therefore, First term a₁ = 2 &  Second term a₂ = S₂ - S₁ = 8 - 2 = 6 and d = 6 - 2 = 4.

Now, 5th term is given by :-

∵ $a_n$ = a + ( n - 1 ) d .

⇒ a₅ = 2 + ( 5 - 1 ) × 4 .  

⇒ a₅ = 2 + 16 .

∴ a₅ = 18

Hence, it is solved .

THANKS

#BeBrainly.

Answer 2

Sn = 2(n^2)

Therefore S1 = 2(1^2)= 2

Now sum of first term(S1) = First term (a)

Sum of n terms of an AP = (n/2)[2a + (n-1)d]

2n^2 = (n/2)[2a + (n-1)d]

simplifying by taking n/2 to the other side

4n = 2a + (n-1)d

a = 2
4n = 4 + (n-1)d

taking 4 to other side

4n-4 = (n-1)d

taking 4 common

4(n-1) = (n-1)d

cancelling n-1 on both sides

d = 4

Tn = a + (n-1)d
T5 = 2 + (5-1)*4
T5 = 2 + 16

Therefore T5 = 18