Question

Find the third term of the sequences (AP) for which $S_n$ is given below.

$S_n=4n^2-2n$

• A. $32$
• B. $24$
• C. $52$
• D. $64$

Answer 1

Answer: (B)

Given:

$S_n=4n^2-2n$

$\therefore S_1=4(1{)}^2-2(1)=2$

$\therefore S_2=4(2{)}^2-2(2)=12$

$\therefore S_3=4(3{)}^2-2(3)=30$

$\therefore S_4=4(4{)}^2-2(4)=56$

We know that $n^{th}$ is given by, $T_n=S_n-S_{n-1}$

$\therefore T_1=S_1=2$

$\therefore T_2=S_2-S_1=12-2=10$

$\therefore T_3=S_3-S_2=30-12=18$

Hence, the third term of the AP is $18$.