Question

lf the sum to $n$ terms of an AP is $\frac{4n^2-3n}{4}$ then the $n^{th}$ term of the AP is equal to

• A. $\frac{5n-1}{4}$
• B. $\frac{8n-7}{4}$
• C. $\frac{3n^2-2}{4}$
• D. $\frac{7n-8}{4}$

Answer 1

The correct option is B $\frac{8n-7}{4}$

$S_n=\frac{4n^2-3n}{4}$ .....(1)

$S_{n-1}=\frac{4(n-1{)}^2-3(n-1)}{4}$

$S_{n-1}=\frac{4(n^2-2n+1)-3(n-1)}{4}$

$S_{n-1}=\frac{4n^2-8n+4-3n+3}{4}$

$S_{n-1}=\frac{4n^2-11n+7}{4}$ .....(2)

$n^{th}$ term of an $A.P=T_n$

$T_n=S_n-S_{n-1}$

$=\frac{4n^2-3n}{4}-\left(\frac{4n^2-11n+7}{4}\right)$

$=\frac{4n^2-3n-4n^2+11n-7}{4}$

$=\frac{8n-7}{4}$

Hence, option $B$ is correct.