Question

The nth terms of an A.P. is $\frac{3+n}{4}$, then the sum of first 105 terms is

• A. 270
• B. 735
• C. 1409
• D. 1470

Answer 1

Answer: (D)

1470

Given that:

$t_n=\frac{3+n}{4}$

To find:

$S_{105}=?$

$t_1=\frac{3+1}{4}=1$ and

$t_{105}=\frac{3+105}{4}=27$

Now, $S_{105}=\frac{105}{2}(1+27)=105\times 14=1470$

$\because S_n=\frac{n}{2}(a+l)$ where, $a$=first term and $l$=last term

Hence, D is the correct option.