Question

If the $n^{th}$ term of an A.P. is $t_n=3-5n$, then the sum of the first $n$ terms is

• A. $\frac{n}{2}[1-5n]$
• B. $n(1-5n)$
• C. $\frac{n}{2}(1+5n)$
• D. $\frac{n}{2}(1+n)$

Answer 1

The correct option is A $\frac{n}{2}[1-5n]$

It is given that the $n$th term od A.P is $t_n=3-5n$, therefore, the first term $a=t_1$ will be obtained by substituting $n=1$ in $t_n=3-5n$ as follows:

$t_1=3-(5\times 1)=3-5=-2$

We know that the sum of an arithmetic progression with first term $a$ and last term $t_n$ is $S_n=\frac{n}{2}(a+t\_n)$, therefore, the sum is:

$S_n=\frac{n}{2}(a+t\_n)=\frac{n}{2}(-2+3-5n)=\frac{n}{2}(1-5n)$

Hence, the sum of the first $n$ terms is $\frac{n}{2}(1-5n)$.