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

Answer: (A)

$\frac{n}{2}[1-5n]$

Given, $t_n=3-5n$

$t_1=3-5=-2$

$t_2=3-5\left(2\right)=3-10=-7$

$t_3=3-5\left(3\right)=3-15=-12$

$t_4=3-5\left(4\right)=3-20=-17$

Thus, $a=-2,d=-5$

Now, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}\left[2\right(-2)+(n-1\left)\right(-5\left)\right]$

$=\frac{n}{2}[-4-5n+5]$

$=\frac{n}{2}[1-5n]$