Question

If the ${n^{th}}$ term of an AP is ${t_n} = 3 - 5n$, then the sum of first n terms is:
A. $\dfrac{n}{2}(1 - 5n)$
B. $n(1 - 5n)$
C. $\dfrac{n}{2}(1 + 5n)$
D. $\dfrac{n}{2}(1 + n)$

Answer 1

Hint: Here we will find the ‘a’ and ‘d ‘values from the given ${t_n} = 3 - 5n$ and then by using the sum of n terms formula in AP the value can be calculated.

Complete step-by-step answer:
Given ${t_n} = 3 - 5n$
Putting some value of n to get AP series
$
  n = 1 \\
   \Rightarrow {t_1} = 3 - 5 \times 1 = - 2 \\
  n = 2 \\
   \Rightarrow {t_2} = 3 - 5 \times 2 = - 7 \\
  n = 3 \\
   \Rightarrow {t_3} = 3 - 5 \times 3 = - 12 \\
  n = 4 \\
   \Rightarrow {t_4} = 3 - 5 \times 4 = - 17 \\
$
The A.P series will be
$ \Rightarrow - 2, - 7 - 12, - 17,......$
Thus,
$a = - 2,d = - 5$
We know that sum of first n numbers in A.P
\[
  {S_n} = \dfrac{n}{2}[2a + (n - 1)d] \\
  {S_n} = \dfrac{n}{2}[2( - 2) + (n - 1) - 5] \\
  {S_n} = \dfrac{n}{2}[ - 4 - 5n + 5] \\
  {S_n} = \dfrac{n}{2}[1 - 5n] \\
\]
The correct option will be A.

Note: These problems can also be solved by taking the nth term and applying summation. The formula of summation of n terms must be known. It will fetch us the same result.