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Question: If the \({n^{th}}\) term of an AP is \({t_n} = 3 - 5n\), then the sum of first n terms is: A. \(\d...

If the nth{n^{th}} term of an AP is tn=3−5n{t_n} = 3 - 5n, then the sum of first n terms is:
A. n2(1−5n)\dfrac{n}{2}(1 - 5n)
B. n(1−5n)n(1 - 5n)
C. n2(1+5n)\dfrac{n}{2}(1 + 5n)
D. n2(1+n)\dfrac{n}{2}(1 + n)

Explanation

Solution

Hint: Here we will find the ‘a’ and ‘d ‘values from the given tn=3−5n{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 tn=3−5n{t_n} = 3 - 5n
Putting some value of n to get AP series
n=1 ⇒t1=3−5×1=−2 n=2 ⇒t2=3−5×2=−7 n=3 ⇒t3=3−5×3=−12 n=4 ⇒t4=3−5×4=−17  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
⇒−2,−7−12,−17,......\Rightarrow - 2, - 7 - 12, - 17,......
Thus,
a=−2,d=−5a = - 2,d = - 5
We know that sum of first n numbers in A.P

Sn=n2[2a+(n−1)d] Sn=n2[2(−2)+(n−1)−5] Sn=n2[−4−5n+5] Sn=n2[1−5n]  {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.