Question

If the sum of n terms of an A.P. be 3n² + n and its common difference is 6, then its first term is
A. 2
B. 3
C. 1
D. 4​

Answer 1

Answer:

The 1st term is 4.

Step-by-step explanation:

Sₙ = 3n² + n

Since we need first term,

n = 1

S₁ = 3(1)² + 1

S₁ = 4

In this question, the common difference is not required to solve the problem.

Answer 2

The formula for the sum of an A.P. is $\frac{n}{2}(2a+(n-1)d)$

So,

$\frac{n}{2}(2a+(n-1)d)=3n^2+n\\\\\frac{n}{2}(2a+(n-1)6)=n(3n+1)\\\\\frac{1}{2}(2a+6n-6)=3n+1\\\\a+3n-3=3n+1\\\\a-3=1\\\\a=4$

(I substituted "d" with 6 because the value is given in the question)

Thus, the value of the first term is 4.