Question

The p th term of an AP is (3p-1)/6. The sum of the first n terms of the AP is​

Answer 1

Step-by-step explanation:

Given: pth term of an AP = (3p - 1)/6

⇒ a + (p - 1) * d = (3p - 1)/6

1st term:

⇒ a = [3(1) - 1]/6

⇒ a = 2/6

⇒ a = 1/3

nth term:

⇒ a + (n - 1) * d = [3(n) - 1]/6

⇒ a + (n - 1) * d = (3n - 1)/6

⇒ (1/3) + (n - 1) * d = (3n - 1)/6

⇒ (n - 1) * d = [(3n - 1)/6] - (1/3)

⇒ (n - 1) * d = (n - 1)/2

Then:

Sum of n terms = (n/2)[2a + (n - 1) * d]

                          = (n/2)[2(1/3) + (n - 1)/2]

                          = (n/2)[4 + 3n - 3]/6

                          = (n/2)[3n + 1/6]

                      = (n/2)[(3n + 1)/12]

Hope it helps!

Answer 2

The pth term of an AP is (3p-1)/6. The sum of the first n terms of the AP is?

Sum Formula: S(n) = (n/2)(a(1)+a(n)

a(1) = (3-1)/6 = 1/2

a(n) = (3n-1)/6

S(n) = (n/2)[(1/2)+(3n-1)/6]

(n/2)[(3+3n-1)/6]

(n/2)(3n+2)/6

(3n^2+2n)/12