Question

The sum of first 'n' terms. Sn of a particular Arithmetic Progression is given by Sn=12n - 2n^2. Find the first term and the common difference

Answer 1

Let a be the first term and d the common difference between consecutive terms. Then the next few terms in the sequence are

a + d

a + 2d

a + 3d

and so on, up to the k-th term

a + (k - 1) d

The sum of the first n terms of this sequence is

`\displaystyle S_n = \sum_(k=1)^n(a+(k-1)d) = 12n-2n^2`

Expanding the sum, we have

`\displaystyle S_n = \sum_(k=1)^n (a+(k-1)d) \\\\ S_n = \sum_(k=1)^n(a-d+dk) \\\\ S_n = (a-d)\sum_(k=1)^n1+d\sum_(k=1)^nk \\\\ S_n = (a-d)n+\frac{d}2n(n+1) \\\\ S_n = (a-d)n+\frac{d}2(n^2+n) \\\\ S_n = \left(a-\frac{d}2\right)n+\frac{d}2n^2`

It follows that

a - d/2 = 12

d/2 = -2

Solve these equations for a and d.

d/2 = -2 ==> d = -4

a - d/2 = a + 2 = 12 ==> a = 10

So the sequence is

10, 6, 2, -2, -6, -10, -14, …