arithmetic series problem

Question

An arithme-c series has first term a and common difference d.

The sum of the first 29 terms is 1102.

(a) Show that a + 14d = 38.

(b) The sum of the second term and the seventh term is 13.

Find the value of a and the value of d.

Answer 1

Series: a, a+d, a+2d, ..., a+(n-1)d up to the nth term.

Sum S_n of first n terms can be found by taking pairs of terms:

(a+a+(n-1)d)+(a+d+a+(n-2)d)+...

The first pair is the sum of the first and last terms: 2a+(n-1)d.

The second pair is the sum of the 2nd term a+d and the penultimate term: 2a+(n-1)d.

So all the pairs have the same sum: 2a+(n-1)d, and because there are n terms in the series there must be n/2 pairs.

Therefore S_n=(n/2)(2a+(n-1)d)=an+dn(n-1)/2.

S₂₉=29a+d(29)(28)/2=29a+406d=1102.

Divide through by 29: a+14d=38 (proves (a))

(b) 2nd term is a+d and the 7th term is a+6d. The sum is 2a+7d=13.

We know that a+14d=38, so a=38-14d. Substitute for a in 2a+7d=13: 76-28d+7d=13, 63=21d, d=63/21=3, so a=38-42=-4.

Therefore a=-4 and d=3.