2860.61 – Arithmetic Sequence I

The initial terms of an arithmetic series is 1. If the sum of the first twenty terms is four times the sum of the first twelve terms, then what's the second term?

Solution

We recollect: an arithmetic sequence of $n$ terms is

$a_1, a_2, a_3, ..., a_n$ where $a_2 = a_1 + d, d$ being the constant difference between consecutive terms.

So $a_3 = a_1 + 2d$ , etc., and $a_n = a_1 + (n - 1)d$ .

We recollect further: the sum of the first $n$ terms, $S_n$ , is $\frac{n}{2}(a_1 + a_n) = \frac{n}{2}(a_1 + a_1 + (n-1))d$ .

The sum of the first 12 terms is $\frac{12}{2}(1 + 1 + 11d) = 6(2 + 11d ) = 12 + 66d$ .

The sum of the first 20 terms is $\frac{20}{2}(1 + 1 + 19d) = 10(2 + 19d) = 20 + 190d$ .

We're told that $20 + 190d = 4(12 + 66d) \rightarrow 20 + 19d = 48 + 264d \rightarrow -74d = 28$ .

Thus $d = \frac{-28}{74} = \frac{-14}{37}$ .

So the second term is $1 - \frac{14}{37} = \frac{23}{37}$ .