Question

Find the sum of the first 100 terms of the arithmetic sequence. a₁ 15 and a100=307

Answer 1

The sequence is arithmetic, so there is a constant difference d between consecutive terms such that

$a_{100} = a_{99} + d$

$a_{100} = (a_{98} + d) + d = a_{98} + 2d$

$a_{100} = (a_{97} + d) + 2d = a_{97} + 3d$

and so on, down to

$a_{100} = a_1 + 99d$

(notice the pattern of 100 = 99 + 1 = 98 + 2 = 97 + 3 = … = 1 + 99)

Solve for d :

$307 = 15 + 99d \implies 99d = 292 \implies d = \dfrac{292}{99}$

Now, the sum of the first 100 terms of this sequence is

$\displaystyle \sum_{n=1}^{100} a_n = \sum_{n=1}^{100} \left(15 + \frac{292}{99}(n-1)\right) = \boxed{16100}$

which follows from the well-known sums

$\displaystyle \sum_{n=1}^N 1 = N$

$\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2$