Question

# Find the sum of the first n terms of the arithmetic sequence.−9 + (−12) + (−15)+ ... (to 10 terms)

Polynomial arithmetic

Find the sum of the first n terms of the arithmetic sequence.$$−9 + (−12) + (−15)+ ...$$ (to 10 terms) Find the sum of the first 100 terms of an arithmetic sequence with 15th term of 86 and first term of 2.

2020-11-08

Step 1 Given: $$-9 + (-12) + (-15)+....$$ It is an arithmetic series with $$a = \text{first term} = -9$$
$$f = \text{common difference} = -12 - (-9) = -3$$ Now sum of the first n terms is, $$S_n = \frac{n}{2} [2a + (n - 1)d]$$
$$\Rightarrow S_n = n/2 [2(-9) + (n - 1)(-3)]$$
$$\Rightarrow S_n = \frac{n}{2} [-18 - 3 n + 3]$$
$$\Rightarrow S_n = \frac{n}{2} [-15 - 3n]$$
$$\Rightarrow S_n = \frac{3n}{2} [5+n]$$

Step 2 Given $$a = 2\ and\ a_{15} = 86$$ Consider, $$a_{15} = 86$$
$$\Rightarrow a+(15-1)d = 86 [ \because a_n=a+(n-1)d]$$
$$\Rightarrow 2+14d = 86$$
$$\Rightarrow 14d = 84$$
$$\Rightarrow d = 6$$ Therefore the sum of the first 100 terms is, $$S_{100} = \frac{100}{2} [2(2) + (100 - 1)(6)] [ \because S_n = \frac{n}{2} [2a + (n - 1)d]]$$
$$= 50(4 + 594)$$
$$=29900$$