Question

The arithmetic sequence a_ia i ​ a, start subscript, i, end subscript is defined by the formula: a_1 = 15a 1 ​ =15a, start subscript, 1, end subscript, equals, 15 a_i = a_{i - 1} -7a i ​ =a i−1 ​ −7a, start subscript, i, end subscript, equals, a, start subscript, i, minus, 1, end subscript, minus, 7 Find the sum of the first 660660660 terms in the sequence.

Answer 1

Answer:

-1,512,390

Step-by-step explanation:

Given

a1 = 15

$a_i = a_{i-1} -7$

Let us generate the first three terms of the sequence

$a_2 = a_{2-1}-7\\a_2 = a_1 - 7\\a_2 = 15-7\\a_2 = 8$

For $a_3$

$a_3 = a_{3-1}-7\\a_3 = a_2 - 7\\a_3 = 8-7\\a_3 = 1$

Hence the first three terms ae 15, 8, 1...

This sequence forms an arithmetic progression with;

first term a = 15

common difference d = 8 - 15 = - -8 = -7

n is the number of terms = 660 (since we are looking for the sum of the first 660 terms)

Using the formula;

$S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the given values;

$S_{660} = \frac{660}{2}[2(15) + (660-1)(-7)]\\S_{660} = 330[30 + (659)(-7)]\\S_{660} = 330[30 -4613]\\S_{660} = 330[-4583]\\S_{660} = -1,512,390$

Hence the sum of the first 660 terms of the sequence is  -1,512,390