Question

The third term of an arithmetic sequence is-8, and the sum of the first 10 terms of the sequence is -230. Find:

Answer 1

Answer:

The first term of the sequence is a = 4

The sum of the first 13 terms is -480

Step-by-step explanation:

Given that:

The third term is: a + 2d = - 8 --- (1)

$The \ sum \ of \ the \ first \ 10 \ terms \ is:$

$-230 = \dfrac{10}{2}[2a+ (10-1)d]$

$-230 =5[2a+ (9)d]$

-230 = 10a + 45d

10a + 45d = -230     ----(2)

From (1) and (2)

a + 2d = - 8

10a + 45d = -230

From (1), let a = -8 - 2d

Replace, the value of a in (1) to equation (2)

10(-8 -2d) + 45d = -230

-80 -20d +45d =-230

25d = -230 + 80

25d = -150

d = -150/25

d = -6

From equation (1);

a + 2(-6) = -8

a -12 = -8

a = -8 + 12

a = 4

Thus,the first term of the sequence a = 4

$The \ sum \ of \ the \ first \ 13 \ terms \ is:$

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

$S_n = \dfrac{13}{2}[2(4)+(13-1)*-6]$

$S_n =7.5 [8+(12)\times -6]$

$S_n =7.5 [8-72]$

= -480