Question

The third term of an arithmetic sequence is equal to 9 and the sum of the first 8 term is 42. Find the first term and the common difference

Answer 1

Answer:

The first term is 14

The common difference is -2.5

Step-by-step explanation:

we know that

The rule to calculate the an term in an arithmetic sequence is

$a_n=a_1+d(n-1)$

where

d is the common difference

a_1 is the first term

we have that

The third term of an arithmetic sequence is equal to 9

so

$a_3=9$

$n=3$

substitute

$9=a_1+d(3-1)$

$9=a_1+2d$ ----> equation A

The rule to find the sum of the the first n terms of the arithmetic sequence is equal to

$S=\frac{n}{2} [2a_1+(n-1)d]$

we have

The sum of the first 8 term is 42

so

$S=42$

$n=8$

substitute

$42=\frac{8}{2} [2a_1+(8-1)d]$

$42=4[2a_1+7d]$

$10.5=2a_1+7d$ ----> equation B

Solve the system of equations

$9=a_1+2d$ ----> equation A

$10.5=2a_1+7d$ ----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

using a graphing tool

the solution is (14,-2.5)

see the attached figure

therefore

$a_1=14\\d=-2.5$

The first term is 14

The common difference is -2.5