Question

The third term of an arithmetic sequence is 21, and the eighth term is 56. The first term is _____.

***So i have no idea how to solve hahaa please help and explain! :)

Answer 1

so we have two equations
\[21=a+2d\]
\[56=a+7d\]

Answer 2

isn't 21 the 3rd term? And 56 the 8th term? how come it says 2d and 7d?

Answer 3

using \[a _{n}=a+(n-1)d\]
\[a _{3}=a+(3-1)d=21\]
\[a _{8}=a+(8-1)d\]

Answer 4

=56

Answer 5

ohhhh i see :) so how do i solve for the first term?

Answer 6

you can subtract (1) from (2)

Answer 7

eliminating a

Answer 8

i'm confused :(

Answer 9

you can re-write
\[a=21-2d\]
and substitute in
\[56=a+7d\]



\[56=(21-2d)+7d\]

Answer 10

and this is to solve for the 1st term? @Jonask

Answer 11

the last equation is in terms of b so it solves b

Answer 12

and b is the first term?

Answer 13

sorry d

Answer 14

oh okay and d is the first term theN??

Answer 15

d is the common difference

Answer 16

and when i simplify the equation i get 56=21+5d right??

Answer 17

oh okay so how do i solve the first term?

Answer 18

yes get d first

Answer 19

okay and here is to solve for d...