T7=a+6d=23
Tn=a+[n-1]d=43
T2n=a+[2n-1]=91
subtract equ 2 from equ 3 give
nd=48---equ 4
from equ2 nd=43+d-a
:- 48=43+d-a
5=d-a ---equ6 recall from equ1 a=23-6d put the value of a into equ 6
d=4 from equ4 nd=48 since d=4, n=12
put the value of d into equ1 :- a=-1
(n=12,d=4,a=-1]
Given that T7=23,Tn=43 and T2n=91.for an A.P.find a,d and n...help please
6 answers
Thank you
What happened to the d in equation 3
You didn't include the d in equation 3
Why?
Why?
I don't understand how you work it
a+6d=23
a+[n-1]d=43
a+[2n-1]d=91
subtract #2 from #3 to get
nd = 48
Now the first two equations become
a+6d = 23
a+48-d = 43
subtract to get 7d = 28, so
d = 4, so a = -1, n=12
a+[n-1]d=43
a+[2n-1]d=91
subtract #2 from #3 to get
nd = 48
Now the first two equations become
a+6d = 23
a+48-d = 43
subtract to get 7d = 28, so
d = 4, so a = -1, n=12
1 answer