Asked by Umaru Monsurat
the first term of an arithmetic progression is 3 and the fifth term is is 9.find the number of terms in the progression if the last term is 81
Answers
Answered by
Ms. Sue
Isn't your school subject math?
Answered by
Steve
well, since
T5-T1 = 4d, d=3/2
we know a=3, so
Sn = n/2 (2*3+(n-1)(3/2))
3/4 (n^2+3n) = 81
n^2+3n-108 = 0
(n+12)(n-9) = 0
n=9
and the sequence is
3 4.5 6 7.5 9 10.5 12 13.5 15
The sum is 81
T5-T1 = 4d, d=3/2
we know a=3, so
Sn = n/2 (2*3+(n-1)(3/2))
3/4 (n^2+3n) = 81
n^2+3n-108 = 0
(n+12)(n-9) = 0
n=9
and the sequence is
3 4.5 6 7.5 9 10.5 12 13.5 15
The sum is 81
Answered by
Anonymous
Thank you
Answered by
Ayobami
It is math