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
4 answers
Isn't your school subject math?
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
Thank you
It is math
4 answers