The third term of a Go is 9 and the fifth term is 16 find the 4th term and the sum of the first four terms
3 answers
9 * r^2 = 16 ... r^2 = 16/9 ... r = ± 4/3
a ar ar^2 ....a r^(n-1)
a r^2 = 9
a r^4 = 16
a = 16/r^4 = 9/r^2
16/9 = r^2
r = 4/3
then a = 9/(16/9) = 81/16
check
ar^4 = 81/16 * (16*16 / 9*9 ) = 16 sure enough
so
a r^3 = fourth term = 81/16 * (4^3/3^3)
= (81/16)(16*4 / 9*3)= 3*4 = 12
a r^2 = 9
a r^4 = 16
a = 16/r^4 = 9/r^2
16/9 = r^2
r = 4/3
then a = 9/(16/9) = 81/16
check
ar^4 = 81/16 * (16*16 / 9*9 ) = 16 sure enough
so
a r^3 = fourth term = 81/16 * (4^3/3^3)
= (81/16)(16*4 / 9*3)= 3*4 = 12
for sum of first n terms google math is fun geometric sequence
a (1-r^n) / (1-r)
a = 81/16
r = 4/3
n = 4
so
(81/16) [ 1 - (4/3)^4 ] / [ 1 - (4/3) ]
a (1-r^n) / (1-r)
a = 81/16
r = 4/3
n = 4
so
(81/16) [ 1 - (4/3)^4 ] / [ 1 - (4/3) ]