105k views
2 votes
Find S18 for geometric series given a5=-6 and a2= -48

1 Answer

2 votes

an = a1r^(n-1)

a5 = a1 r^(5-1)

-6 =a1 r^4


a2 = a1 r^(2-1)

-48 = a1 r


divide

-6 =a1 r^4

---------------- yields 1/8 = r^3 take the cube root or each side

-48 = a1 r 1/2 = r


an = a1r^(n-1)

an = a1 (1/2)^ (n-1)

-48 = a1 (1/2) ^1

divide by 1/2

-96 = a1


an = -96 (1/2)^ (n-1)


the sum

Sn = a1[(r^n - 1/(r - 1)]

S18 = -96 [( (1/2) ^17 -1/ (1/2 -1)]

=-96 [ (1/2) ^ 17 -1 /-1/2]

= 192 * [-131071/131072]

approximately -192

User Kalpana C
by
6.4k points