The sum of an infinitely decreasing G.P. is equal to 4 and the sum of the cubes of its terms is equal to `64//7`. Then `5^(th)` term of the progression is

Options

A. `1/4`
B. `1/8`
C. `1/16`
D. `1/32`

Answer

B

Explanation

Let the G.P. be, `a,ar,ar^(2)`
`rArra/(1-r)=4` ….(1)
Also, `a^(3)+(ar)^(3)+` ……..=`(a^(3))/(1-r^(3))rArr(a^(3))/(1-r^(3))=64/7`
`rArr7a^(3)=64(1-r^(3))` …(2)
Using (1) in (2), we have
`7xx64(1-r)^(3)=64(1-r^(3))rArr2r^(2)-5r+2=0rArrr=1//2,2`
`therefore` G.P. is decreasing `rArr r=1//2` and a=2
`a_(5)=2xx1/(2^(4))=1/8`