The sum of an infinite geometric series is 162 and the sum of its first `n` terms is 160. If the inverse of its common ratio is an integer, then which of the following is not a possible first term? `108` b. `144` c. `160` d. none of these

Options

A. 108
B. 120
C. 144
D. 160

Answer

A, C, D

Explanation

`S_(oo)=a/(1-r)=162`
`S_(n)=(a(1-r^(n)))/(1-r)=160`
Solving these equations, we get
`1-r^(n)=160/162=80/81`
`rArr1-80/81=r^(n)`
`rArrr^(n)=1/81`
or `(1/r)^(n)=81`
Now, it is given that `1/r` is an integer and n is also an integer.
Hence, the relation (1) implies that `1/r`=3,9 or 81 so that n=4,2 or 1 accordingly.
Therefore, (i) `a=162(1-1/3)=108`
(ii) `a=162(1-1/9)=144`
(iii) `a=162(1-1/81)=160`