In a n increasing G.P. , the sum of the first and the last term is 66, the product of the second and the last but one is 128 and the sum of the terms is 126. How many terms are there in the progression?

Options

A. 64
B. 128
C. 256
D. 729

Answer

B

Explanation

Let a be the term and r the common ratio of the given G.P. Further, let there be n terms in the given G.P. Then
`a_(1)+a_(n)=66rArra+ar^(n-1)=66`
`a_(2)xxa_(n-1)=128`
`rArrarxxar^(n-2)=128`
or `a^(2)r^(n-1)=128`
or `axx(ar^(n-1))=128` or `ar^(n-1)=128/a`
Putting this value of `ar^(n-1)` in (1), we get
`a+128/a=66`
or `a^(2)-66a+128=0`
or `(a-2)(a-64)=0`
or a=2,64
Putting a=2 in (1), we get
`2+2xxr^(n-1)=66` or `r^(n-1)=32`
Putting a=64 in (1), we get
`64+64r^(n-1)=66` or `r^(n-1)=1/32`
For an increasing G.P., `rgt1`. Now,
`S_(n)=126`
`rArr2((r^(n)-1)/(r-1))=126`
or `(r^(n)-1)/(r-1)=63`
or `(r^(n-1)xxr-1)/(r-1)=63`
or `(32r-1)/(r-1)=63`
or r=2
`thereforer^(n-1)=32`
`rArr2^(n-1)=32=2^(5)`
`rArrn-1=5`
`rArrn=6`
For decreasing G.P., a=64 and r=1/2. Hence, the sum of infiite terms is 64/{1-(1/2)}=128.
For a=2,r=2, terms are 2,4,8,16,32,64. For a=64,r=1/2 terms are 64,32,16,8,4,2. Hence, difference is 62.