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?

Explanation

Let the given GP contain n terms. Let a be the first term and r be the common ratio of this GP.
Since the given GP is increasing, we have `r gt 1`.
Now, `T_(1)T_(n)=66 rArr a+ar^((n-1))=66`. ...(i)
And, `T_(2)xxT_(n-1)=128 rArr ar xx ar^((n-2))=128`
`rArr a^(2)r^((n-1))=128 rArr ar^((n-1))=128/a`. ...(ii)
Using (ii) in (i), we get
`a+128/a=66 rArr a^(2)-66a+128=0`
`rArr a^(2)-2a-64a+128=0`
`rArr a(a-2)-64(a-2)=0`
`rArr (a-2)(a-64)=0`
`rArr a=2 or a=64`.
Putting `a=2` in (ii), we get
`r^((n-1))=128/a^(2)=128/4=32`. ..(iii)
Putting `a=64` in (ii), we get
`r^((n-1))=128/a^(2)=128/(64xx64)=1/32`, which is rejected, sice `r gt 1`.
Thus, `a=2` and `r^((n-1))=32`.
Now, `S_(n)=126 rArr (a(r^(n)-1))/((r-1))=126`
`rArr 2((r^(n)-1)/(r-1))=126 rArr (r^(n)-1)/(r-1)=63`
`rArr (r^((n-1))xxr-1)/(r-1)=63 rArr (32r-1)/(r-1)=63`
`rArr 32r-1=63 r-63 rArr 31 r=62 rArr r=2`.
`:. r^((n-1))=32=2^(5) rArr n-1=5rArr n=6`.
Hence, there are 6 terms in the given GP.