Correct Answer - C
Let the first term be a and common difference be d for the first A.P. and the first term be b and common difference be e for the second A.P. and let the numbers of terms be n. Then
`(a+(n-1)d)/b=(b+(n-1)e)/a=4` (1)
`(n/2[2a+(n-1)d])/(n/2[2b+(n-1)e])=2` (2)
From (1) and (2), we get
a-4b+(n-1)d=0 (3)
b-4a+(n-1)e=0 (4)
2a-4b+(n-1)d-2(n-1)e=0 (5)
`4xx(3)+(4)` gives
`-15b+4(n-1)d+(n-1)e=0` (6)
`(4)+2xx(5)` gives
`-7b+2(n-1)d-3(n-1)e=0` (7)
Further, `15xx(7)-7xx(6)` gives
`2(n-1)d-52(n-1)e=0`
or d=26e `(becausengt1)`
`therefored//e=26`
Putting d=26e in (3) and solving it with (4), we get
a=2(n-1)e,b=7(n-1)e
Then, the ratio of their `n^(th)` terms is
`(2(n-1)+(n-1)26e)/(7(n-1)e+(n-1)e)=7/2`