Two arithmetic progressions have the same numbers. The reatio of the last term of the first progression to the first term of the second progression is equal to the ratio of the last term of the second progression to the first term of first progression is equal to 4. The ratio of the sum of the n terms of the first progression to the sum of the n terms of teh first progression to the sum of the n terms of the second progerssion is equal to 2.
The ratio of their common difference is

Options

A. 12
B. 24
C. 26
D. 9

Answer

C

Explanation

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`