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Let a, b and c be the 7th, 11th and 13th...

Let a, b and c be the 7th, 11th and 13th terms, respectively, of a non-constant A.P.. If these are also the three consecutive terms of a G.P., then `(a)/(c )` is equal to

A

2

B

`(7)/(13)`

C

4

D

`(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Let A be the Ist term of AP and d be the common difference.
`:.` 7th term `= a = A + 6d " " [ :' nth " term " = A + (n -1) d]`
11th term `= b + A + 10d`
13th term `= c = A + 12d`
`:' a, b, c` are also in GP
`:. B^(2) = ac`
`rArr (A + 10d)^(2) = (A + 6d) (A + 12d)`
`rArr A^(2) + 20Ad + 100d^(2) = A^(2) + 18Ad + 72d^(2)`
`rArr 2Ad + 28d^(2) = 0`
`rArr 2d (A + 14d) = 0`
`rArr d = 0 or A + 14d = 0`
But `d != 0` [`:'` the series is non constant AP]
`rArr A = - 14d`
`:. a = A + 6d = - 14d + 6d = - 8d`
and `c = A + 12d = - 14d + 12d = - 2d`
`rArr (a)/(c) = (-8d)/(-2d) = 4`
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