Let a, b and c be in GP with common ratio, r where `a !=0 and 0 lt r le (1)/(2)`. If 3a, 7b and 15 c are the first three terms of an AP, then the 4th term of this AP is

Options

A. 5a
B. `(2)/(3) a`
C. a
D. `(7)/(3)a`

Answer

C

Explanation

Key Idea Use `n^(th)` term of AP i.e., `a_(n) = a + (n -1) d`, if a, A, b are in AP, then `2A = a + b and n^(th)` term of G.P. i.e., `a_(n) = ar^(n-1)`
It is given that, the terms a, b, c are in GP with common ratio r, where `a != 0 and 0 lt r le (1)/(2)`.
So, lt, `b = ar and c = ar^(2)`
Now, the terms 3a, 7b and 15c are the first three terms of an AP, then
`2(7b) = 3a + 15c`
`rArr 14ar = 3a + 15ar^(2) [" as " b = ar, c = ar^(2)]`
`rArr 14r = 3 + 15r^(2) [ " as " a != 0]`
`rArr 15r^(2) - 14r + 3 = 0`
`rArr 15r^(2) - 5r - 9r + 3 = 0`
`rArr 5r ( 3r -1) - 3 (3r - 1) = 0`
`rArr (3r -1) (5r -3) = 0`
`rArr r = (1)/(3) or (3)/(5)`
as, `r in (0, (1)/(2)], " so " r = (1)/(3)`
Now, the common difference of AP `= 7b - 3a`
`7ar - 3a = a ((7)/(3) - 3) = - (2a)/(3)`
So, `4^(th)` term of AP `= 3a + 3 ((-2a)/(3)) = a`