Let a, b and c be in G.P.

Question

Let $a$, b and c be in G.P. with common ratio r, where $a$ $$ \ne $$ 0 and 0 < r $$ \le $$ $${1 \over 2}$$
. If 3$a$, 7b and 15c are the first three
terms of an A.P., then the 4th term of this A.P. is :

Accepted Answer

a = a, b = ar and c = ar2

3a, 7b, 15c $$ \to $$ A.P.

14b = 3a + 15c

14(ar) = 3a + 15(ar2)

15r2 -- 14r + 3 = 0

$$ \Rightarrow r = {1 \over 3},{3 \over 5}(rejected)$$

Common difference = 7b -- 3a

= 7ar -- 3a

$$ \Rightarrow $$ $${{7a} \over 3} - 3a =  - {2 \over 3}a$$

4th term is $$ \Rightarrow $$ $$15c - {2 \over 3}a = {{15} \over 9}a - {2 \over 3}a = a$$

Solutions · 1