Three numbers are in an increasing geometric progression with common ratio r.

Question

Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r2, then r2 - d is equal to :

Accepted Answer

Let numbers be $${a \over r}$$, a, ar $$\to$$ G.P.

$${a \over r}$$, 2a, ar $$\to$$ A.P. $$\Rightarrow$$ 4a = $${a \over r}$$ + ar $$\Rightarrow$$ r + $${1 \over r}$$ = 4

r = 2 $$\pm$$ $$\sqrt 3 $$

4th form of G.P. = 3r2 $$\Rightarrow$$ ar2 = 3r2 $$\Rightarrow$$ a = 3

r = 2 + $$\sqrt 3 $$, a = 3, d = 2a - $${a \over r}$$ = 3$$\sqrt 3 $$

r2 - d = (2 + $$\sqrt 3 $$)2 - 3$$\sqrt 3 $$

= 7 + 4$$\sqrt 3 $$ - 3$$\sqrt 3 $$

= 7 + $$\sqrt 3 $$

Solutions · 1