JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 5)
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 :
7 $$-$$ 7$$\sqrt 3 $$
7 + $$\sqrt 3 $$
7 $$-$$ $$\sqrt 3 $$
7 + 3$$\sqrt 3 $$
Explanation
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 + $$\sqrt 3 $$
$${a \over r}$$, 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 + $$\sqrt 3 $$
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