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 :
Solution
Let numbers be ${a \over r}$, a, ar $\to$ G.P.<br><br>${a \over r}$, 2a, ar $\to$ A.P. $\Rightarrow$ 4a = ${a \over r}$ + ar $\Rightarrow$ r + ${1 \over r}$ = 4<br><br>r = 2 $\pm$ $\sqrt 3$<br><br>4<sup>th</sup> form of G.P. = 3r<sup>2</sup> $\Rightarrow$ ar<sup>2</sup> = 3r<sup>2</sup> $\Rightarrow$ a = 3<br><br>r = 2 + $\sqrt 3$, a = 3, d = 2a $-$ ${a \over r}$ = 3$\sqrt 3$<br><br>r<sup>2</sup> $-$ d = (2 + $\sqrt 3$)<sup>2</sup> $-$ 3$\sqrt 3$<br><br>= 7 + 4$\sqrt 3$ $-$ 3$\sqrt 3$<br><br>= 7 + $\sqrt 3$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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