JEE Mains · Maths · STD 11 - 8. sequence and series
Three numbers are in an increasing geometric progression with common ratio \(\mathrm{r}\). If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference \(\mathrm{d}\). If the fourth term of GP is \(3 \mathrm{r}^{2}\), then \(\mathrm{r}^{2}-\mathrm{d}\) is equal to:
- A \(7-7 \sqrt{3}\)
- B \(7+\sqrt{3}\)
- C \(7-\sqrt{3}\)
- D \(7+3 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(7+\sqrt{3}\)
Step-by-step Solution
Detailed explanation
Let numbers be \(\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar} \rightarrow\) \(G.P.\)…
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