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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
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
    $7-7 \sqrt{3}$
  • ✓
    $7+\sqrt{3}$
  • C
    $7-\sqrt{3}$
  • D
    $7+3 \sqrt{3}$
✓

Answer

Correct option: B.
$7+\sqrt{3}$
b
Let numbers be $\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar} \rightarrow$ $G.P.$

$\mathrm{a}_{\mathrm{r}}, 2 \mathrm{a}, \mathrm{ar} \rightarrow \mathrm{A} \cdot \mathrm{P} \Rightarrow 4 \mathrm{a}=\frac{\mathrm{a}}{\mathrm{r}}+\mathrm{ar} \Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}=4$

$\mathrm{r}=2 \pm \sqrt{3}$

$4^{\text {th }}$ form of G.P $=3 \mathrm{r}^{2} \Rightarrow \mathrm{ar}^{2}=3 \mathrm{r}^{2} \Rightarrow \mathrm{a}=3$

$\mathrm{r}=2+\sqrt{3}, \mathrm{a}=3, \mathrm{~d}=2 \mathrm{a}-\frac{\mathrm{a}}{\mathrm{r}}=3 \sqrt{3}$

$\mathrm{r}^{2}-\mathrm{d}=(2+\sqrt{3})^{2}-3 \sqrt{3}$

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

$=7+\sqrt{3}$

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