Inverse Trigonometric Functions Ans 5

Q5 - 31 January - Shift 1

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

(1) 7

(2) $\frac{9}{2}$

(3) 3

(4) 14

Show Answer

Answer: (1)

Solution:

Formula: Properties of obtain terms of G.P.

$a, ar, ar^{2}, ar^{3}(a, r>0)$

$a^{4} r^{6}=1296$

$a^{2} r^{3}=36$

$a=\frac{6}{r^{3 / 2}}$

$a+a r+a r^{2}+a r^{3}=126$

$\frac{1}{r^{3 / 2}}+\frac{r}{r^{3 / 2}}+\frac{r^{2}}{r^{3 / 2}}+\frac{r^{3}}{r^{3 / 2}}=\frac{126}{6}=21$

$(r^{-3 / 2}+r^{3 / 2})+(r^{1 / 2}+r^{-1 / 2})=21$

$r^{1 / 2}+r^{-1 / 2}=A$

$r^{-3 / 2}+r^{3 / 2}+3 A=A^{3}$

$A^{3}-3 A+A=21$

$A^{3}-2 A=21$

$A=3$

$\sqrt{r}+\frac{1}{\sqrt{r}}=3$

$r+1=3 \sqrt{r}$

$r^{2}+2 r+1=9 r$

$r^{2}-7 r+1=0$

$\therefore \text{The ; roots ; are} ; \frac{7 \pm \sqrt{3}i}{2}$

The sum of common ratios of all such GPs is 7.