Inverse Trigonometric Functions Question 5
Question 5 - 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.
