5^{text {th }}, 8^{text {th }} and 11^{text {th }} terms of a G.P. are p, q and s, respectively. Sho

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8. Sequences and Series

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The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a $G.P.$ are $p, q$ and $s,$ respectively. Show that $q^{2}=p s$

Option A

Option B

Option C

Option D

Solution

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$ According to the given condition,

$a_{5}=a r^{5-1}=a r^{4}=p$ ………$(1)$

$a_{8}=a r^{8-1}=a r^{7}=q$ ………$(2)$

$a_{11}=a r^{11-1}=a r^{10}=s$ ………$(3)$

Dividing equation $(2)$ by $(1),$ we obtain

$\frac{a r^{7}}{a r^{4}}=\frac{q}{p}$

$r^{3}=\frac{q}{p}$ ………$(4)$

Dividing equation $(3)$ by $(2),$ we obtain

$\frac{a r^{10}}{a r^{7}}=\frac{s}{q}$

$\Rightarrow r^{3}=\frac{s}{q}$ …….$(5)$

Equating the values of $r^{3}$ obtained in $(4)$ and $(5),$ we obtain

$\frac{q}{p}=\frac{s}{q}$

$\Rightarrow q^{2}=p s$

Thus, the given result is proved.

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