The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.


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


Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

$a_{5}=a r^{5-1}=a r^{4}=p \ldots(1)$

$a_{8}=a r^{8-1}=a r^{7}=q \ldots(2)$

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

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

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

$r^{3}=\frac{q}{p}$ $\ldots(4)$

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

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

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

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


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

Thus, the given result is proved.

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