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Q.
If, in a G.P., the $(p+q)$ th term is $a$ and the $(p-q)$ th term is $b$, then $p$ th term is
Sequences and Series
Solution:
Let the G.P. be $x, x y, x y^{2}, \ldots$
Then $t_{p+q}=x y^{p+q-1}=a \,\,\,\, (1)$
and $t_{p-q}=x y^{p-q-1}=b\,\,\,\, (2)$
Dividing Eqs (1) by (2),
$y^{2 q}=\frac{a}{b}$
$\therefore y=\left(\frac{a}{b}\right)^{\frac{1}{2 q}}$
From Eq. (1),
$x=a\left(\frac{b}{a}\right)^{\frac{p+q-1}{2 q}}$
$\therefore t_{p}=x y^{p-1}=a \cdot\left(\frac{b}{a}\right)^{\frac{p+q-1}{2 q}}\left(\frac{a}{b}\right)^{\frac{p-1}{2 q}}$
$=1-\frac{p+q-1}{2 q}+\frac{p-1}{2 q} \times b^{\frac{p+q-1}{2 q}-\frac{p-1}{2 q}}$
$=a^{\frac{1}{2}} b^{\frac{1}{2}}=\sqrt{a b}$