$a_{n} = ar^{n - 1}$

where a= first term

common ratio = r

number of terms = n

first term = a = 1

∴ $a_{3} = 1 \times r^{3 - 1}$ ⇒ $a_{3} = r^{2}$

similarly, $a_{5} = 1 \times r^{5 - 1}$ ⇒ $a_{5} = r^{4}$

As $a_{3} +a_{5}$ = 90

$r^{2} + r^{4}$ = 90

$r^{4} + r^{2}$ - 90 = 0

Considering this as a quadratic equation so we will find the factors of it

$\left ( r^{2} \right )^{2} + 10 r^{2} -9 r^{2}$- 90 = 0

$r^{2}( r^{2}$ + 10) -9($r^{2}$ + 10) = 0

∴ ( $r^{2}$ + 10)( $r^{2}$ - 9)

∴ $r^{2}$ + 10 = 0 and $r^{2}$ - 9 = 0

$r^{2}$ = - 10 $r^{2}$ = 9

r =$\pm\sqrt{-10}$ r = $\pm$ 3

The common ratio never be imaginary(complex number)

∴ common ratio = r = $\pm$ 3

2) If the first and the nth terms of G.P. are 'a' and 'b' respectively. If 'p' is the product of the first and the nth term then prove that : $p^{2} = (ab)^{n}$

$a_{n} = ar^{n - 1}$

nth term = b

∴ b = $ar^{n - 1}$

∴ $r^{n - 1} = \frac{b}{a}$

∴ r =$\left ( \frac{b}{a} \right )^\frac{1}{n - 1}$ -----------(1)

Now, the product of n terms = p

∴ p = a $\times$ ar $\times ar^{2} \times$ ...a$r^{n - 1}$

= $a^{n}\times r^{1 + 2 + 3 + ...+(n -1)}$

= $a^{n}\times r^{\frac {n(n -1)}{2}}$

=$a^{n}\left ( \frac{b}{a} \right )^{\frac{n(n-1)}{2}\times \frac{1}{(n -1)}}$

= $a^{n}\left ( \frac{b}{a} \right )^\frac{n}{2}$

= $a^{n}\left ( \frac{b^\frac{n}{2}}{a^\frac{n}{2}} \right )$

p = $a^{\frac{n}{2}} . b^{\frac{n}{2}}$

∴ p = $\left ( ab \right )^{\frac{n}{2}}$

∴ $p^{2} = \left ( ab \right )^{n}$

3) If the $p^{th}, q^{th} and r^{th}$ terms of G.P. are 'a' , 'b' and 'c' respectively then prove that

$a^{q-r}. b^{r -p} . c^{p -q}$ = 1

$a_{n} = ar^{n - 1}$

n = p,q and r and first term = A and common ratio = R

∴ $a_{p} = AR^{p - 1}$ $a_{q} = AR^{q - 1}$

$a_{r} = AR^{r - 1}$

$p^{th}$ term = a , $q^{th}$ = b and $r^{th}$ = c

∴ a = $A R^{p - 1}$ b = $A R^{q - 1}$ c =$A R^{r - 1}$

We have , $a^{q-r}. b^{r -p} . c^{p -q}$

Substitute the values of a,b and c we get

L.H.S = $\left ( AR^{(p -1)} \right )^{(q-r)}.\left ( AR^{(q -1)} \right )^{(r-p)}.\left ( AR^{(r -1)} \right )^{(p-q)}$

= $A^{(q -r +r - p + p -q)}. R^{(p-1)(q-r) +(q -1)(r-p)+(r -1)(p-q)}$

= $ A^{0} . R^{(pq -pr -q + r + qr-pq - r + p +rp - rq - p +q)}$

= 1. $R^{0}$

= 1 ---------Proved

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