# Solved sums on Geometric Progression

In this section, ask-math has given some solved sums on geometric progression. These problems help the students to learn how to solve the difficult questions on G.P. This page is based on the problems on nth terms of G.P. For that we use the following formula$a_{n} = ar^{n - 1}$

where a= first term

common ratio = r

number of terms = n

## solved sums on geometric progression

1) The first term fo G.P is 1 and $a_{3} +a_{5}$ = 90. Find 'r'.**Solution :**We know that, $a_{n} = ar^{n - 1}$ and

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 : $p^{2} = (ab)^{n}$

**Solution :**We know that,

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

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

**Solution :**We know that,

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

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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