Question

#Quality Question

@Sequence And Series

Let a1 , a2 , a3 , .... , a10 be a GP. If
$\frac{a_3}{a_1} = 25 \: then \: find \: \frac{a_9}{a_5}$
​

Answer 1

Answer:

625

Step-by-step explanation:

Given :-

a1 , a2 , a3 , ..., a10 are in the GP

and a3/a1 = 25

To find :-

Find the value of a9 / a5 ?

Solution :-

Given that :

a1 , a2 , a3 , ..., a10 are in the GP

We know that

The General term of a GP = an = a×r^(n-1)

a1 = a is the first term

a3 = a×r^(3-1)

=> a3 = a×r²

=> a3 = ar²

Now

a3/a1

=> ar²/a

=> r²

Given that

a3/a1 = 25

=> r² = 25

=> r = ±√25

=> r = ±5

=> r = 5 or -5

Now , a9/a5

=> a×r^(9-1)/a×r^(5-1)

=> ar⁸/ar⁴

=> r⁸/r⁴

=> r^8-4

=> r⁴

if r = 5 then r⁴ = 5⁴ = 5×5×5×5 = 625

If r = -5 then r⁴ = (-5)⁴=-5×-5×-5×-5 = 625

so, a9/a5 = 625

Answer:-

The value of a9/a5 for the given problem is 625

Used formulae:-

The General term of a GP = an = a×r^(n-1)

Where ,a = First term

r = Commo ratio

• a^m/a^n = a^(m-n)

Answer 2

$\large\underline{\sf{Solution-}}$

It is provided that

$\rm :\longmapsto\:a_1, \: a_2, \: a_3, - - - ,a_{10} \: are \: in \: GP$

and

$\rm :\longmapsto\:\dfrac{a_3}{a_1} = 25$

Let assume that

First term of GP = a

Common ratio = r

So,

nth term of GP is given by

$\rm :\longmapsto\:a_n = {ar}^{n - 1}$

Therefore,

$\rm :\longmapsto\:\dfrac{a_3}{a_1} = 25$

can be reduced to

$\rm :\longmapsto\:\dfrac{a {r}^{2} }{a} = 25$

$\bf\implies \: {r}^{2} = 25$

Now,

$\rm :\longmapsto\:\dfrac{a_9}{a_5}$

$\rm \: = \: \dfrac{ {ar}^{8} }{ {ar}^{4} }$

$\rm \: = \: {r}^{4}$

$\rm \: = \: {( {r}^{2})}^{2}$

$\rm \: = \: {25}^{2}$

$\rm \: = \: 625$

Therefore,

$\rm :\longmapsto\:\dfrac{a_9}{a_5} = 625$