Question

If the sum of an infinite GP be 3 and the sum
of the squares of its term is also 3, then find
its first term and common ratio​

Answer 1

Answer:

The first term is $\frac{3}{2}$ & common ratio is $\frac{1}{2}$,.

Step-by-step explanation:

Given :

The sum of an infinite G.P is 3,

The sum of the squares of it's term is also 3,.

Find the first term & the common ratio,.

Solution :

Let the first term be a,

Let the common ratio be r,

Then,.

The G.P will be,

a , ar , ar² ,......

We know that,

Sum of an infinite G.P is given by,

⇒ $S_n = \frac{a}{1-r}$ = 3 (given)  ... (case 1)

The squared term G.P would be,

(a)² , (ar)² , (ar²)² ,.....

Their first term = a²,

Common ratio = r²

Their sum would be,

⇒ $S_n = \frac{a^2}{1-r^2}$ = 3 (given) ... (case 2)

case 1 :

$\frac{a}{1-r} = 3$

⇒ $a = 3(1-r)$ ....(i)

⇒ $a = 3 - 3r$ ...(ii)

___

case 2 :

$\frac{a^2}{1-r^2} = 3$

⇒ $a^2 = 3(1-r^2)$ ..(iii)

By dividing (iii) by (ii)

We get,

$\frac{a^2}{a} = \frac{3(1-r^2)}{(1-r)}$

⇒ $a = \frac{(1-r)(1+r)}{1-r} = 1 + r$ by (a² - b²) formula

⇒ $a = 1 + r$ ...(iv)

By equating (ii) & (iv)

We get,

⇒ $3 - 3r = 1 + r$

⇒ $3 - 1 = r + 3r$

⇒ $2 = 4r$

⇒ $r=\frac{1}{2}$

By substituting value of r in (ii),

We get,

⇒ $a = 1 + \frac{1}{2}$

⇒ $a = \frac{2+1}{2} = \frac{3}{2}$

∴ The first term is $\frac{3}{2}$ & common ratio is $\frac{1}{2}$,.

Answer 2

Step-by-step explanation:

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