Question

the sum of the first two terms of a g.p is 5/3 and the sum to infinity of the series is 3. the common ratio is

Answer 1

If G.P. is 5/3 & the series is 3 ,then.

S(infinity) = a/1-r

= 3 =a/1-r

3(1-r)= a

a-ar=5/3

3-3r+(3-3r)(r)=5/3

3^2-3r+3r-3r^2=5/3

9-9r^2=5

9-5=9r^2

= 4 = 9r^2

4/9=r^2

r^2=2/3

Answer 2

Answer:

$\text{The value of r is }\pm\frac{2}{3}$  

Step-by-step explanation:

$\text{Given that the sum of the first two terms of a g.p is }\frac{5}{3}\text{ and the sum to infinity of the series is 3. }$

we have to find the common ratio r.

The G.P series is

$a, ar, ar^2, ar^3,...$  

$\text{The sum of first two terms of G.P is }\frac{5}{3}$

$a+ar=\frac{5}{3}$

$a(1+r)=\frac{5}{3}$     →   (1)

Also the sum to infinity of the series is 3.

$S_{\infty}=3$

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

$a=3(1-r)$       →    (2)

Using 2 in 1

$3(1-r)(1+r)=\frac{5}{3}$  

$3(1-r^2)=\frac{5}{3}$  

$9-9r^2=5$

$9r^2=4$

$r^2=\frac{4}{9}$

$r=\pm\frac{2}{3}$

$\text{The value of r is }\pm\frac{2}{3}$