Question

the sum of an infinite G.P is 5. The sum of the squares of these terms is 15. Find the G.p​

Answer 1

Answer:  The required G.P. is

$\dfrac{15}{4},\dfrac{15}{16},\dfrac{15}{64},\dfrac{15}{256},~~.~~.~~.$

Step-by-step explanation:  Given that the sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15.

We are to find the G.P.

Let a be the first term and r be the common ratio of the G.P.

Then, the G.P. will be a, ar, ar², ar³,  .  .  .

According to the given information, we have

$\dfrac{a}{1-r}=5\\\\\Rightarrow a=5(1-r)\\\\\Rightarrow a^2=25(1-r)^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and the sum of the squares of the terms is 15 So, we have

$\dfrac{a^2}{1-r^2}=15\\\\\Rightarrow a^2=15(1-r^2)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Comparing equations (i) and (ii), we get

$25(1-r)^2=15(1-r^2)\\\\\Rightarrow 5(1-2r+r^2)=3(1-r^2)\\\\\Rightarrow 5-10r+5r^2=3-3r^2\\\\\Rightarrow 8r^2-10r+2=0\\\\\Rightarrow 4r^2-5r+1=0\\\\\Rightarrow 4r^2-4r-r+1=0\\\\\Rightarrow 4r(r-1)-1(r-1)=0\\\\\Rightarrow (4r-1)(r-1)=0\\\\\Rightarrow 4r-1=0,~~~~r-1=0\\\\\Rightarrow r=\dfrac{1}{4},1.$

Since |r| should be less than 1, so r ≠ 1.

So, the common ratio of the G.P. is

$r=\dfrac{1}{4}.$

From equation (i), we get

$\dfrac{a}{1-\frac{1}{4}}=5\\\\\\\Rightarrow a=\dfrac{3}{4}\times5\\\\\\\Rightarrow a=\dfrac{15}{4}.$

Therefore, the G.P. is

$\dfrac{15}{4},\dfrac{15}{4}\times\dfrac{1}{4},\dfrac{15}{4}\times\left(\dfrac{1}{4}\right)^2,\dfrac{15}{4}\times\left(\dfrac{1}{4}\right)^3,~~.~~.~~.\\\\\\\\=\dfrac{15}{4},\dfrac{15}{16},\dfrac{15}{64},\dfrac{15}{256},~~.~~.~~.$

Thus, the required G.P. is

$\dfrac{15}{4},\dfrac{15}{16},\dfrac{15}{64},\dfrac{15}{256},~~.~~.~~.$