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 G.P. is $5,~~0,~~0,~~.~~.~~.$ or $\dfrac{15}{4},~~\dfrac{15}{16},~~\dfrac{15}{64},~~.~~.~~.$

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

We are given to find the G.P.

Let a be the first term and r be the common ratio of the given geometric progression.

Then, according to the given information, we have

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

and

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

Squaring equation (i) on both sides and comparing with equation (ii), we get

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

If r = 1, then from equation (i), we get

$a=5-5\times1=0.$

If $r=\dfrac{1}{4},$ then from equation (i), we get

$a=5-5\times\dfrac{1}{4}=\dfrac{15}{4}.$

Therefore, the required G.P. is

$5,~~5\times0,~~5\times 0^2,~~.~~.~~.~~=5,~~0,~~0,~~.~~.~~.$

or

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

Thus, the G.P. is $5,~~0,~~0,~~.~~.~~.$ or $\dfrac{15}{4},~~\dfrac{15}{16},~~\dfrac{15}{64},~~.~~.~~.$

Answer 2

Answer:15/4,15/16,15/64

Step-by-step explanation:

Solution:

Sum of an infinite =5..............(given)

Sum of the square of those term

That Is:

t1^2+t2^2+t3^2=15

a^2+a^2r^2+a^2r^4=15

We know

Sum of the infinity=5

Sum of the infinity exist for r^2

Therefore a^2*1/1-r^2=15................1

Sum of the infinity =5

Sum of the infinity =a/1-r.............formula

a/1-r=15

a=5(1-r)......................2

Put eq 2 in 1

[5(1-r)]^2/1-r^2=15

5(1-r)^2/(1-r)(1+r)=3

5(1-r)/(1+r)=3

5-5r=3+3r

5-3=5r+3r

2=8r

r=1/4

Put 1/4 in eq 2

a=5(1-1/4)

a=5(3/4)

a=15/4

Value of a=15/4 and r=1/4

Therefore G.P is given by

a,ar,ar^2

15/4,15/16,15/64

Finial answer is

15/4,15/16,15/64