Question

the sum of an infinite G.P is 57 and the sum of the cubes is 9747,find the common ratio of G.P

Answer 1

hope this solves your problem out...

Answer 2

Answer: Common ratio is $\frac{3}{2}$

Step-by-step explanation:

Since we have given that

Sum of an infinite G.P. = 57

Here, a denotes first term.

r denotes common ratio.

so, it becomes,

$\frac{a}{1-r}=57\\\\\implies a=57(1-r)---------------(1)$

Sum of the cubes = 9747

i.e. sequence of g.p. will be

$a^3,a^3r^3,................$

So, it becomes,

$\frac{a^3}{1-r^3}=9747------------------(2)$

So, put the value of Eq(2) in Eq(1), we get,

$\frac{(1-r)^3}{1-r^3}=\frac{9747}{57^3}=\frac{1}{19}\\\\\frac{(1-r)(1-r)^2}{(1-r)(1+r^2+r)}=\frac{1}{19}\\\\\frac{1+r^2-2r}{1+r^2+r}=\frac{1}{19}\\\\19(1+r^2-2r)=1+r^2+r\\\\19+19r^2-38r=1+r^2+r\\\\18r^2-39r+18=0\\\\6r^2-13r+6=0\\\\6r^2-9r-4r+6=0\\\\3r(2r-3)-2(2r-3)=0\\\\(3r-2)(2r-3)=0\\\\r=\frac{2}{3},\frac{3}{2}$

r must be greater than 1.

So, Common ratio is $\frac{3}{2}$