Question

The sum to infinity of G.P with common ratio r is 4 . The sum of the infinite G.P obtained by squaring the terms of this G.P is 16/3 then the first term and common ratio of given G.P is?

Answer 1

formula for sum of infinite GP is:
\[\frac{a}{1-r}\]
where a is first term and r<1
now set up 2 equations using given info
\[\frac{a}{1-r} = 4\]
\[\frac{a^{2}}{1-r} = \frac{16}{3}\]
solve for a,r using substitution

Answer 2

no the common ratio must be less than 1 since the sum converges

Answer 3

The answer should be a=2 r =1/2

Answer 4

yup @dumbcow is correct

Answer 5

that answer doesn't work for 2nd sequence ?

Answer 6

what @dumbcow ,said is correct,except a^2/(1-r^2)=16/3
did u get that??

Answer 7

oh yeah...oops the ratio is squared
sorry

Answer 8

u need explanation??i m getting r as 0.5,a=2......

Answer 9

Yup

Answer 10

a=4(r-1)
\[a^2= 16/3*(r^2-1)\]
squaring 1st equation and put a^2 in 2nd
\[16(r-1)^2=16/3*(r^2-1)\]
simplifying we get r=1/2

Answer 11

but it should be (1-r)^2

Answer 12

still u will get same result,i will be cancelling r-1 on both sides,u will be cancelling 1-r on both sides

Answer 13

i mean the second eq

Answer 14

The sum of the infinite G.P obtained by squaring the terms of this G.P is 16/3

Answer 15

@hartnn

Answer 16

a^2/(1-r)^2

Answer 17

have u tried the same thing with (1-r)???i will write again using (1-r),but both things are essentially same.....
a=4(1-r)
a2=16/3∗(1-r2)
squaring 1st equation and put a^2 in 2nd
16(1-r)2=16/3∗(1-r2)
simplifying we get r=1/2

Answer 18

it should be a^2=16/3∗(1-r)^2

Answer 19

i think

Answer 20

and no it should not be (1-r)^2
it must be (1-r^2)
because the common ratio now is r^2
instead of r

Answer 21

ok.....thxxxx

Answer 22

1st term in new series is a^2,and common ratio is r^2
and formula is (1st term)/(1-common ratio)