Question

i have solved this question but I wanted to share this with OS it took me 1/2 hour
Consider two different GP with their sum S1and S2
S1=a+ar+ar^2.......infinite=1
S2=b+bR+bR^2+bR^3.....infinite=1
S1=S2=1,ar=bR and ar^2=1/8

Answer 1

ques
Sum of their first terms=?
Common ratio of first G.P=?
Common Ration of Second GP=?

Answer 2

note aris not equal to br

Answer 3

looks interesting...

Answer 4

@ganeshie8 is solved it ans1)1
Ans2)\[1-\sqrt5/4\]
Ans3)\[3-\sqrt5/4\]

Answer 5

good questions

Answer 6

guess i would start here :
\[\large \dfrac{a}{1-r} = \dfrac{b}{1-R} = 1\]

Answer 7

so u'll get equation a=1-r and substitute in ar^2

Answer 8

check that u'll get relation R+r=1

Answer 9

\(\large ar^2=1/8\)

\(\large (1-r)r^2=1/8\)

like this ?

Answer 10

in the equtaion u have wriitten

Answer 11

ya

Answer 12

how did u get R+r = 1 ?

Answer 13

ar=bR

Answer 14

R-R^2=r-r^2

Answer 15

i think u'll get using a^2-b^2

Answer 16

not sure, let me grab paper and pen lol this looks bit involved...

Answer 17

*** ( Doing this as i am doing sequences and series and so that i get notified)

Answer 18

S1=a+ar+ar^2.......infinite=1
sub a+ar from both sides
S1-(a+ar)=1-(a+ar)
S1-(a+ar)=a+ar+ar^2.......infinite= 1-(a+ar)

S1-(a+ar) = \dfrac{ar^2}{1-r} = 1-a-ar
mmm does this work ?

Answer 19

\( S_1-(a+ar) = \dfrac{ar^2}{1-r} = 1-a-ar\)