OA: A
point 1: a + ar + ar^2.........infinity = a/ (1-r) = 27 -----(1)
point 2: a^2 + a^2 r^2......infinity = a^2/(1-r^2) = 243
point 2 can be written as a/(1-r) * a/(1+r) = 243
27 * a/(1+r) = 243
so, a/(1+r) = 9 ------(2)
divide (1) by (2)
(1+r)/(1-r) = 3
on solving this you will get r = 1/2 = common ratio
1+r = 3-3r
4r = 2
r = 1/2
Sequences
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Another option:
a + ar + ar² + ... = 27
a² + a²r² + a²r� + ... = 243
so ...
a * (1 + r + r² + ...) = 27
a² * (1 + r² + r� + ...) = 243
We know that if r < 1, we can write (1 + r + r² + ...) as 1 / (1 - r).
Assuming this to be true (otherwise the series diverges), we can write the first series as
a/(1 - r) = 27
In the second GP, since we have ((r²)� + (r²)¹ + (r²)² + ...), our ratio is r², so we have (1 + r² + r� + ...) = 1 / (1 - r²), and
a²/(1 - r²) = 243
Using the difference of squares, this gives us
a/(1 + r) = 9
So we have a/(1 + r) = 9 and a/(1 - r) = 27, or
9(1 + r) = 27(1 - r), or r = 1/2
a + ar + ar² + ... = 27
a² + a²r² + a²r� + ... = 243
so ...
a * (1 + r + r² + ...) = 27
a² * (1 + r² + r� + ...) = 243
We know that if r < 1, we can write (1 + r + r² + ...) as 1 / (1 - r).
Assuming this to be true (otherwise the series diverges), we can write the first series as
a/(1 - r) = 27
In the second GP, since we have ((r²)� + (r²)¹ + (r²)² + ...), our ratio is r², so we have (1 + r² + r� + ...) = 1 / (1 - r²), and
a²/(1 - r²) = 243
Using the difference of squares, this gives us
a/(1 + r) = 9
So we have a/(1 + r) = 9 and a/(1 - r) = 27, or
9(1 + r) = 27(1 - r), or r = 1/2
