# Geometric Sequence Question?

the first two terms of a geometric progression are 1 and 1/3 tan^2 theta respectively, where 0<theta<1/2 pi

i) find the set of values of theta for which the progression is convergent.

ii) find the exact value of sum to infinity when theta=1/6 pi (ii is easy as it involves substituting the values in the sum to infinity equation, but its tan^2 theta i dont know how to calculate that.)

### 1 Answer

- 2 months agoFavorite Answer
a * r = 1

a * r^2 = (1/3) * tan(t)^2

r = (1/3) * tan(t)^2

S = ar + ar^2 + ar^3 + .... + ar^n

Sr = ar^2 + ar^3 + ... + ar^n + ar^(n + 1)

S - Sr = ar - ar^(n + 1)

S * (1 - r) = ar * (1 - r^n)

S = ar * (1 - r^n) / (1 - r)

S = 1 * (1 - (tan(t)^2 / 3)^n) / (1 - (tan(t)^2 / 3))

If -1 < r < 1, and there are an infinite number of terms, then the sum is convergent

-1 < (1/3) * tan(t)^2 < 1

-3 < tan(t)^2 < 3

tan(t)^2 is always positive or 0, so this can be rewritten:

0 </= tan(t)^2 < 3

-sqrt(3) < tan(t) </= 0 and 0 </= tan(t) < sqrt(3)

or

-sqrt(3) < tan(t) < sqrt(3)

arctan(-sqrt(3)) < t < arctan(sqrt(3))

-pi/3 + pi * k < t < pi/3 + pi * k

k is an integer

-pi/3 < t < pi/3

2pi/3 < t < 4pi/3

5pi/3 < t < 7pi/3

8pi/3 < t < 10pi/3

and so on

t = pi/6

S = 1 * (1 - (tan(t)^2 / 3)^n) / (1 - (tan(t)^2 / 3))

S = 1 * (1 - (tan(pi/6)^2 / 3)^n) / (1 - (tan(pi/6)^2 / 3))

S = (1 - (sqrt(3)/9)^n) / (1 - (sqrt(3)/9))

(sqrt(3)/9)^inf = 0, so this simplifies to:

S = (1 - 0) / ((9 - sqrt(3)) / 9)

S = 1 / ((9 - sqrt(3)) / 9)

S = 9 / (9 - sqrt(3))

S = 9 * (9 + sqrt(3)) / (81 - 3)

S = 9 * (9 + sqrt(3)) / 78

S = 3 * (9 + sqrt(3)) / 26

S = (27 + 3 * sqrt(3)) / 26

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