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Infinite Sum of a GP - Questions

Question 1
124479

Consider the geometric series \(2+\dfrac{2}{3}+\dfrac{2}{9}+\dfrac{2}{27}+\cdots\)

i) Explain why the series has a limiting sum.

ii) Find the limiting sum.

i) as \(-1<r\left(\dfrac{1}{3}\right)<1\), the series will have a limiting sum.

ii) \(3\)

Question 2
11993

Find the difference between the limiting sum and the sum to the first 6 terms of 256 + 128 + 64 + ...

\(8\)

Question 3
12010

The limiting sum of \(1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + \) .... is?

1.5

Question 4
12011

The limiting sum of \(1 - \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} + \) ... is?

\(\dfrac{2}{3}\)

Question 5
12012

The sum to infinity of a G.P. is three times the first term. The common ratio is?

\(r = \dfrac{2}{3}\)

Question 6
12013

The sum to infinity of a G.P. is 4 and the 2nd term is 1. The first term is?

\(a = 2\)

Question 7
12014

The 2nd term of a G.P. is 24 and its sum to infinity is 100. The two possible values of the common ratio are?

\(\dfrac{3}{5}{\rm{ or }}\dfrac{1}{5}\)

Question 8
24684

The first three terms of a geometric series are \(\dfrac{2}{\omega },\,\dfrac{2}{{{\omega ^2}}},\,\dfrac{2}{{{\omega ^3}}}\). Given that \(\omega > 1\)

(i)  Find an expression in \(\omega \) for the limiting sum.

(ii)  Find the value of \(\omega \) when the limiting sum is \(\omega \)

(i)  \({S_\infty } = \dfrac{2}{{\omega - 1}}\)

(ii)  Since \(\omega > 1\) then \(\omega = 2\) is the only valid solution

Question 9
24685

For the geometric series \(1 - 3x + 9{x^2} - ...\)

(i)  For what values of x will the limiting sum exist?

(ii)  Find the limiting sum when \(x = \dfrac{1}{4}\)

(i)  \( - \dfrac{1}{3} < x < \dfrac{1}{3}\)

(ii)  \({S_\infty } = \dfrac{4}{7}\)

Question 10
24686

The first term of a geometric series is 25 and the fourth term is \(\dfrac{1}{5}\)

(i)  Find the common ratio.

(ii)  Find the limiting sum of the series.

(i)  \(r = \dfrac{1}{5}\)

(ii)  \({S_\infty } = 31.25\)

Question 11
24687

Consider the geometric series \(1 - (\sqrt 3 - 1) + {(\sqrt 3 - 1)^2} - ...\)

(i) Explain why the geometric series has a limiting sum.

(ii)  Find the exact value of the limiting sum. Write your answer with a rational denominator.

(i)  \( - 1 < r < 1\) therefore, G.P has a limiting sum

(ii)  \({S_\infty } = \dfrac{\sqrt 3}{3}  \)

Question 12
25870

For the geometric series: \((\sqrt 3 + \sqrt 2 ) + (\sqrt 3 - \sqrt 2 ) + \dfrac{{{{(\sqrt 3 - \sqrt 2 )}^2}}}{{\sqrt 3 + \sqrt 2 }} + \ldots \) find

(i) the common ratio (1 mark)

(ii) the sum of the infinite series (3 marks)

(i) \(r = 5 - 2\sqrt 6 \)

(ii) \(\dfrac{1}{4}\{ 4\sqrt 3 + 5\sqrt 2 \} \)

Question 13
21383

The sum to infinity of a geometric series is 16 and the second term is 4. The first term is?

\(a=8\)