Home / IB DP Math AA: Topic : SL 1.8: The sum of infinite geometric sequences: IB style Questions HL Paper 1

IB DP Math AA: Topic : SL 1.8: The sum of infinite geometric sequences: IB style Questions HL Paper 1

[Maximum mark: 18]

Question:

Consider the series ln x p + + ln x + \(\frac{1}{3}In x + ….,\)  where x ∈ R, x > 1 and p ∈ R, p ≠ 0.

(a) Consider the case where the series is geometric.

(i) Show that p = ± \(\frac{1}{\sqrt{3}}\)

▶️Answer/Explanation

Ans: EITHER
attempt to use a ratio from consecutive terms

\(\frac{p In x}{In x} = \frac{\frac{1}{3}In x}{p In x} OR \frac{1}{3}In x = (In x)r^{2} OR p In x = In x \left ( \frac{1}{3p} \right )\)

Note: Candidates may use \(In x^{1} + In x^{p} + Inx^{\frac{1}{3}} + …..\) and consider the powers of x in geometric sequence. 

Award M1 for \(\frac{p}{1} = \frac{\frac{1}{3}}{p}.\)

OR

r = p  and r2 =  \(\frac{1}{3}\)

THEN

\(p^{2}= \frac{1}{3} OR r = \pm \frac{1}{\sqrt{3}}\)

\(p = \pm \frac{1}{\sqrt{3}}\)

Note: Award M0A0 for \(r^{2} = \frac{1}{3} or p^{2} = \frac{1}{3}\)   with no other working seen. 

(ii) Hence or otherwise, show that the series is convergent.

▶️Answer/Explanation

Ans: EITHER

\(since, \left | p \right |= \frac{1}{\sqrt{3}} and \frac{1}{\sqrt{3}}<1\)

OR

\(since, \left | p \right |= \frac{1}{\sqrt{3}} and -1<p<1\)

THEN

⇒the geometric series converges.

Note: Accept r instead of p .
          Award RO if both values of p not considered.

(iii) Given that p > 0 and S = 3 + \(\sqrt{3}\) , find the value of x .

▶️Answer/Explanation

Ans: \(\frac{In x}{1-\frac{1}{\sqrt{3}}} (=3 + \sqrt{3})\)

\(In x = 3 – \frac{3}{\sqrt{3}}+ \sqrt{3} – \frac{\sqrt{3}}{\sqrt{3}}\)   OR \(In x = 3 – \sqrt{3} + \sqrt{3} – 1 (\Rightarrow In x =2)\)

x = e2

(b) Now consider the case where the series is arithmetic with common difference d.

(i) Show that p = \(\frac{2}{3}\)

▶️Answer/Explanation

Ans: METHOD 1
attempt to find a difference from consecutive terms or from u2 correct equation 

\(p In x – In x = \frac{1}{3}In x – p In x OR \frac{1}{3}In x = In x +2 (p In x – In x)\)

Note: Candidates may use \(In x^{1} + In x^{p} + In x^{\frac{1}{3}} + ……\)    and consider the powers of x in arithmetic sequence.

Award M1A1 for   \(p – 1 = \frac{1}{3} – p.\)

\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)

\(p = \frac{2}{3}\)

METHOD 2
attempt to use arithmetic mean \(u_{2} = \frac{u_{1}+u_{3}}{2}\)

\(p In x = \frac{In x + \frac{1}{3}In x}{2}\)

\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)

\(p = \frac{2}{3}\)

(ii) Write down d in the form k ln x , where k ∈ R.

▶️Answer/Explanation

Ans: \(d = -\frac{1}{3}In x\)

(iii) The sum of the first n terms of the series is ln \(\left ( \frac{1}{x^{3}} \right )\)

Find the value of n.

▶️Answer/Explanation

Ans: METHOD 1

\(S_{n} = \frac{n}{2}\left [ 2 In x +(n-1)\times \left ( -\frac{1}{3}In x \right ) \right ]\)

attempt to substitute into Sn and equate to \(In \left ( \frac{1}{x^{3}} \right )\)

\(\frac{n}{2}\left \lfloor 2In x + (n-1)\times \left ( -\frac{1}{3} In x \right ) \right \rfloor = In \left ( \frac{1}{x^{3}} \right )\)

\(In\left ( \frac{1}{x^{3}} \right ) = -In x^{3}\left ( = In x^{-3} \right )\)

= −3ln x

correct working with Sn (seen anywhere)

\(\frac{n}{2}\left \lfloor 2 In x – \frac{n}{3} In x + \frac{1}{3}In x\right \rfloor OR nIn x – \frac{n(n-1)}{6}In x OR \frac{n}{2}\left ( In x + \left ( \frac{4-n}{3} \right )In x\right )\)

correct equation without ln x

\(\frac{n}{2}\left ( \frac{7}{3} – \frac{n}{3} \right ) = -3 OR n-\frac{n(n-1)}{6} = -3\) (or equivalent)

Note: Award as above if the series \(1 + p + \frac{1}{3}+ ….\)   is considered leading to \(\frac{n}{2}\left ( \frac{7}{3} -\frac{n}{3}\right ) = -3.\)

attempt to form a quadratic = 0

n2 -7n − 18 = 0

attempt to solve their quadratic

(n – 9) (n + 2) = 0 

n = 9

Question

[Maximum mark: 8]
Consider the geometric sequence 10, 5, 2.5, 1.25, …
(a) Express the general term nu in terms of n .                                                                                                                                                                           [1]

▶️Answer/Explanation

Ans: 10 x 0.5n-1 (= 20 x 0.5n )

(b) Find the first term which is smaller than 10-3 = 0.001.                                                                                                                                                     [3]

▶️Answer/Explanation

Ans: 0.000610

(c) Find the sum of the first 20 terms correct to 6 decimal places.                                                                                                                                   [2]

▶️Answer/Explanation

Ans: 19.999981

(d) Find the sum of the infinite series.

▶️Answer/Explanation

Ans: 20

Question

[Maximum mark: 6]
Find the sum of each of the following infinite geometric series

(i)  \( 1+\frac{2}{5}+\frac{4}{25}+\frac{8}{125}+…\)     

▶️Answer/Explanation

Ans: 5/3

(ii)    \(1-\frac{2}{5}+\frac{4}{25}-\frac{8}{25}+…\)

▶️Answer/Explanation

Ans: 5/7

Question

[Maximum mark: 6]
Calculate the following sums by using the appropriate formulas

(i)        \(\sum_{k=1}^{6}4^{k}\)           

▶️Answer/Explanation

Ans: \(\frac{4(4^{6-1})}{4-1}=5460\)

(ii)      \(\sum_{k=11}^{6}(0.25)^{k}\)  (correct to 6 dp)             

▶️Answer/Explanation

Ans: \(\frac{0.25(1-0.25^{6})}{1-0.25}=0.333252\)

(iii)   \(\sum_{k=11}^{+\infty }(0.25)^{k}\)

▶️Answer/Explanation

Ans: \(\frac{0.25}{1-0.25}=\frac{1}{3}\)

Question

Maximum mark: 5]
Consider the infinite geometric sequence 3, 3(0.9), 3(0.9)2, 3(0.9)3, … .
(a) Write down the 10th term of the sequence. Do not simplify your answer. [1]

▶️Answer/Explanation

Ans: \(u^{10}=3(0.9)^{9}\)

(b) Find the sum of the infinite sequence. [4]

▶️Answer/Explanation

Ans: \(S=\frac{3}{1-0.9}=\frac{3}{0.1}=30\)

Question

[Maximum mark: 6]
Consider the infinite geometric sequence 25, 5, 1, 0.2, … .
(a) Find the common ratio. [1]

▶️Answer/Explanation

Ans: \(\frac{1}{5}(=0.2)\)

(b) Find
(i)  the 10th term;           (ii) an expression for the nth term. [3]

▶️Answer/Explanation

Ans: u10 \(=25\left ( \frac{1}{5} \right )^{9}=0.0000128\)

(c) Find the sum of the infinite sequence. [2]

▶️Answer/Explanation

Ans: \(S=\frac{125}{4}=31.25\)

 

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