Geometric Sequence and Series

(a) First term and common ratio

\(u_1 = 5\)

\(r = \frac{15}{5} = 3\)

(b) Find the \(8^\text{th}\) term

Use the formula: \(u_n = u_1 \cdot r^{n – 1}\)

\(u_8 = 5 \cdot 3^{7}\)

\(u_8 = 5 \cdot 2187\)

\(u_8 = 10935\).

(c) Find the sum of the first 8 terms

Use the formula: \(S_n = u_1 \cdot \frac{r^n – 1}{r – 1}\)

\(S_8 = 5 \cdot \frac{3^8 – 1}{3 – 1}\)

\(S_8 = 5 \cdot \frac{6561 – 1}{2}\)

\(S_8 = 5 \cdot \frac{6560}{2}\)

\(S_8 = 5 \cdot 3280\)

\(S_8 = 16400\)

(d) General term

\(u_n = u_1 \cdot r^{n – 1}\)

So, \(u_n = 5 \cdot 3^{n – 1}\)

**(e) Find \(n\) such that \(u_n = 3645\)

\(u_n = 5 \cdot 3^{n – 1} = 3645\)

Divide both sides by 5: \(3^{n – 1} = \frac{3645}{5} = 729\)

Now solve: \(3^{n – 1} = 729 = 3^6\)

So, \(n – 1 = 6 \Rightarrow n = 7\)

(a) First term and common ratio

\(u_1 = 81\)

\(r = \frac{27}{81} = \frac{1}{3}\)

(b) Find the 8th term

\(u_n = u_1 \cdot r^{n-1}\)

\(u_8 = 81 \cdot \left(\frac{1}{3}\right)^{7}\)

\(u_8 = 81 \cdot \frac{1}{2187}\)

\(u_8 = \frac{81}{2187}\)

\(u_8 = \frac{1}{27}\)

Final Answer: \(u_8 = \frac{1}{27}\)

(c) Sum of the first 8 terms

\(S_n = u_1 \cdot \frac{1 – r^n}{1 – r}\)

\(S_8 = 81 \cdot \frac{1 – \left(\frac{1}{3}\right)^8}{1 – \frac{1}{3}}\)

\(S_8 = 81 \cdot \frac{1 – \frac{1}{6561}}{\frac{2}{3}}\)

\(S_8 = 81 \cdot \frac{\frac{6560}{6561}}{\frac{2}{3}}\)

\(S_8 = 81 \cdot \frac{6560}{6561} \cdot \frac{3}{2}\)

\(S_8 = \frac{81 \cdot 3 \cdot 6560}{2 \cdot 6561}\)

\(S_8 = \frac{159120}{4374}\)

\(S_8 \approx 36.38\)

Final Answer: \(S_8 \approx 36.38\)

(d) General term

\(u_n = u_1 \cdot r^{n-1}\)

\(u_n = 81 \cdot \left(\frac{1}{3}\right)^{n-1}\)

(e) Find \(n\) when \(u_n = 1\)

\(81 \cdot \left(\frac{1}{3}\right)^{n-1} = 1\)

\(\left(\frac{1}{3}\right)^{n-1} = \frac{1}{81}\)

\(\left(\frac{1}{3}\right)^{n-1} = \left(\frac{1}{3}\right)^4\)

So, \(n – 1 = 4\)

\(n = 5\)

Final Answer: \(n = 5\)

(f) Infinite sum and convergence

Since the common ratio is \(r = \frac{1}{3}\) and \(|r| < 1\), the geometric series converges.

Sum to infinity:

\(S_\infty = \frac{u_1}{1 – r}\)

\(S_\infty = \frac{81}{1 – \frac{1}{3}}\)

\(S_\infty = \frac{81}{\frac{2}{3}}\)

\(S_\infty = 121.5\)

Final Answer: \(S_\infty = 121.5\)

(a) First 4 terms

\(u_1 = 13\)

\(u_2 = 8 \cdot u_1 = 8 \cdot 13 = 104\)

\(u_3 = 8 \cdot u_2 = 8 \cdot 104 = 832\)

\(u_4 = 8 \cdot u_3 = 8 \cdot 832 = 6656\)

First 4 terms: \(13,\ 104,\ 832,\ 6656\)

(b) General term \(u_n\)

We observe this is a geometric sequence with:

• First term \(u_1 = 13\)

• Common ratio \(r = 8\)

General term of a geometric sequence: \(u_n = u_1 \cdot r^{n-1}\)

So, \(u_n = 13 \cdot 8^{n-1}\)

Final Answer: \(u_n = 13 \cdot 8^{n-1}\)

Let the first term be \(a\) and the common ratio be \(r\).

We are given:

\(u_2 = ar = 2\)

\(u_4 = ar^3 = 0.5\)

(a) Find \(r\) and \(a\)

Divide the equations:

\(\frac{ar^3}{ar} = \frac{0.5}{2}\)

\(r^2 = 0.25\)

\(r = 0.5 \quad \text{(since all terms are positive)}\)

Substitute into \(ar = 2\):

\(a \cdot 0.5 = 2\)

\(a = 4\)

(b) Find the sum to infinity

The formula for the sum to infinity (when \(|r| < 1\)) is: \(S_\infty = \frac{a}{1 – r}\)

Substitute the values:

\(S_\infty = \frac{4}{1 – 0.5}\)

\(S_\infty = \frac{4}{0.5}\)

\(S_\infty = 8\)

Final Answers: • \(a = 4,\ r = 0.5\) ; • \(S_\infty = 8\)

Let the first term be \(a\) and the common ratio be \(r\).

We are given: \(u_3 = ar^2 = 7.16\) ;  \(u_6 = ar^5 = 57.28\)

(a) Find the common ratio

Divide the equations:

\(\frac{ar^5}{ar^2} = \frac{57.28}{7.16}\)

\(r^3 = 8\)

\(r = 2\)

(b) Find the first term

Substitute \(r = 2\) into \(ar^2 = 7.16\):

\(a \cdot 2^2 = 7.16\)

\(a \cdot 4 = 7.16\)

\(a = \frac{7.16}{4}\)

\(a = 1.79\)

(c) Find the sum of the first 20 terms

Use the geometric series sum formula: \(S_n = \frac{a(1 – r^n)}{1 – r}\)

Substitute values:

\(S_{20} = \frac{1.79(1 – 2^{20})}{1 – 2}\)

\(S_{20} = \frac{1.79(1 – 1048576)}{-1}\)

\(S_{20} = \frac{1.79(-1048575)}{-1}\)

\(S_{20} = 1.79 \cdot 1048575\)

\(S_{20} \approx 1876987.25\)

Final Answers: \(r = 2\) ; \(a = 1.79\) ;  \(S_{20} \approx 1876987 \) (to the nearest whole number)

(a) Show that \(k^2 – 7k – 60 = 0\)

For a geometric series, the ratio of consecutive terms is constant. So:

\(\frac{k}{k + 4} = \frac{2k – 15}{k}\)

Cross-multiply:

\(k \cdot k = (k + 4)(2k – 15)\)

\(k^2 = 2k^2 – 15k + 8k – 60\)

\(k^2 = 2k^2 – 7k – 60\)

Bring all terms to one side:

\(k^2 – 2k^2 + 7k + 60 = 0\)

\(-k^2 + 7k + 60 = 0\)

Multiply both sides by \(-1\):  \(k^2 – 7k – 60 = 0\)

(b) Hence show that \(k = 12\)

Solve the quadratic: \(k^2 – 7k – 60 = 0\)

Factor: \((k – 12)(k + 5) = 0\)

So, \(k = 12\) or \(k = -5\)

But \(k\) is a positive constant, so: \(k = 12\) ✅

(c) Find the common ratio

First three terms:

\(u_1 = k + 4 = 16\)

\(u_2 = k = 12\)

\(u_3 = 2k – 15 = 24 – 15 = 9\)

Common ratio:

\(r = \frac{u_2}{u_1} = \frac{12}{16} = \frac{3}{4}\). or

\(r = \frac{u_3}{u_2} = \frac{9}{12} = \frac{3}{4}\)

(d) Find the sum to infinity

Use the formula: \(S_\infty = \frac{a}{1 – r}\)

\(a = u_1 = 16\), \(r = \frac{3}{4}\)

\(S_\infty = \frac{16}{1 – \frac{3}{4}}\)

\(S_\infty = \frac{16}{\frac{1}{4}}\)

\(S_\infty = 16 \cdot 4 = 64\)

Final Answers:  • \(k = 12\) ; • \(r = \frac{3}{4}\) ; • \(S_\infty = 64\)

(a) Find the 4th term

\(u_1  = -272\)

\(u_2 =  -\frac{1}{2} \cdot u_1  =  -\frac{1}{2} \cdot (-272)  =  136\)

\(u_3 =  -\frac{1}{2} \cdot u_2  =  -\frac{1}{2} \cdot 136  =  -68\)

\(u_4  =  -\frac{1}{2} \cdot u_3  =  -\frac{1}{2} \cdot (-68)  =  34\)

(b) Type of sequence

Each term is obtained by multiplying the previous term by \(-\frac{1}{2}\).

Therefore, this is a geometric sequence with common ratio \(r = -\frac{1}{2}\).

(c) Convergence and sum to infinity

For a geometric series to converge, \(|r| < 1\).

Here, \(|r| = \frac{1}{2} < 1\), so the series converges.

Use the sum to infinity formula:

\(S_{\infty} = \frac{u_1}{1 – r}\)

\(S_{\infty} = \frac{-272}{1 – (-\frac{1}{2})}\)

\(S_{\infty} = \frac{-272}{1 + \frac{1}{2}}  =  \frac{-272}{\frac{3}{2}}  =  -272 \cdot \frac{2}{3} =  -\frac{544}{3}\)

Final Answer: • \(u_4 = 34\) ; • Type: Geometric sequence ; • \(S_{\infty} = {-\frac{544}{3}}v\)

(a) Value after 6 years

Let \(V_n\) represent the value of the laptop in year \(n\).

\(V_0 = 24000\)

This is a geometric sequence with \(r = 0.82\)

So, \(V_6 = 24000 \cdot (0.82)^6\)

\(V_6 = 24000 \cdot 0.32740741\)

\(V_6 = £7,857.79\)

**(b) Find \(n\) such that \(V_n < 5000\)

We solve: \(24000 \cdot (0.82)^n < 5000\)

Divide both sides by 24000: \((0.82)^n < \frac{5}{24}\)

Take logarithms:

\(\log(0.82)^n < \log\left(\frac{5}{24}\right)\)

\(n \log(0.82) < \log(0.2083)\)

\(n > \frac{\log(0.2083)}{\log(0.82)}\)

\(n > \frac{-0.6812}{-0.0864} \approx 7.88\)

Therefore, \(n = 8\)

(c) Cost of warranty in 8th year

This is a geometric sequence with: \(a = 500,\quad r = 1.12\)

\(C_8 = 500 \cdot (1.12)^{7}\)

\(C_8 = 500 \cdot 2.2107 = £1,105.35\)

(d) Total cost for first 15 years

Sum of first \(n\) terms in a geometric sequence:

\(S_{15} = \frac{a(1 – r^{15})}{1 – r}\)

\(S_{15} = \frac{500(1 – 1.12^{15})}{1 – 1.12}\)

\(S_{15} = \frac{500(1 – 5.473)}{-0.12} = \frac{500(-4.473)}{-0.12}\)

\(S_{15} = £18,637.50\)

(a) Finding \(r\)
Using the formula \(u_2 = u_1 \cdot r\):

\(3.52 = 3.2 \cdot r\)

\(r = \frac{3.52}{3.2} = 1.1\)

(b) Finding \(u_{10}\)

Use the geometric term formula: \(u_{10} = u_1 \cdot r^{9} = 3.2 \cdot (1.1)^9\)

\(u_{10} \approx 3.2 \cdot 2.35795 \approx 7.5454\)

\(u_{10} \approx 7.55\)

(c) Find the least \(n\) such that \(S_n > 8000\)

Use the geometric sum formula: \(S_n = \frac{u_1 (r^n – 1)}{r – 1}\)

Substitute values:

\(\frac{3.2(1.1^n – 1)}{0.1} > 8000\)

\(3.2(1.1^n – 1) > 800\)

\(1.1^n – 1 > 250\)

\(1.1^n > 251\)

Take logarithms:

\(n \log(1.1) > \log(251)\)

\(n > \frac{\log(251)}{\log(1.1)} \approx \frac{2.3997}{0.0414} \approx 57.94\)

So, the least integer \(n = 58\)

Final answers:

(a) \(r = 1.1\)

(b) \(u_{10} \approx 7.55\)

(c) \(n = 58\)

 (a) Find the common ratio

Use the logarithmic identity:
\(\ln(x^n) = n \ln x\)

So the terms become:
\(u_1 = 12 \ln x\),
\(u_2 = 6 \ln x\),
\(u_3 = 3 \ln x\)

Now find the common ratio:
\(r = \frac{u_2}{u_1} = \frac{6 \ln x}{12 \ln x} = \frac{1}{2}\)

(b) Evaluate the infinite sum

The expression is:
\(\sum_{k=1}^\infty 2^{6-k} \ln x\)

This is a geometric series with first term:
\(u_1 = 2^5 \ln x = 32 \ln x\)

Common ratio:
\(r = \frac{2^{5}}{2^{6}} = \frac{1}{2}\)

Apply the infinite sum formula:
\(S_\infty = \frac{u_1}{1 – r} = \frac{32 \ln x}{1 – \frac{1}{2}} = \frac{32 \ln x}{\frac{1}{2}} = 64 \ln x\)

Set equal to 96:
\(64 \ln x = 96\)
\(\ln x = \frac{96}{64} = 1.5\)

Solve for \(x\):
\(x = e^{1.5}\)

Final Answers:

• (a) \(r = \frac{1}{2}\)
• (b) \(x = e^{1.5}\)