Sequences and Series Questions and Answers

Answer: A

The common ratio is \(\frac{1}{2}\).

The twentieth term is \(ar^{19}\).

\[ \begin{aligned} 512\,r^{19} &= \frac{1}{1024} \\ r^{19} &= \frac{1}{2^{9} \times 2^{10}} = \frac{1}{2^{19}} \\ r &= \frac{1}{2} \end{aligned} \]

Since 19 is odd, the real root is unique, so -1/2 is not possible. A ratio of 2 or 4 would make the terms grow, not shrink, and 1/4 or 1/8 would make the twentieth term far smaller than 1/1024.

Answer: D

The possible ratios are \(\frac{3}{2}\) and \(-\frac{2}{3}\).

For three terms of a G.P., the middle term squared equals the product of the outer two.

\[ \begin{aligned} 36 &= k(k + 5) \\ k^{2} + 5k - 36 &= 0 \\ (k + 9)(k - 4) &= 0 \end{aligned} \]

If k = 4, the terms are 4, 6, 9 and r = \(\frac{6}{4} = \frac{3}{2}\). If k = -9, the terms are -9, 6, -4 and r = \(\frac{6}{-9} = -\frac{2}{3}\).

Option E inverts both ratios, and the others do not satisfy \(6^{2} = k(k + 5)\) for any k.

Answer: E

There are 7 terms.

The sequence is a G.P. with first term 2 and common ratio 3, so the \(n\)th term is \(2 \cdot 3^{n-1}\).

\[ \begin{aligned} 2 \cdot 3^{n-1} &= 1458 \\ 3^{n-1} &= 729 = 3^{6} \\ n &= 7 \end{aligned} \]

Check by listing: 2, 6, 18, 54, 162, 486, 1458, which is seven terms. 8 counts one too many, and the larger options treat the sequence as arithmetic.

Answer: C

The odd-numbered terms \(t_1, t_3, t_5, \ldots, t_{2n-1}\) are in G.P. with common ratio \(r^{2}\).

Each term is \(t_k = t_1 r^{k-1}\). Moving from one odd-numbered term to the next skips two places:

\[ \begin{aligned} \frac{t_{k+2}}{t_k} &= \frac{t_1 r^{k+1}}{t_1 r^{k-1}} \\ &= r^{2} \end{aligned} \]

So the ratio is \(r^{2}\), not \(r\), which rules out the first option. Terms \(t_1, t_4, t_7, \ldots\) skip three places and have ratio \(r^{3}\), not \(r^{2}\). The even-numbered terms also have ratio \(r^{2}\), not \(r^{3}\) or \(r^{5}\).

Answer: D

\(n = \frac{-1}{2}\).

The geometric mean of 4 and 16 is \(\sqrt{4 \times 16} = 8\). Write \(a = 4^{n}\), so that \(16^{n} = a^{2}\):

\[ \begin{aligned} \frac{4a + 16a^{2}}{a + a^{2}} &= 8 \\ 4 + 16a &= 8 + 8a \quad (a \gt 0) \\ a &= \frac{1}{2} \end{aligned} \]

Then \(4^{n} = \frac{1}{2}\), so \(2^{2n} = 2^{-1}\) and \(n = -\frac{1}{2}\).

Check: \(4^{1/2} + 16^{1/2} = 6\) and \(4^{-1/2} + 16^{-1/2} = \frac{3}{4}\), and \(6 \div \frac{3}{4} = 8\). The option \(\frac{1}{2}\) has the right size but the wrong sign; it gives \(\frac{8 + 64}{2 + 4} = 12\).

Answer: B

The common ratio is 2.

For a G.P. with first term \(a\), last term \(l\) and ratio \(r\), the sum is \(S = \frac{lr - a}{r - 1}\). With \(a = 7\), \(l = 448\), \(S = 889\):

\[ \begin{aligned} \frac{448r - 7}{r - 1} &= 889 \\ 448r - 7 &= 889r - 889 \\ 441r &= 882 \Rightarrow r = 2 \end{aligned} \]

Check: 7, 14, 28, 56, 112, 224, 448 add up to 889. A ratio of 4 gives 7, 28, 112, 448 with sum 595, and ratios below 1 cannot go from 7 up to 448.

Answer: B

\(\left(\frac{t_{4}}{t_{2}}\right)^{3}\) equals \(\left(\frac{t_{7}}{t_{4}}\right)^{2}\).

In a G.P. with common ratio \(r\), \(\frac{t_{m}}{t_{n}} = r^{m - n}\). So

\[ \begin{aligned} \left(\frac{t_{4}}{t_{2}}\right)^{3} &= (r^{2})^{3} = r^{6} \\ \left(\frac{t_{7}}{t_{4}}\right)^{2} &= (r^{3})^{2} = r^{6} \end{aligned} \]

The others give different powers: \(\left(\frac{t_{7}}{t_{3}}\right)^{2} = r^{8}\), \(\left(\frac{t_{6}}{t_{4}}\right)^{2} = r^{4}\), \(\left(\frac{t_{7}}{t_{2}}\right)^{2} = r^{10}\) and \(\left(\frac{t_{9}}{t_{8}}\right)^{2} = r^{2}\).

Answer: E

The common ratio is \(\dfrac{1}{100}\).

Rewrite the sum in the standard form \(S_{n} = \frac{a(1 - r^{n})}{1 - r}\):

\[ \begin{aligned} s_{n} &= 100 - 100 \cdot 100^{-n} \\ &= 100\left(1 - \left(\tfrac{1}{100}\right)^{n}\right) \end{aligned} \]

The term raised to the power \(n\) is the common ratio, so \(r = \frac{1}{100}\). Check: \(a = s_{1} = 100 - 1 = 99\), and \(\frac{a}{1 - r} = \frac{99}{0.99} = 100\), matching the factor in front. The other fractions do not reproduce \(s_{1} = 99\).

Answer: A

The value is \((6 + a_{2})^{4}\).

Since \(a_{3} = 6\), the bracket factorises:

\[ \begin{aligned} 6a_{1} + 6a_{2} + a_{1}a_{4} + a_{2}a_{4} &= (a_{1} + a_{2})(6 + a_{4}) \\ &= (a_{1} + a_{2})(a_{3} + a_{4}) \end{aligned} \]

In a G.P., \(a_{2} = a_{1}r\), \(a_{3} = a_{2}r\) and \(a_{4} = a_{3}r\), so \(a_{3} + a_{4} = r^{2}(a_{1} + a_{2})\) and \(a_{2} + a_{3} = r(a_{1} + a_{2})\). Hence

\[ \begin{aligned} (a_{1} + a_{2})(a_{3} + a_{4}) &= r^{2}(a_{1} + a_{2})^{2} \\ &= (a_{2} + a_{3})^{2} \end{aligned} \]

Squaring gives \((a_{2} + a_{3})^{4} = (6 + a_{2})^{4}\). The other options pair 6 with the wrong neighbour.

Answer: A

\(g_{1}^{3} + g_{2}^{3}\) equals 160.

Inserting two geometric means makes 2, \(g_{1}\), \(g_{2}\), 8 a G.P., so \(2r^{3} = 8\) and \(r^{3} = 4\).

\[ \begin{aligned} g_{1}^{3} &= (2r)^{3} = 8 \times 4 = 32 \\ g_{2}^{3} &= (2r^{2})^{3} = 8 \times 16 = 128 \\ g_{1}^{3} + g_{2}^{3} &= 160 \end{aligned} \]

There is no need to find \(r = 4^{1/3}\) itself; cubing the means turns \(r^{3}\) and \(r^{6}\) into 4 and 16. A neat check: \(g_{1}g_{2} = 2 \times 8 = 16\).