Sequences and Series Questions and Answers

Answer: E

The first term is 12.

Write the terms as \(a, ar, ar^{2}, ar^{3}\). Dividing the product of four terms by the product of three leaves the fourth term:

\[ \begin{aligned} ar^{3} &= \frac{324}{216} = \frac{3}{2} \\ a^{3}r^{3} &= 216, \quad ar = 6 \\ r^{2} &= \frac{ar^{3}}{ar} \\ &= \frac{1}{4}, \quad r \\ &= \frac{1}{2} \\ a &= \frac{6}{r} = 12 \end{aligned} \]

Check: the terms are 12, 6, 3, 1.5, with products \(12 \times 6 \times 3 = 216\) and \(216 \times 1.5 = 324\). 6 is the second term \(ar\), and 3 is the third.

Answer: D

The product of the second and third terms is \(\frac{1}{32}\).

In a G.P. \(a, ar, ar^{2}, ar^{3}\), terms equidistant from the ends have equal products: \(T_{1}T_{4} = a^{2}r^{3} = T_{2}T_{3}\). So the product of all four is \((T_{2}T_{3})^{2}\):

\[ \begin{aligned} (T_{2}T_{3})^{2} &= \frac{1}{1024} \\ T_{2}T_{3} &= \frac{1}{32} \end{aligned} \]

(The positive root is taken, as the options are all positive.) The key step is the square root: \(\sqrt{1024} = 32\). \(\frac{1}{64}\), \(\frac{1}{16}\) and \(\frac{1}{128}\) are other powers of 2 whose squares are not \(\frac{1}{1024}\), and \(\frac{1}{28}\) is not a power of 2 at all.

Answer: A

The G.M. is \(8\sqrt{3}\).

The arithmetic mean gives the second number, then the geometric mean is the square root of the product:

\[ \begin{aligned} \frac{a + c}{2} &= 16, \quad c = 32 - 8 = 24 \\ \text{G.M.} &= \sqrt{ac} = \sqrt{8 \times 24} \\ &= \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3} \end{aligned} \]

A check: \(8\sqrt{3} \approx 13.9\), which is less than the A.M. 16, as the A.M.-G.M. inequality requires. \(4\sqrt{3}\) is half the answer, from using \(\sqrt{ac}/2\) or taking \(c = 6\).

Answer: C

The common ratio is 2.

With first term \(a\) and ratio \(r\):

\[ \begin{aligned} a + ar &= 12, \quad ar^{2} = 16 \\ \frac{r^{2}}{1 + r} &= \frac{16}{12} = \frac{4}{3} \\ 3r^{2} - 4r - 4 &= 0 \\ (3r + 2)(r - 2) &= 0 \end{aligned} \]

Since \(r \gt 0\), \(r = 2\), which gives \(a = 4\) and the terms 4, 8, 16. Check: \(4 + 8 = 12\) and the third term is 16. The other root, \(-\frac{2}{3}\), is excluded by the condition \(r \gt 0\), and none of the other positive options satisfies the quadratic.

Answer: B

The smallest number is 64.

Let the four terms be \(a, ar, ar^{2}, ar^{3}\). The fourth term is 8 times the first, so \(r^{3} = 8\) and \(r = 2\).

\[ \begin{aligned} a(1 + 2 + 4 + 8) &= 960 \\ 15a &= 960 \\ a &= 64 \end{aligned} \]

The terms are 64, 128, 256 and 512, which add to 960. Since the ratio is greater than 1, the first term is the smallest. The option 128 is the second term, and 240 is \(960 \div 4\), the average rather than a term.

Answer: A

The first term is \(\frac{1}{2\sqrt{6}}\).

In a geometric series the square of the middle of three consecutive terms equals the product of its neighbours:

\[ \begin{aligned} (\sqrt{2}\sin\theta)^{2} &= \cos\theta \cdot \sqrt{3}\tan\theta \\ 2\sin^{2}\theta &= \sqrt{3}\sin\theta \\ \sin\theta &= \tfrac{\sqrt{3}}{2}, \quad \theta = \tfrac{\pi}{3} \end{aligned} \]

So the second, third and fourth terms are \(\frac{1}{2}\), \(\frac{\sqrt{6}}{2}\) and 3, with common ratio \(\sqrt{6}\).

\[ \begin{aligned} \text{first term} &= \frac{1/2}{\sqrt{6}} \\ &= \frac{1}{2\sqrt{6}} \end{aligned} \]

The option \(\frac{1}{\sqrt{6}}\) forgets the factor \(\frac{1}{2}\) in \(\cos\frac{\pi}{3}\).

Answer: B

The ratio is 82:9.

Take \(A = 5k\) and \(G = 3k\). Then \(a + b = 2A = 10k\) and \(ab = G^{2} = 9k^{2}\).

\[ \begin{aligned} a^{2} + b^{2} &= (a + b)^{2} - 2ab \\ &= 100k^{2} - 18k^{2} = 82k^{2} \\ a^{2} + b^{2} : ab &= 82 : 9 \end{aligned} \]

The option 83:9 is a one-off slip in the subtraction. The condition \(a \gt b\) only fixes which root is which; it does not change \(a^{2} + b^{2}\) or \(ab\).

Answer: E

The first term is 135.

For a geometric series, \(S_{n} = \frac{a(1 - r^{n})}{1 - r}\). With \(r = \frac{1}{3}\) and n = 4:

\[ \begin{aligned} 200 &= a \cdot \frac{1 - \frac{1}{81}}{\frac{2}{3}} = a \cdot \frac{80}{81} \cdot \frac{3}{2} \\ 200 &= \frac{40a}{27} \\ a &= 135 \end{aligned} \]

Check: 135 + 45 + 15 + 5 = 200. Using the infinite-sum formula \(\frac{a}{1 - r}\) would give \(a = \frac{400}{3}\), not an option.

Answer: A

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

Write the terms as \(ar, ar^{2}, ar^{3}\):

\[ \begin{aligned} ar(1 + r) &= 8, \quad ar^{3} = 4 \\ \frac{1 + r}{r^{2}} &= 2 \\ 2r^{2} - r - 1 &= 0 \\ (2r + 1)(r - 1) &= 0 \end{aligned} \]

So r = 1 or \(r = -\frac{1}{2}\), and since \(r \neq 1\), \(r = -\frac{1}{2}\). Check: a = -32 gives terms -32, 16, -8, 4; 16 + (-8) = 8 and the fourth term is 4.

Answer: D

n = 11.

Use \(\log(5^{k}) = k\log 5\):

\[ \begin{aligned} \sum_{k=1}^{n} k\log_{10}5 &= \frac{n(n + 1)}{2}\log_{10}5 \\ \frac{n(n + 1)}{2} &= 66 \\ n(n + 1) &= 132 = 11 \times 12 \end{aligned} \]

So n = 11. Check the neighbours: n = 10 gives 55 and n = 12 gives 78, so only 11 makes the sum 66 times \(\log 5\).

The key step is the log rule \(\log(a^{k}) = k\log a\), which turns the sum of logarithms into \(\log 5\) times the sum 1 + 2 + ... + n.