Sequences and Series Questions and Answers

Answer: C

The fifth term is \(\frac{c^{2}}{a}\).

With first term a and ratio r, the third term is \(c = ar^{2}\) and the fifth is \(ar^{4}\).

\[ ar^{4} = \frac{(ar^{2})^{2}}{a} = \frac{c^{2}}{a} \]

Equivalently, the 1st, 3rd and 5th terms are themselves in G.P. with ratio \(r^{2}\), so \(c^{2} = a \times T_{5}\). The options with 2a or 3a in the denominator, or c instead of \(c^{2}\), do not satisfy this; for example with a = 1, r = 2 the terms are 1, 2, 4, 8, 16 and \(\frac{c^{2}}{a} = 16\).

Answer: C

The value is n = 10.

The sum of n terms of a G.P. with first term a and ratio r is \(\frac{a(r^{n} - 1)}{r - 1}\).

\[ \begin{aligned} \frac{1(2^{n} - 1)}{2 - 1} &= 1023 \\ 2^{n} &= 1024 = 2^{10} \end{aligned} \]

So n = 10. n = 11 would give 2047 and n = 9 would give 511; the slip is usually to take \(2^{n} = 1023\) and round up.

Answer: D

The common ratio is \(\left(\frac{n}{l}\right)^{1/4}\).

Inserting three geometric means gives the five-term G.P. \(l, G_{1}, G_{2}, G_{3}, n\), so n is the fifth term and is reached after four ratios.

\[ n = lr^{4} \Rightarrow r = \left(\frac{n}{l}\right)^{1/4} \]

The exponent 1/3 counts the three means instead of the four gaps between five terms, and 1/2 is the ratio for a single mean. \(\frac{n}{l}\) and its square are the ratio of the end terms, not of consecutive terms.

Answer: D

\(a = \frac{1}{2}\).

The second term is \(ar = \frac{3}{4}\). The product of the first four terms is \(a \cdot ar \cdot ar^{2} \cdot ar^{3} = a^{4}r^{6}\), which can be written using \(ar\):

\[ \begin{aligned} a^{4}r^{6} &= \frac{(ar)^{6}}{a^{2}} = \frac{3^{6}}{4^{6}a^{2}} \\ \frac{3^{6}}{4^{6}a^{2}} &= \frac{3^{6}}{4^{5}} \Rightarrow a^{2} = \frac{1}{4} \end{aligned} \]

With \(r \gt 0\) and \(ar \gt 0\), \(a = \frac{1}{2}\) (then \(r = \frac{3}{2}\)). The other values do not satisfy the product condition.

Answer: E

\(\alpha^{2} + \beta^{2}\) equals \(2\alpha\beta\).

If \(\lambda\) is both the A.M. and the G.M. of \(\alpha\) and \(\beta\), then

\[ \begin{aligned} \frac{\alpha + \beta}{2} &= \sqrt{\alpha\beta} \\ (\alpha + \beta)^{2} &= 4\alpha\beta \\ \alpha^{2} + \beta^{2} &= 4\alpha\beta - 2\alpha\beta = 2\alpha\beta \end{aligned} \]

(In fact this forces \((\alpha - \beta)^{2} = 0\), so \(\alpha = \beta\).) \(4\alpha\beta\) is \((\alpha + \beta)^{2}\), not \(\alpha^{2} + \beta^{2}\); the other multiples do not follow.

Answer: C

\(\frac{y}{x}\) equals 9.

In a geometric progression every term is the previous one times the common ratio r. From the last two terms, \(r = \frac{162}{54} = 3\). Working back: \(y = \frac{54}{3} = 18\), then \(6 = 18/3\) checks, and \(x = \frac{6}{3} = 2\).

\[ \frac{y}{x} = \frac{18}{2} = 9 = r^{2} \]

Since y is two steps after x, \(\frac{y}{x}\) is always \(r^{2}\). 3 is the ratio r itself, and 18 is the value of y.

Answer: D

\(\sqrt[3]{abc}\) equals 4.

With first term 1 and common ratio r, the terms are \(1, r, r^{2}, r^{3}, r^{4}\). So \(r^{4} = 16\) and \(a = r\), \(b = r^{2}\), \(c = r^{3}\):

\[ \begin{aligned} abc &= r^{6} = (r^{2})^{3} \\ \sqrt[3]{abc} &= r^{2} = \sqrt{16} = 4 \end{aligned} \]

This holds whether r is 2 or -2, since only \(r^{2}\) appears. Equivalently, b is the geometric mean of 1 and 16, so b = 4 and \(abc = b^{3}\). 2 is the common ratio, and 8 is the fourth term.

Answer: A

The sum is \(1023\sqrt{3}\).

The terms are \(\sqrt{3}, 2\sqrt{3}, 4\sqrt{3}, \ldots\), since \(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{48} = 4\sqrt{3}\). So the first term is \(\sqrt{3}\) and the common ratio is 2:

\[ \begin{aligned} S_{10} &= \sqrt{3}\,\frac{2^{10} - 1}{2 - 1} \\ &= 1023\sqrt{3} \end{aligned} \]

\(1024\sqrt{3}\) forgets the -1 in \(r^{n} - 1\), and \(511\sqrt{3}\) and \(512\sqrt{3}\) use 9 terms instead of 10.

Answer: E

The numbers are 1 and 25.

Let the arithmetic mean be A and the geometric mean G. From A + G = 18 and A - G = 8, A = 13 and G = 5. So for the two numbers a and b:

\[ \begin{aligned} a + b &= 2A = 26 \\ ab &= G^{2} = 25 \end{aligned} \]

They are the roots of \(t^{2} - 26t + 25 = 0\), that is \((t - 1)(t - 25) = 0\), giving 1 and 25. Check: AM 13, GM 5, sum 18, difference 8.

The other pairs fail the check. For 8 and 18, the AM is 13 but the GM is 12; for 2 and 24 the GM is not even an integer.

Answer: C

The 100th term is \(2^{100}\).

Let the G.P. have first term \(a\) and ratio \(r\). Terms 4 to 6 are terms 1 to 3 multiplied by \(r^{3}\), so the ratio of the two sums gives \(r^{3}\) directly:

\[ \begin{aligned} r^{3} &= \frac{112}{14} = 8, \quad r = 2 \\ a(1 + 2 + 4) &= 14, \quad a = 2 \\ T_{100} &= a r^{99} = 2 \times 2^{99} = 2^{100} \end{aligned} \]

\(2^{99}\) is \(r^{99}\) without the factor \(a = 2\) (or the formula with \(r^{n}\) and \(a = 1\) mixed up); \(2^{101}\) uses \(r^{100}\). The options with \(-1\) and \(+1\) look like sum formulas, not a single term.