Sequences and Series Questions and Answers

Answer: E

The 21st term is 4096.

With first term \(a = 4\) and common ratio \(r\), the third term is \(ar^{2} = 8\), so \(r^{2} = 2\).

\[ \begin{aligned} t_{21} &= ar^{20} = 4\,(r^{2})^{10} \\ &= 4 \times 2^{10} = 4 \times 1024 = 4096 \end{aligned} \]

2048 is \(2 \times 2^{10}\), using the wrong first term; 4012, 4064 and 4098 are not of the form \(4 \times 2^{k}\), so they cannot be terms of this G.P.

Answer: D

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

In a G.P. the product of terms placed symmetrically about a middle term is a power of that middle term:

\[ \begin{aligned} a_{1}a_{2}a_{3} &= a_{2}^{3} = 64 \Rightarrow a_{2} = 4 \\ a_{1}a_{2}a_{3}a_{4}a_{5} &= a_{3}^{5} = 32 \Rightarrow a_{3} = 2 \\ r &= \frac{a_{3}}{a_{2}} = \frac{2}{4} = \frac{1}{2} \end{aligned} \]

Check: the G.P. is 8, 4, 2, 1, 1/2, with products 64 and 32. The values 1/4 and 1/8 would make the five-term product far smaller than 32.

Answer: C

\(n = \) 4.

Set the general term equal to 5 and clear the denominator:

\[ \begin{aligned} n(n + 6) &= 5(n + 4) \\ n^{2} + n - 20 &= 0 \\ (n + 5)(n - 4) &= 0 \end{aligned} \]

Since \(n\) is a natural number, \(n = 4\). Check: \(t_{4} = \frac{4 \times 10}{8} = 5\). For the other options, \(t_{2} = \frac{16}{6}\), \(t_{3} = \frac{27}{7}\), \(t_{5} = \frac{55}{9}\) and \(t_{6} = \frac{72}{10}\), none equal to 5.

Answer: A

The third term is 2.

Write the five terms as \(\frac{a}{r^{2}}, \frac{a}{r}, a, ar, ar^{2}\), where \(a\) is the third term. Their product is \(a^{5}\), because the powers of \(r\) cancel:

\[ a^{5} = 32 = 2^{5} \Rightarrow a = 2 \]

The ratio does not matter (the word increasing only says \(r \gt 1\)). 4 and 8 would give products \(4^{5}\) and \(8^{5}\); 1/2 and 1/8 would give products less than 1.

Answer: C

The sum is \(2^{20} - 1\).

The terms \(1, 2, 4, \ldots, 2^{19}\) form a G.P. with first term 1 and ratio 2, so

\[ \begin{aligned} \sum_{n=1}^{20} 2^{n-1} &= \frac{2^{20} - 1}{2 - 1} \\ &= 2^{20} - 1 \end{aligned} \]

The same follows from binary: twenty 1s in a row is \(2^{20} - 1\). The other options divide by powers of 2 or \(\sqrt{2}\) that do not appear in the sum, and none of them equals \(2^{20} - 1\).

Answer: B

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

Each pair is a multiple of the first pair: \(a_{9} + a_{10} = r^{8}(a_{1} + a_{2})\). So

\[ \begin{aligned} r^{8} &= \frac{3/128}{6} = \frac{1}{256} \\ &= \left(\frac{1}{2}\right)^{8} \end{aligned} \]

giving \(r = \pm\frac{1}{2}\); only \(\frac{1}{2}\) is offered. Check: \(a_{1}(1 + r) = 6\) gives \(a_{1} = 4\), and \(a_{9} + a_{10} = 4\cdot\frac{1}{256} + 4\cdot\frac{1}{512} = \frac{3}{128}\). Ratios of \(\frac{1}{3}\) or \(\frac{1}{4}\) give \(r^{8}\) far smaller, and 2 or 3 make the terms grow.

Answer: C

\(a^{2} + b^{2} + c^{2} =\) 91.

In a G.P. \(b^{2} = ac\), so \(abc = b^{3} = 27\) gives \(b = 3\) and \(ac = 9\). Then

\[ \begin{aligned} a^{2} + c^{2} &= (a + c)^{2} - 2ac = 100 - 18 = 82 \\ a^{2} + b^{2} + c^{2} &= 82 + 9 = 91 \end{aligned} \]

(The numbers are 1, 3, 9.) 82 forgets to add \(b^{2}\), and 81, 92 and 99 come from arithmetic slips with \(ac\) or \(b^{2}\).

A general identity used here: \(a^{2} + c^{2} = (a + c)^{2} - 2ac\), with \(ac = b^{2}\) for terms in G.P.

Answer: D

\(\lambda =\) 72.

A geometric mean of 6 means \(\sqrt{\alpha\beta} = 6\), so \(\alpha\beta = 36\). For \(2x^{2} - 25x + \lambda = 0\) the product of the roots is \(\frac{\lambda}{2}\):

\[ \frac{\lambda}{2} = 36 \;\Rightarrow\; \lambda = 72 \]

Check: the roots are then \(\frac{25 \pm 7}{4}\), that is 8 and 4.5, both positive with product 36. 36 forgets the leading coefficient 2, 6 is the geometric mean itself, and 12 and 48 have no basis.

Answer: E

The determinant is 0.

In a G.P., \(a_{2} = ra_{1}\), \(a_{4} = ra_{3}\) and \(a_{6} = ra_{5}\). So the second column is \(r\) times the first column, and a determinant with two proportional columns is zero:

\[ \begin{vmatrix} a_{1} & ra_{1} & 1 \\ a_{3} & ra_{3} & 1 \\ a_{5} & ra_{5} & 1 \end{vmatrix} = r\begin{vmatrix} a_{1} & a_{1} & 1 \\ a_{3} & a_{3} & 1 \\ a_{5} & a_{5} & 1 \end{vmatrix} = 0 \]

The values \(a_{1} = 2\) and \(r = \frac{1}{2}\) do not matter; any G.P. gives 0, so the non-zero options are wrong.

Answer: D

The 25th term is \(\frac{1}{3^{22}}\).

The sequence is a G.P. with first term 9 and common ratio \(\frac{1}{3}\).

\[ \begin{aligned} T_{25} &= 9\left(\frac{1}{3}\right)^{24} \\ &= \frac{3^{2}}{3^{24}} = \frac{1}{3^{22}} \end{aligned} \]

\(\frac{1}{3^{24}}\) forgets that the first term is 9, not 1, and \(\frac{1}{3^{25}}\) or \(\frac{1}{3^{23}}\) use the wrong number of ratios (the nth term has n - 1 factors of r).