Sequences and Series Questions and Answers

Answer: B

The common ratio is \(\frac{\sqrt{5} - 1}{2}\).

Let the term be \(a\), so the next two are \(ar\) and \(ar^{2}\). Then

\[ \begin{aligned} a &= ar + ar^{2} \\ r^{2} + r - 1 &= 0 \\ r &= \frac{-1 \pm \sqrt{5}}{2} \end{aligned} \]

The terms are positive, so \(r \gt 0\), which leaves \(r = \frac{\sqrt{5} - 1}{2} \approx 0.618\). \(\frac{\sqrt{5} + 1}{2}\) is the reciprocal (the golden ratio), which would make each term the sum of the two before it rather than the two after it.

Answer: A

The common ratio is 4.

With an even number of terms, the terms pair up as (1st, 2nd), (3rd, 4th), and so on. Each even-placed term is \(r\) times the odd-placed term before it, so

\[ \begin{aligned} S_{\text{even}} &= r\,S_{\text{odd}} \\ S_{\text{all}} &= S_{\text{odd}} + r\,S_{\text{odd}} = (1 + r)S_{\text{odd}} \end{aligned} \]

Given \(S_{\text{all}} = 5S_{\text{odd}}\), \(1 + r = 5\) and \(r = 4\).

\(\frac{1}{5}\) and \(\frac{4}{5}\) invert the ratio of the sums, and 2 would make the total three times the odd-term sum.

Answer: B

The ratio \(a : b : c\) is 1:2:3.

Since \(a, b, c\) are in AP, \(b - a = c - b = d\), with \(d \ne 0\) because the numbers are unequal. The terms \(b - a, c - b, a\) are then \(d, d, a\), and for these to be in GP:

\[ \begin{aligned} d^{2} &= d \cdot a \\ d &= a \end{aligned} \]

So \(b = a + a = 2a\) and \(c = 3a\), giving \(a : b : c = 1 : 2 : 3\). Check: 1, 2, 3 is an AP, and \(1, 1, 1\) is a GP. The other ratios are not even arithmetic progressions, except 3:4:5, for which \(1, 1, 3\) is not a GP.

Answer: E

The three numbers are 4, 8, 16.

Inserting three geometric means gives a GP of five terms: 2, \(2r\), \(2r^{2}\), \(2r^{3}\), 32.

\[ \begin{aligned} 2r^{4} &= 32 \\ r^{4} &= 16, \quad r = 2 \end{aligned} \]

So the means are 4, 8 and 16, and the sequence 2, 4, 8, 16, 32 doubles at each step. 6, 8, 10 and 4, 6, 8 are arithmetic, not geometric, and 4, 8, 32 repeats the end term instead of reaching it.

Answer: C

The first term is 243.

For a G.P. with first term a and ratio r, the n-th term is \(ar^{n-1}\).

\[ \begin{aligned} ar^{2} &= 108,\quad ar^{5} = -32 \\ r^{3} &= \frac{-32}{108} = -\frac{8}{27} \Rightarrow r = -\frac{2}{3} \\ a &= \frac{108}{r^{2}} = 108 \times \frac{9}{4} = 243 \end{aligned} \]

Check: \(243 \times \frac{4}{9} = 108\). The sign of r is negative, but a depends on \(r^{2}\), so a is positive; -243 is the sign trap.

Answer: E

The geometric means are 12, 36, 108.

Inserting three geometric means between 4 and 324 makes a G.P. of five terms with first term 4 and fifth term 324.

\[ \begin{aligned} 4r^{4} &= 324 \\ r^{4} &= 81 \Rightarrow r = 3 \end{aligned} \]

So the terms are 4, 12, 36, 108, 324. Each of the other options has one wrong term: 46, 128, 14 and 98 do not continue a common ratio of 3.

Answer: A

\((ac)^{4}\) equals \(b^{8}\).

The 3rd, 7th and 11th terms are equally spaced (four steps apart), so they themselves form a G.P., and the middle one is the geometric mean of the other two.

\[ \begin{aligned} a &= Ar^{2},\ b = Ar^{6},\ c = Ar^{10} \\ ac &= A^{2}r^{12} = b^{2} \\ (ac)^{4} &= b^{8} \end{aligned} \]

\(b^{4}\) is \((ac)^{2}\), and \(b^{2}\) is ac itself, so those options stop a step short.

Answer: A

\((x + y)^{2}\) equals \((a + x)(y + b)\).

Let the common ratio be r, so \(x = ar\), \(y = ar^{2}\), \(b = ar^{3}\).

\[ \begin{aligned} (x + y)^{2} &= a^{2}r^{2}(1 + r)^{2} \\ (a + x)(y + b) &= a(1 + r) \cdot ar^{2}(1 + r) \\ &= a^{2}r^{2}(1 + r)^{2} \end{aligned} \]

The two agree. A quick check with 1, 2, 4, 8: \((2 + 4)^{2} = 36\) and \((1 + 2)(4 + 8) = 36\), while \((8 + 2)(4 + 1) = 50\) rules out the second option and the doubled terms rule out the rest.