Geometric Progression

The terms are (7,14,28,56,112,224), so (224) is the sixth term. For small questions listing terms is quick.

In (2,4,8,15,\ldots), the ratios (2,2,\frac{15}{8}) are not equal. A geometric progression must have a constant ratio.

The first term is (6), and multiplying by (2) each time gives (6,12,24,48). In a geometric progression each next term is formed by multiplication.

In a geometric sequence, the common ratio is found by dividing any term by the preceding term. Here, \(r=\frac{-6}{2}=-3\). Checking further, \(\frac{18}{-6}=-3\) and \(\frac{-54}{18}=-3\). Hence, the common ratio is \(-3\). Option \(3\) is a close distractor, but it does not account for the alternating signs. Exam tip: divide consecutive terms and verify the ratio with another pair.

The common ratio is (3), so the next term is (36\cdot3=108). Multiply the previous term by the ratio for the next term.

For the middle term (x), (x^2=3\cdot27=81), so (x=9). For positive terms take the positive geometric mean.

The first four terms are 5, 15, 45, and 135. Therefore, their sum is \(5+15+45+135=200\). Hence, 200 is correct. The value 180 is the sum of only the first three terms, so it is a close but incorrect option. Exam tip: when there are only a few terms, direct addition is usually the quickest and safest method.

In a geometric progression, the second term is \(a_2=a_1r\). Thus, \(10=a_1\times2\), so \(a_1=10/2=5\). Taking \(10\) as the first term is incorrect because it is given as the second term. Exam tip: To find a preceding term, divide the next term by the common ratio.

The general term of a GP is
\(a_n=a_1r^{n-1}\). Thus,
\(a_4=4r^3=108\), so
\(r^3=27\) and
\(r=3\). If the ratio were 2, the fourth term would be
\(4\times2^3=32\), not 108. Exam tip: from
\(a_m\) to
\(a_n\), the exponent of the ratio is
\(n-m\).

This is a geometric sequence because each term is twice the preceding term, so the common ratio is \(r=2\). The terms are \(\frac{1}{2},1,2,4,8,16\); therefore, the 6th term is \(16\). Note that \(8\) is the 5th term, making it a close but incorrect option. Exam tip: Start counting terms from the first given term.

The common ratio is (\frac{1}{2}), so the fifth term is (4). In a decreasing geometric progression multiply by the ratio repeatedly.

The \(n\)th term of a geometric progression is \(a_n=ar^{n-1}\). Thus, \(a_4=3\times5^{4-1}=3\times125=375\). \(125\) is only \(5^3\); it misses multiplication by the first term, \(a=3\). Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\), not \(n\).

In (64,16,4,1,\ldots), (\frac{16}{64}=\frac{1}{4}). Divide consecutive terms to find the ratio.

In this GP, the first term is 1 and the common ratio is 3. Its terms are 1, 3, 9, 27, 81; therefore, 81 is the fifth term. The fourth term is 27, so it is a close but incorrect option. Exam tip: list the terms in order or recognise \(81=3^4\).

For the third term, substitute \(n=3\): \(a_3=9\left(\frac{1}{3}\right)^{3-1}=9\left(\frac{1}{3}\right)^2=9\times\frac{1}{9}=1\). The option \(3\) results from incorrectly using the exponent as 1. Exam tip: after substituting the term number in \(a_n\), calculate \(n-1\) first.

The governing property is that consecutive terms of a geometric progression have one constant ratio. From the first two terms, r = 6 ÷ 2 = 3. The missing third term must therefore be x = 6 × 3 = 18. We can verify the answer using the final term: 18 × 3 = 54, so the complete sequence is 2, 6, 18, 54. Hence option B is correct. Option A would give a ratio of 2 between the second and third terms and would produce 36 rather than 54 next. Option C would give a ratio of 4, and option D would give a ratio of 5; neither agrees with the established ratio 3. Both forward construction and backward checking lead to 18.

A negative common ratio means that each term is obtained by multiplying the previous term by a negative number, so the signs alternate when the terms are nonzero. In option C, \(\frac{-10}{5}=-2\), \(\frac{20}{-10}=-2\), and \(\frac{-40}{20}=-2\). Thus its common ratio is negative, and option C is correct. Option A has ratio 2, option B has ratio 3, and option D has ratio \(\frac12\); all three ratios are positive. A decreasing sequence does not automatically have a negative ratio, as option D demonstrates.

The first four terms are 10, 20, 40, and 80. Therefore, their sum is 10+20+40+80=150. This is a GP with first term 10 and common ratio 2; using the formula also gives \(S_4=10(2^4-1)/(2-1)=150\). A result such as 160 comes from an incorrect addition. Exam tip: for a few terms, verify the answer by direct addition.

In a GP, each term is obtained by multiplying the previous term by the same constant, so consecutive-term ratios remain equal. For 3, 6, 12, the ratio is 2. Equal differences describe an AP. Exam tip: check ratios first.

The first term is (6) and the ratio is (3), so (a_n=6\cdot3^{n-1}). Use both first term and ratio in the general term.