Geometric Progression

The nth term of a geometric progression is \(a_n=a_1r^{n-1}\). Therefore, \(a_3=7\times5^{3-1}=7\times25=175\). The value 35 is only the second term, since \(a_2=7\times5\). Exam tip: for the third term, use \(r^2\) in the general-term formula.

This is of the form (3\cdot2^{n-1}) so the tenth term is (3\cdot2^9=1536). In exams identify powers to solve quickly.

Given \(a_n=2\cdot5^{n-1}\), substitute \(n=4\): \(a_4=2\cdot5^{4-1}=2\cdot5^3=2\cdot125=250\). Hence, option C is correct. Option D can result from an error in multiplying after evaluating \(5^3\). Exam tip: substitute the term number first, and then evaluate the exponent carefully.

The first term is (60) and the ratio is \(\frac{1}{2}\). In exams put the correct fractional ratio in \(ar^{n-1}\).

The square of the middle term is (10\times90=900) so the positive (x=30). In exams for three GP terms the square of the middle term equals the product of the outer terms.

The common ratio is (3) so the fifth term is (486\times3=1458). In exams multiply the last known term by the ratio.

A geometric progression with first term 8 and common ratio 3 must begin with 8, and each later term must be three times the preceding term. Option A satisfies both requirements: 24 = 8 × 3, 72 = 24 × 3, and 216 = 72 × 3. Therefore option A is correct. Option B has ratio 3 but begins with 3, so its first term is wrong. Option C begins with 8 but adds 3 each time, making it an arithmetic progression with common difference 3. Option D follows ratio 3 but begins with 24, which would be the second term of the required progression. Both the starting value and repeated multiplication must be checked.

The nth term of a geometric progression is \(T_n=ar^{n-1}\). Therefore, \(T_6=12\times2^{6-1}=12\times2^5=12\times32=384\). The value \(288\) can result from an error in evaluating the power or multiplication. Exam tip: for the sixth term, always use the exponent \(6-1=5\).

The third term is (a_3=a_2\times r=15\times3=45). In exams multiply the previous term by the ratio to get the next term.

The third term of a GP is \(a_3=a_1r^{3-1}\). Therefore, \(a_3=6\times4^2=6\times16=96\). The value 72 would result from applying the ratio only once, which gives the second term. Exam tip: use \(a_n=a_1r^{n-1}\) to find the \(n\)th term of a GP.

The common ratio is (\frac{25}{125}=\frac{1}{5}). In a decreasing geometric progression the ratio can be less than (1).

Given \(a_n=5\cdot3^{n-1}\), substitute \(n=5\): \(a_5=5\cdot3^{5-1}=5\cdot3^4=5\cdot81=405\). Hence, 405 is correct. The value 135 may result from incorrectly using \(3^3\). Exam tip: substitute the term number first and simplify the exponent \(n-1\) carefully.

Direct answer: 256 is the sixth term, so option C is correct. Each term is twice the preceding term, so the common ratio is 2. Count the terms carefully: first 8, second 16, third 32, fourth 64, fifth 128, and sixth 256. The general-term equation gives the same result: aₙ = 8 × 2ⁿ⁻¹. Setting aₙ = 256 gives 8 × 2ⁿ⁻¹ = 256, so 2ⁿ⁻¹ = 32 = 2⁵. Therefore n − 1 = 5 and n = 6. Option A is 64, the fourth term. Option B is 128, the fifth term. Option C is 256, the sixth term. Option D would be 512, the seventh term. A common counting mistake is to call 8 the zeroth term; in this question the first displayed term is term 1.

The first term is (10), and multiplying by (2) each time gives (10,20,40,80). In a geometric progression each next term is formed by multiplication.

In a geometric progression, the common ratio is found by dividing any term by the preceding term. Here, \(r=\frac{-14}{7}=-2\). This is confirmed by \(\frac{28}{-14}=-2\) and \(\frac{-56}{28}=-2\), so the common ratio is \(-2\). Option 2 is a close distractor, but it does not account for the alternating signs of the terms. Exam tip: Always include the sign when calculating a common ratio.

The common ratio is (5), so the next term is (150\cdot5=750). Multiply the previous term by the ratio to get the next term.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of the first and third terms. Thus, \(x^2=9\times81=729\). Hence \(x=\sqrt{729}=27\), since the terms are stated to be positive. A value such as \(18\) does not give a common ratio. Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\).

The first four terms are 8, 24, 72, and 216. Therefore, \(8+24+72+216=320\). Hence, 320 is the correct answer. Leaving out 216 would give 104, which is the sum of only the first three terms. Exam tip: When only a few terms are given, add every listed term directly to verify the answer.

In a geometric progression, the second term is \(a_2=a_1r\). Thus, \(28=a_1\times4\), so \(a_1=28\div4=7\). Multiplying 28 by 4 gives 112, which would be the next term \(a_3\), not the first term. Exam tip: To find a previous term in a GP, divide by the common ratio.

The general term of a GP is \(a_n=a_1r^{n-1}\). Therefore, \(a_4=3r^3=81\), so \(r^3=27\). Since the common ratio is positive, \(r=3\). If \(r=9\), the fourth term would be \(3\times9^3\), not 81. Exam tip: in \(a_n\), the exponent of \(r\) is always \(n-1\).