Geometric Progression

The nth term of a geometric progression is \(a_n=ar^{n-1}\). Hence, \(a_8=5\times3^{8-1}=5\times3^7=5\times2187=10935\). Therefore, option C is correct. \(7290\) is not correct because it is not the value of \(5\times3^7\). Exam tip: the exponent of the common ratio in the nth term is always \(n-1\); for the eighth term, use \(r^7\).

Here \(r=\frac{1}{2}\) so \(a_9=256\cdot\left(\frac{1}{2}\right)^8=1\). In exams use the fractional ratio in decreasing progressions.

From (4\cdot3^{n-1}=2916), (3^{n-1}=729=3^6) so (n=7). In exams first equate the given term to (a_n).

In a GP, the \(n\)th term is \(a_n=ar^{n-1}\). Therefore, \(486=6r^4\), so \(r^4=81\). Since \(r\) is stated to be positive, \(r=3\). For example, if \(r=2\), the fifth term would be \(6\times2^4=96\), not 486. Exam tip: for the fifth term, the power of \(r\) is \(4\), because the exponent is always \(n-1\).

The first term is (10) and the ratio is (3) so (a_n=10\cdot3^{n-1}). In exams keep (a) and (r) correct in (ar^{n-1}).

The sum of the first (6) terms is (2(3^6-1)/(3-1)=728). In exams use (S_n=\frac{a(r^n-1)}{r-1}) for (r>1).

For the middle term (x^2=8\cdot72=576), so (x=24). In exams the square of the middle GP term equals the product of outer terms.

In a GP, each term is obtained by multiplying the previous term by the common ratio. Multiplication by a negative ratio reverses the sign, as in 3, −6, 12, −24. Exam tip: always check the sign of the ratio.

In a GP, \(y=xr\) and \(z=yr=xr^2\). Hence \(y^2=(xr)^2=xz\). The relation \(x+z=2y\) belongs to an AP, not necessarily to a GP. Exam tip: square the middle term to test three terms.

The first term is (81) and the ratio is \(\frac{1}{3}\), so the correct rule is \(81\cdot\left(\frac{1}{3}\right)^{n-1}\). In exams write the fractional ratio in a decreasing GP.

The nth term of a GP is












a_n=ar^{n-1}. Thus,












_6=4r^5=972, so r^5=243=3^5. Since r is positive, r=3. If r=2, the sixth term would be 128, not 972. Exam tip: for the sixth term, the exponent of r is 5, not 6.

Direct answer: 3072 is the 6th term, so A is correct. First identify the pattern: 12 ÷ 3 = 4, 48 ÷ 12 = 4, and 192 ÷ 48 = 4. Thus this is a geometric progression with first term a = 3 and common ratio r = 4. The nth-term formula is aₙ = arⁿ⁻¹. Put the required value into the formula: 3 × 4ⁿ⁻¹ = 3072. Dividing by 3 gives 4ⁿ⁻¹ = 1024. Since 1024 = 4⁵, n − 1 = 5, so n = 6. A direct check gives 3 × 4⁵ = 3 × 1024 = 3072. B is wrong because the 7th term is 3072 × 4 = 12288. C and D are even later terms and therefore are still larger. Memory cue: in a GP, count powers from the first term; the first term has power zero, not power one.

\(a_7=96\cdot\left(\frac{1}{2}\right)^6=\frac{3}{2}\). In exams apply powers of fractional ratios carefully.

The governing concept is the sum formula for the first n terms of a geometric progression when r ≠ 1: Sₙ = a(rⁿ − 1)/(r − 1). Substituting a = 2, r = 3 and n = 5 gives S₅ = 2(3⁵ − 1)/(3 − 1). Since 3⁵ = 243, this becomes 2(242)/2 = 242. Therefore the statement claiming S₅ = 242 is true, so option A is correct. Direct checking gives the same result: the first five terms are 2, 6, 18, 54 and 162, whose sum is 242. Option B is short by 2, option C is greater by 2, and option D is exactly half the correct sum.

This uses the finite geometric-series sum formula. For a GP with first term a and ratio r ≠ 1, Sₙ = a(rⁿ − 1)/(r − 1). In this sequence, a = 1, r = 3 and n = 7. Thus S₇ = 1(3⁷ − 1)/(3 − 1) = (2187 − 1)/2 = 2186/2 = 1093. Therefore option A is correct. A useful verification is to add the terms directly: 1 + 3 + 9 + 27 + 81 + 243 + 729 = 1093. Option B is one less than the correct total and may result from subtracting incorrectly; option C has no valid calculation behind it, while option D is only the seventh term, not the sum of all seven terms.

(\frac{a_6}{a_3}=r^3=\frac{160}{20}=8), so (r=2). In exams the ratio of distant terms gives a power based on position gap.

In a GP, consecutive ratios are equal: \(q/p=r/q\). Cross-multiplying gives \(q^2=pr\). The condition \(2q=p+r\) belongs to an AP. Exam tip: square the middle term to test three terms.

The position gap is (3) and (r=3), so the ratio is (3^3=27). In exams use (\frac{a_m}{a_n}=r^{m-n}).

Here (a=7) and (r=2), so (S_6=7(2^6-1)=441). In exams identify (a) and (r) from the general term.

From (2\cdot5^{n-1}=1250), (5^{n-1}=625=5^4) so (n=5). In exams compare powers to find the term number.