Geometric Progression

The sum of the first four terms is (3+18+108+648=777). In exams, direct addition is safe when the number of terms is small.

There are (6) steps from the fourth to the tenth term, so (a_{10}=96\cdot2^6=6144). In exams, count the position gap.

Each term is multiplied by (\frac{1}{3}), so the terms are (810,270,90,30,10). In exams, you can also check a decreasing GP in order.

(S_5=\frac{10(3^5-1)}{3-1}=1210). In exams, confirm statement-type questions with the formula.

The governing concept is the general term of a geometric progression. The first term is a = 4 and the common ratio is r = 16/4 = 4. The nth term is a_n = ar^(n-1). Therefore a_5 = 4 × 4^4 = 4 × 256 = 1024, and a_6 = 4 × 4^5 = 4 × 1024 = 4096. Adding these two required terms gives a_5 + a_6 = 1024 + 4096 = 5120. Hence option B is correct. Option A is only a_6, so it omits a_5. Options C and D are larger sums that do not follow from the specified two terms. Finding each requested term before adding prevents confusing a term with the sum of terms.

From \(162\left(\frac{1}{3}\right)^{n-1}=2\), \(\left(\frac{1}{3}\right)^{n-1}=\frac{1}{81}\), so \(n=5\). In exams, simplify fractions first.

Here, the first term is \(a=7\), the common ratio is \(r=2\), and \(n=9\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_9=\frac{7(2^9-1)}{2-1}=7(512-1)=7\times511=3577\). Therefore, option A is correct. A close value such as 3584 can result from an error while evaluating the power or subtraction. Exam tip: when \(r=2\), calculate \(2^n\) carefully before substituting.

If the common ratio is \(r\), then \(b=ar\) and \(c=ar^2\). Hence \(b^2=(ar)^2=a\cdot ar^2=ac\). The relation \(2b=a+c\) belongs to an AP. Exam tip: square the middle term.

For consecutive GP terms, \(b/a=c/b\). Cross-multiplying gives \(b^2=ac\), so A is necessary. The relation \(a+c=2b\) is the condition for an AP. Exam tip: square the middle term and compare it with the product of the extremes.

The terms are (128,64,32,16,8), and the sum is (248). In exams, direct addition is also easy for small decreasing terms.

From (\frac{648}{24}=27=r^3), (r=3) and (a_1=24\div3=8). In exams, find (r) first and then (a_1).

The governing concept is an explicit, or general, rule for a geometric progression. The formula a_n = 4 × 5^(n−1) gives any term directly, with first term 4 and ratio 5. For n = 4, a_4 = 4 × 5^3 = 4 × 125 = 500. For n = 5, a_5 = 4 × 5^4 = 4 × 625 = 2500. Consequently, a_4 + a_5 = 500 + 2500 = 3000, so option B is correct. Option C is 5^5 rather than the requested sum, option A represents only a_5, and option D does not follow from substituting either index. The safest method is to substitute each index carefully and add the resulting terms.

From (18\cdot2^{n-1}=2304), (2^{n-1}=128=2^7), so (n=8). In exams, factor first and compare powers.

From (216=8r^3), (r^3=27) and (r=3), so (a_5=216\cdot3=648). In exams, find the ratio first.

Here, the first term is \(a=20\), the common ratio is \(r=2\), and \(n=8\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_8=\frac{20(2^8-1)}{2-1}=20(256-1)=5100\). Therefore, 5100 is correct. The nearby answer 5120 can result from using \(2^8\) instead of \(2^8-1\). Exam tip: when \(r\ne1\), always include the denominator \(r-1\).

(1792=7r^7) gives (r^7=256), which does not match any option exactly. In exams, check (n-1) and powers carefully.

Each term is multiplied by (\frac{1}{3}), and the sixth term is (\frac{50}{81}). In exams, fractional terms can also be checked in order.

(a_5=5\cdot4^4=1280) and (a_6=5120), so the sum is (6400). In exams, calculate large powers separately.

(S_n=\frac{6(3^n-1)}{3-1}=3(3^n-1)), and (3(3^n-1)=2184) gives (n=6). In exams, simplify the sum and identify the power.

In option A, the successive ratios are \(6/(-3)=-2\), \((-12)/6=-2\), and \(24/(-12)=-2\), so it is a GP. In option C, the ratios change. Exam tip: compare ratios of consecutive terms.