Geometric Progression

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 direct answer is A, 248. A geometric progression starts with (a=128) and has common ratio (r=\frac{1}{2}). The first five terms are found by multiplying by one-half each time: first 128, second 64, third 32, fourth 16, and fifth 8. Adding them gives 128+64+32+16+8=248. The geometric-sum formula gives the same result: (S_n=a\frac{1-r^n}{1-r}), so (S_5=128\frac{1-(1/2)^5}{1-1/2}=248). Option A is correct. Option B, 250, is close but results from an arithmetic error. Option C, 252, is also not the sum of the listed terms. Option D, 256, incorrectly ignores the decrease or treats the total as a power of two rather than a sum. The safest check is to list the five terms and add them. Memory cue: ratio one-half means each next term is half the previous one.

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.

The direct answer is option C: \(a_5=648\). In a geometric progression, the term formula is \(a_n=a_1r^{n-1}\). Since \(a_1=8\) and \(a_4=216\), we use \(a_4=8r^3\), because the fourth term is three ratio steps after the first. Thus \(216=8r^3\), so \(r^3=27\). The question says that \(r\) is positive, therefore \(r=3\), not a negative cube root. The fifth term is one more multiplication by 3: \(a_5=a_4r=216\times3=648\). Option A, 432, would use a ratio of 2 and does not satisfy the given fourth term. Option B, 486, does not follow from the required ratio. Option C is correct because it uses the verified ratio 3. Option D, 729, is another unsupported value and would require a different preceding term or ratio. The safe method is to find the ratio from the known terms before calculating the requested term. Memory cue: from \(a_1\) to \(a_4\), use \(r^3\), then multiply once more for \(a_5\).

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.

For a geometric progression, moving from a_2 to a_6 involves four ratio-steps, so a_6 = a_2r^4. Substituting the given values gives 8192 = 32r^4, hence r^4 = 256. The positive-r condition gives r = 4 rather than −4. Since a_2 = a_1r, we have 32 = 4a_1, so a_1 = 8. Therefore option C is correct. The positivity condition is essential: without it, a negative ratio could also satisfy the fourth-power equation, although the requested first term would then differ in sign. Options A, B and D arise from using the wrong exponent or dividing by the wrong ratio step.

The governing relation is a_(m+k) = a_mr^k. From the third term to the fifth term there are two steps, so a_5 = a_3r^2. Substitution gives 288 = 18r^2, hence r^2 = 16. Since r is specified as positive, r = 4. From the fifth term to the seventh term there are again two steps, so a_7 = a_5r^2 = 288 × 16 = 4608. Therefore option C is correct. It is not necessary to calculate every intermediate term; the equal two-step gaps allow the same multiplier r^2 to be used twice. The other options result from using r instead of r^2 or from an arithmetic error.

The governing concept is the general-term rule aₙ = a₁rⁿ⁻¹. With r = 2, the second term is a₂ = a₁ × 2¹ = 2a₁, while the fourth term is a₄ = a₁ × 2³ = 8a₁. The exponent for the fourth term is 3 because three ratio multiplications take us from the first term to the fourth. Using the given relation, 2a₁ + 8a₁ = 170, so 10a₁ = 170 and a₁ = 17. Therefore option C is correct. Checking the alternatives, 12 would produce a sum of 120, 15 would produce 150, and 20 would produce 200. The key error to avoid is replacing a₄ = a₁r³ with 4a₁, which would not represent a geometric progression.

The governing concept is the nth-term formula for a geometric progression: aₙ = arⁿ⁻¹. In the given sequence, the first term is a = 2 and the common ratio is r = 6 ÷ 2 = 3; the same ratio continues because 18 ÷ 6 and 54 ÷ 18 are also 3. For the tenth term, substitute n = 10: a₁₀ = 2 × 3⁹. Since 3⁹ = 19683, the result is 2 × 19683 = 39366. Therefore option B is correct. Option A is only 3⁹ and omits the first-term factor 2; option C corresponds to 3¹⁰, while option D doubles the correct value again. The exponent is n − 1 because the first term has zero ratio multiplications.

The governing concept is the finite sum of a geometric progression. For first term a and ratio r not equal to 1, use Sₙ = a(rⁿ − 1)/(r − 1) when r is greater than 1. Here a = 3, r = 3 and n = 7, so S₇ = 3(3⁷ − 1)/(3 − 1). Because 3⁷ = 2187, the numerator becomes 3(2186) = 6558, and division by 2 gives S₇ = 3279. Hence option B is correct. Directly adding the terms gives 3 + 9 + 27 + 81 + 243 + 729 + 2187 = 3279, which confirms the formula. The nearby alternatives result from arithmetic or endpoint errors; the sum must include exactly seven terms.

The governing concept is the general term of a geometric progression. Each term is obtained by multiplying the previous term by 5, so the first term is a₁ = 4 and the common ratio is r = 5. The nth-term formula is aₙ = arⁿ⁻¹. Therefore a₇ = 4 × 5⁶. Since 5⁶ = 15,625, the product is 4 × 15,625 = 62,500. Equivalently, continuing gives a₆ = 15,625 and a₇ = 78,125? This reveals the listed sequence indexing check: 4,20,100,500 gives a₅=2500, a₆=12500, a₇=62500. Thus option C is correct. The apparent intermediate alternative 15,625 is a₆ only under a mistaken count.