Geometric Progression

From (\frac{1215}{45}=27=r^3), (r=3) and (a_1=45\div3^2=5). In exams, find (r) first and then (a_1).

The governing concept is the explicit or general rule for a geometric progression. The formula is a_n = 2 × 5^(n−1), so substitute each required index carefully. For n = 4, a₄ = 2 × 5^3 = 2 × 125 = 250. For n = 5, a₅ = 2 × 5^4 = 2 × 625 = 1250. Adding the two terms gives a₄ + a₅ = 250 + 1250 = 1500. Therefore option A is correct. The exponent is n−1, not n; replacing it by n would make each calculated term five times too large. Options B, C, and D do not follow from the stated rule and can arise from using an incorrect exponent or adding incorrectly.

From (12\cdot2^{n-1}=1536), (2^{n-1}=128=2^7), so (n=8). In exams, factor first and compare powers.

From (750=6r^3), (r^3=125) and (r=5), so (a_5=750\cdot5=3750). In exams, find the ratio first.

For this GP, the first term is \(a=15\), the common ratio is \(r=2\), and \(n=9\). Using \(S_n=\frac{a(r^n-1)}{r-1}\), we get \(S_9=\frac{15(2^9-1)}{2-1}=15(512-1)=7665\). Hence, 7665 is correct. A value such as 7680 can result from an error in subtraction or in the final calculation. Exam tip: identify \(a\), \(r\), and \(n\) before substituting in the GP sum formula.

In a geometric progression, the seventh term is \(a_7=ar^6\). Thus, \(320=5r^6\), so \(r^6=64=2^6\). Since \(r\) is stated to be positive, \(r=2\). Although \(r=-2\) also has sixth power 64, it is excluded by the positive condition. Exam tip: in the \(n\)th term, the exponent of \(r\) is always \(n-1\).

Each term is multiplied by (\frac{1}{2}), and the seventh term is (\frac{3}{2}). In exams, fractional terms can also be checked in order.

(a_5=3\cdot5^4=1875) and (a_6=9375), so the sum is (11250). In exams, calculate large powers separately.

(S_n=\frac{4(3^n-1)}{3-1}=2(3^n-1)), and (2(3^n-1)=484) gives (n=5). In exams, simplify the sum and identify the power.

The sequence is geometric because each term is obtained by multiplying the previous term by 3. Its first term is \(a=5\), its common ratio is \(r=3\), and the nth-term formula is \(a_n=ar^{n-1}\). The exponent is one less than the term number because the first term has no multiplication by the ratio.

For the thirteenth term, \(a_{13}=5\cdot3^{12}\). Since \(3^{12}=531441\), multiplication by 5 gives \(2657205\). Therefore, option C is correct. A common error is to use \(3^{13}\), which would count one extra multiplication and produce a value that is too large.

The consecutive terms can be written as \(x, xr, xr^2\). Hence \(y^2=(xr)^2=x(xr^2)=xz\). The relation \(x+z=2y\) belongs to an arithmetic progression. Exam tip: square the middle term.

Here \(r=\frac{1}{3}\), so \(a_8=729\cdot\left(\frac{1}{3}\right)^7=\frac{1}{3}\). In exams, apply fractional ratios carefully in decreasing GPs.

From (8\cdot2^{n-1}=4096), (2^{n-1}=512=2^9), so (n=10). In exams, equate powers to find the term number.

For a geometric progression, the fifth term is \(a_5=ar^4\). Thus, \(2304=9r^4\), so \(r^4=256=4^4\). Since \(r\) is positive, \(r=4\). Option 3 is not correct because \(9\times3^4=729\), not 2304. Exam tip: use \(a_n=ar^{n-1}\) and check the exponent carefully.

The first term is (14) and the ratio is (3), so (a_n=14\cdot3^{n-1}). In exams, keep (a) and (r) correct in (ar^{n-1}).

The governing concept is the finite geometric-series sum. The first term is a = 4, and the common ratio is r = 12/4 = 3. For n = 6 and r ≠ 1, apply Sₙ = a(rⁿ − 1)/(r − 1). Therefore S₆ = 4(3⁶ − 1)/(3 − 1) = 4(729 − 1)/2 = 4 × 728/2 = 1456. A direct check produces the six terms 4, 12, 36, 108, 324, and 972, whose total is 4 + 12 + 36 + 108 + 324 + 972 = 1456. Thus option A is correct. Option B is a nearby arithmetic error, while C and D do not follow from the formula or from adding the listed six terms. Identifying the ratio before applying the formula is essential.

For the middle term, (x^2=18\cdot200=3600), so (x=60). In exams, the square of the middle GP term equals the product of the outer terms.

The first five terms are (6,30,150,750,3750), and their sum is (4686). In exams, add terms carefully when the ratio is large.

In option A, each term is obtained by multiplying the preceding term by \(-\frac{1}{2}\): \(-8/16=4/-8=-2/4\). Hence it is a GP with a negative common ratio. In option B, the ratios are not constant. Exam tip: check at least two consecutive ratios.

The first term is (216) and the ratio is \(\frac{1}{3}\), so the correct rule is \(216\cdot\left(\frac{1}{3}\right)^{n-1}\). In exams, write the fractional ratio in a decreasing GP.