Geometric Progression

From (\frac{486}{18}=27=r^3), (r=3) and (a_1=18\div3=6). In exams find (r) first and then (a_1).

The governing concept is an explicit formula for the terms of a geometric progression. Substitute n = 4 into aₙ = 3 × 4ⁿ⁻¹: a₄ = 3 × 4³ = 3 × 64 = 192. Next, substitute n = 5: a₅ = 3 × 4⁴ = 3 × 256 = 768. Therefore a₄ + a₅ = 192 + 768 = 960, so option A is correct. A quick check is that the ratio between consecutive terms is 4, and 768 is four times 192. Option B is larger than the actual sum, option C does not result from the stated formula, and option D is also too large. The exponent must be n − 1; using n instead would shift both terms and produce an incorrect result.

From (11\cdot2^{n-1}=1408), (2^{n-1}=128=2^7), so (n=8). In exams factor first and compare powers.

In a GP, consecutive ratios are equal. From \(q^2=pr\), \(q/p=r/q\); from \(r^2=qs\), \(r/q=s/r\). Hence all ratios are equal. Exam tip: divide only after confirming that the terms are non-zero.

Here, the first term is \(a=9\), the common ratio is \(r=2\), and \(n=12\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_{12}=\frac{9(2^{12}-1)}{2-1}=9(4096-1)=36855\). The option 36864 equals \(9\times4096\); it results from missing the \(-1\). Exam tip: when \(r\ne1\), remember the \(r^n-1\) part of the sum formula.

In a geometric progression, the sixth term is \(a_6=ar^5\). Thus, \(224=7r^5\), so \(r^5=32=2^5\). Since \(r\) is positive, \(r=2\). If \(r=3\), then \(7\times3^5\) is obtained, not 224. Exam tip: in \(a_n=ar^{n-1}\), the exponent of \(r\) is always \(n-1\).

Each term is multiplied by (\frac{1}{3}), and the sixth term is (\frac{5}{27}). In exams fractional terms can also be checked in order.

(a_5=2\cdot4^4=512) and (a_6=2048), so the sum is (2560). In exams calculate large powers separately.

(S_n=2(3^n-1)/(3-1)=3^n-1), and (3^n-1=242) gives (n=5). In exams simplify the sum and identify the power.

From the fourth term to the seventh term, the position gap is 3, so \(a_7=a_4r^3\). Thus, \(1458=54r^3\), giving \(r^3=27\) and hence \(r=3\). If the ratio were 6, then \(54\times6^3\) would not equal 1458. Exam tip: for two terms \(a_m\) and \(a_n\), use \(a_n=a_mr^{n-m}\).

In 3, 6, 12, 24, the ratios of consecutive terms are 6/3 = 12/6 = 24/12 = 2, so it is a geometric progression. In option B, the differences change. Exam tip: identify a GP by checking equal consecutive ratios.

The common ratio is (-3), and the terms are (5,-15,45,-135,405,-1215). With a negative ratio, signs alternate.

For three consecutive terms of a geometric progression, the square of the middle term equals the product of its neighbouring terms. Thus, \(x^2=4\times100=400\). Hence \(x=\pm20\), but the progression is stated to be positive, so \(x=20\). Taking \(40\) does not give a common ratio. Exam tip: for consecutive GP terms \(a,b,c\), use \(b^2=ac\) directly.

(a_5=a_1r^4), so (243=3r^4) and (r^4=81), hence (r=3). When positive ratio is asked, take the positive root.

(a_6=a_3r^3), so (648=24r^3) and (r=3), hence (a_4=72). First find the ratio and then form the next term.

In (6,18,54,162,486,\ldots), (a_2=18) and (a_5=486). Check each option up to the required positions.

Given \(a_n=2\cdot3^{n-1}\), \(a_4=2\cdot3^3=54\) and \(a_6=2\cdot3^5=486\). Hence, \(a_4+a_6=54+486=540\). Option 486 is only the value of \(a_6\), not the sum of both terms. Exam tip: substitute the term number carefully into the exponent \(n-1\).

The terms are (160,80,40,20,10,5,\frac{5}{2}), so it is the seventh term. In a decreasing progression, multiply repeatedly by the ratio.

The governing concept is the finite sum of a geometric progression. With first term 32 and ratio one-half, successive terms are found by halving: \(32,16,8,4\). Adding the four required terms gives \(32+16+8+4=60\). The formula \(S_n=a(1-r^n)/(1-r)\) provides an independent check: \(S_4=32[1-(1/2)^4]/[1-1/2]=32(15/16)/(1/2)=60\). Hence option D is correct. The distractors 50, 54, and 58 do not equal the complete sum; they can arise from omitting a term, using an incorrect power, or making an addition error. The direct method and formula agree exactly.

For the middle term, ((x+4)^2=(x-2)(2x+8)), which gives (x=8). In three GP terms, the square of the middle term equals the product of the extremes.