Geometric Progression

The \(n\)th term of a GP is \(a_n=ar^{n-1}\). Thus, \(a_6=3r^5=96\), so \(r^5=32=2^5\). Since \(r\) is positive, \(r=2\). If \(r=3\), then \(3\times3^5\) is not equal to 96. Exam tip: for the sixth term, the exponent of \(r\) is \(6-1=5\).

From (5\cdot4^{n-1}=5120), (4^{n-1}=1024=4^5), so (n=6). In exams, equate powers to find (n).

\(a_6=160\cdot\left(\frac{1}{2}\right)^5=5\). In exams, apply powers of fractional ratios carefully.

The governing concept is the finite sum formula for a geometric progression: Sₙ = a(rⁿ − 1)/(r − 1) when r ≠ 1. Here a = 3, r = 3, and n = 5. Substitution gives S₅ = 3(3⁵ − 1)/(3 − 1) = 3(243 − 1)/2 = 3 × 242/2 = 363. Direct verification gives the five terms 3, 9, 27, 81, and 243, whose sum is also 363. Therefore the statement is true and option A is correct. Option B is too small and option C is too large. Option D gives only the fifth term, 243, rather than the sum of all five terms, so it does not correctly justify the statement even though it uses a genuine term of the progression.

The governing idea is the finite-sum formula for a geometric progression. The first term is a = 1, the common ratio is r = 4, and the number of terms is n = 6. Since r ≠ 1, Sₙ = a(rⁿ − 1)/(r − 1). Therefore S₆ = 1(4⁶ − 1)/(4 − 1) = (4096 − 1)/3 = 4095/3 = 1365. A direct calculation confirms this result: 1 + 4 + 16 + 64 + 256 + 1024 = 1365. Thus option A is correct. Options B, C, and D are nearby distractors likely caused by failing to subtract 1, using the wrong number of terms, or making an arithmetic error while dividing 4095 by 3.

(\frac{a_8}{a_4}=r^4=\frac{768}{48}=16), so (r=2). In exams, the ratio of distant terms gives a power based on the position gap.

In a GP, the ratio of every term to its preceding term must remain constant. In option B, \(10/5=2\) and \(20/10=2\), but \(35/20=7/4\). Hence it is not a GP. Exam tip: check all consecutive ratios.

The position gap is (4) and (r=3), so the ratio is (3^4=81). In exams, use (\frac{a_m}{a_n}=r^{m-n}).

Here (a=9) and (r=2), so (S_8=9(2^8-1)=2295). In exams, identify (a) and (r) from the general term.

From (3\cdot5^{n-1}=9375), (5^{n-1}=3125=5^5), so (n=6). In exams, compare powers to find the term number.

(\frac{a_6}{a_2}=r^4=\frac{1458}{18}=81), so (r=3). In exams, a gap of (4) positions gives (r^4).

In a geometric progression, the seventh term is \(a_7=ar^6\). Thus, \(384=6r^6\), so \(r^6=64=2^6\). Since \(r\) is stated to be positive, \(r=2\). Although \(r=-2\) also satisfies \(r^6=64\), it violates the given condition. Exam tip: in \(a_n=ar^{n-1}\), the exponent of \(r\) is always \(n-1\).

Here, the first term is \(a=6\), the common ratio is \(r=3\), and the number of terms is \(n=7\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Therefore, \(S_7=\frac{6(3^7-1)}{3-1}=\frac{6(2187-1)}{2}=6558\). Hence, option A is correct. A nearby value such as \(6560\) can result from an arithmetic error in the final calculation. Exam tip: write down \(a\), \(r\), and \(n\) separately before applying the formula.

(\frac{6250}{50}=125=r^3), so (r=5) and (a_1=\frac{50}{5^2}=2). In exams, find (r) first and then the first term.

The fifth term is (x\cdot5^4=625x=3125), so (x=5). In exams, put the algebraic first term in (ar^{n-1}) too.

(4\cdot4^{n-1}=4096) gives (4^n=4096=4^6), so (n=6). In exams, equate the last term to the general term.

The first four terms are 7, 21, 63, and 189. Therefore, \(S_4=7+21+63+189=280\). This can also be checked using \(S_n=\frac{a(r^n-1)}{r-1}\), where \(a=7\) and \(r=3\). The option 294 may result from an incorrect addition involving the last term. Exam tip: for a small number of terms, add directly and verify with the GP sum formula if needed.

From (2\cdot4^{n-1}=8192), (4^{n-1}=4096=4^6), so (n=7). In exams, equate powers to find the term number.

In a geometric progression, the nth term is given by \(a_n=a_1r^{n-1}\). Therefore, \(a_8=9(-2)^{8-1}=9(-2)^7=9(-128)=-1152\). Hence, option A is correct. \(1152\) results from a sign error, since an odd power of \((-2)\) is negative. Exam tip: When the common ratio is negative, first check whether the exponent is even or odd.

The sum of the first four terms is (5+30+180+1080=1295). In exams, direct addition is safe when the number of terms is small.