Geometric Progression

The first five terms are (4,20,100,500,2500), and the sum is (3124). In exams, add terms carefully when the ratio is large.

In a GP, the ratio of consecutive terms remains fixed: \(a_{n+1}/a_n=r\). Option B describes a constant difference, which identifies an AP. Exam tip: check ratios, not differences, to identify a GP.

The first term is (100) and the ratio is \(\frac{1}{2}\), so the correct rule is \(100\cdot\left(\frac{1}{2}\right)^{n-1}\). In exams, write the fractional ratio in a decreasing GP.

The nth term of a GP is \(a_n=ar^{n-1}\). Thus, \(a_5=9r^4=2304\), so \(r^4=256=4^4\). Since \(r\) is stated to be positive, \(r=4\). If \(r=2\), then \(9\times2^4=144\), not 2304. Exam tip: for the fifth term, use the exponent \(5-1=4\).

The direct answer is option B, the 7th term. This is a geometric progression because every term is multiplied by the same ratio: 18/6=3, 54/18=3, and 162/54=3. Its nth-term formula is a_n=6(3)^(n-1). Put the given value 4374 into it: 6(3)^(n-1)=4374. Dividing by 6 gives 3^(n-1)=729. Since 729=3^6, n-1=6, so n=7. Option A is wrong because the 6th term is 6(3^5)=1458. Option B works because the 7th term is 6(3^6)=4374. Option C is wrong because the 8th term is 4374×3=13122. Option D is wrong because the 9th term is still three times the 8th term, not 4374. The useful method is to identify the first term and common ratio, then compare powers. Memory cue: in a GP, use first term times ratio to the power n-1.

\(a_8=192\cdot\left(\frac{1}{2}\right)^7=\frac{3}{2}\). In exams, apply powers of fractional ratios carefully.

(\frac{a_6}{a_2}=r^4=\frac{224}{14}=16), so (r=2). In exams, the ratio of distant terms gives a power based on the position gap.

In a GP, \(y=xr\) and \(z=xr^2\). Hence \(y^2=(xr)^2=x(xr^2)=xz\). The relation \(x+y=2z\) is associated with an AP, not a GP. Exam tip: square the middle term and compare it with the product of the outer terms.

The position gap is (5) and (r=2), so the ratio is (2^5=32). In exams, use (\frac{a_m}{a_n}=r^{m-n}).

Here (a=5) and (r=3), so (S_6=\frac{5(3^6-1)}{3-1}=1820). In exams, identify (a) and (r) from the general term.

From (4\cdot3^{n-1}=972), (3^{n-1}=243=3^5), so (n=6). In exams, compare powers to find the term number.

(\frac{a_7}{a_3}=r^4=\frac{1458}{18}=81), so (r=3). In exams, a gap of (4) positions gives (r^4).

From \(1024\left(\frac{1}{2}\right)^{n-1}=2\), \(2^{10-(n-1)}=2^1\), so \(n=10\). In exams, count decreasing powers carefully.

In a GP, the sixth term is \(a_6=ar^{6-1}=ar^5\). Thus, \(352=11r^5\), so \(r^5=32=2^5\). Since \(r\) is positive, \(r=2\). Taking \(r=3\) would give \(11\times3^5\), not 352. Exam tip: in \(a_n=ar^{n-1}\), remember that the exponent is \(n-1\).

Here, the first term is \(a=9\), the common ratio is \(r=3\), and the number of terms is \(n=6\). The sum of the first \(n\) terms of a GP is \(S_n=\frac{a(r^n-1)}{r-1}\). Therefore, \(S_6=\frac{9(3^6-1)}{3-1}=\frac{9(729-1)}{2}=3276\). Hence, option A is correct. Exam tip: identify \(a\), \(r\), and \(n\) before substituting into the formula.

(\frac{1024}{16}=64=r^3), so (r=4) and (a_1=\frac{16}{4}=4). In exams, find (r) first and then the first term.

The sixth term is (x\cdot3^5=243x=2430), so (x=10). In exams, put the algebraic first term in (ar^{n-1}) too.

(2\cdot4^{n-1}=2048), so (4^{n-1}=1024=4^5) and (n=6). In exams, equate the last term to the general term.

Here, \(S_4\) means the sum of the first four terms: \(3+15+75+375=468\). Therefore, 468 is correct. Note that 375 is the last term, not the sum; similarly, 475 is not obtained from the required addition. Exam tip: when only a few terms are given, direct addition is often the quickest way to verify the sum.

From (10\cdot4^{n-1}=10240), (4^{n-1}=1024=4^5), so (n=6). In exams, equate powers to find the term number.

In a geometric progression, \(a_n=a_1r^{n-1}\). Therefore, \(a_5=4(-3)^{5-1}=4(-3)^4=4\times81=324\). Hence, \(324\) is correct. \(-324\) would result only if the power of \(-3\) were odd. Exam tip: with a negative common ratio, first check whether the exponent is even or odd.

Direct answer: Option B, 800. The displayed sequence is a GP: each term is multiplied by 7, since 14/2=7, 98/14=7, and 686/98=7. The question asks for the first four terms, which are already shown. Add them carefully: 2+14=16, 16+98=114, and 114+686=800. Therefore option B is correct. Option A, 798, is 2 less than the correct total and may result from an addition mistake. Option C, 802, is 2 more and may result from another arithmetic error. Option D, 806, is also not the sum of the four displayed values. A formula could be used, but direct addition is safer for only four visible terms. Memory cue: when all requested terms are written and few, add them directly and recheck once.

There are (6) steps from the third to the ninth term, so (a_9=72\cdot2^6=4608). In exams, count the position gap.

Each term is multiplied by (\frac{1}{3}), so the terms are (405,135,45,15,5). In exams, you can also check a decreasing GP in order.

For a geometric progression with \(r\ne1\), the sum of the first \(n\) terms is \(S_n=\frac{a(r^n-1)}{r-1}\). Thus, \(S_4=\frac{8(3^4-1)}{3-1}=\frac{8(81-1)}{2}=320\). Hence, the given statement is true. Values such as \(312\) or \(324\) result from an incorrect calculation. Exam tip: substitute \(n=4\) correctly in \(r^n\) before simplifying.