Geometric Progression

The defining feature is a constant ratio: each term is twice the preceding term, so \(a=3\) and \(r=2\). For a geometric progression, the nth term is \(a_n=ar^{n-1}\). Hence \(a_{12}=3\times2^{11}=3\times2048=6144\). The exponent is 11 rather than 12 because the first term already contains the initial factor 3; reaching the twelfth position requires eleven successive multiplications by 2. Thus option B is correct. Option A corresponds to a smaller power, while options C and D result from using an incorrect initial factor or exponent. Substitution into the formula and repeated doubling both verify the answer.

The nth term of a geometric progression is \(a_n=ar^{n-1}\). Hence, \(a_8=5\times3^{8-1}=5\times3^7=5\times2187=10935\). Therefore, option C is correct. \(7290\) is not correct because it is not the value of \(5\times3^7\). Exam tip: the exponent of the common ratio in the nth term is always \(n-1\); for the eighth term, use \(r^7\).

Here \(r=\frac{1}{2}\) so \(a_9=256\cdot\left(\frac{1}{2}\right)^8=1\). In exams use the fractional ratio in decreasing progressions.

From (4\cdot3^{n-1}=2916), (3^{n-1}=729=3^6) so (n=7). In exams first equate the given term to (a_n).

In a GP, the \(n\)th term is \(a_n=ar^{n-1}\). Therefore, \(486=6r^4\), so \(r^4=81\). Since \(r\) is stated to be positive, \(r=3\). For example, if \(r=2\), the fifth term would be \(6\times2^4=96\), not 486. Exam tip: for the fifth term, the power of \(r\) is \(4\), because the exponent is always \(n-1\).

The first term is (10) and the ratio is (3) so (a_n=10\cdot3^{n-1}). In exams keep (a) and (r) correct in (ar^{n-1}).

The sum of the first (6) terms is (2(3^6-1)/(3-1)=728). In exams use (S_n=\frac{a(r^n-1)}{r-1}) for (r>1).

For the middle term (x^2=8\cdot72=576), so (x=24). In exams the square of the middle GP term equals the product of outer terms.

In a GP, each term is obtained by multiplying the previous term by the common ratio. Multiplication by a negative ratio reverses the sign, as in 3, −6, 12, −24. Exam tip: always check the sign of the ratio.

In a GP, \(y=xr\) and \(z=yr=xr^2\). Hence \(y^2=(xr)^2=xz\). The relation \(x+z=2y\) belongs to an AP, not necessarily to a GP. Exam tip: square the middle term to test three terms.

The first term is (81) and the ratio is \(\frac{1}{3}\), so the correct rule is \(81\cdot\left(\frac{1}{3}\right)^{n-1}\). In exams write the fractional ratio in a decreasing GP.

The nth term of a GP is












a_n=ar^{n-1}. Thus,












_6=4r^5=972, so r^5=243=3^5. Since r is positive, r=3. If r=2, the sixth term would be 128, not 972. Exam tip: for the sixth term, the exponent of r is 5, not 6.

\(a_7=96\cdot\left(\frac{1}{2}\right)^6=\frac{3}{2}\). In exams apply powers of fractional ratios carefully.

(\frac{a_6}{a_3}=r^3=\frac{160}{20}=8), so (r=2). In exams the ratio of distant terms gives a power based on position gap.

In a GP, consecutive ratios are equal: \(q/p=r/q\). Cross-multiplying gives \(q^2=pr\). The condition \(2q=p+r\) belongs to an AP. Exam tip: square the middle term to test three terms.

The position gap is (3) and (r=3), so the ratio is (3^3=27). In exams use (\frac{a_m}{a_n}=r^{m-n}).

Here (a=7) and (r=2), so (S_6=7(2^6-1)=441). In exams identify (a) and (r) from the general term.

From (2\cdot5^{n-1}=1250), (5^{n-1}=625=5^4) so (n=5). In exams compare powers to find the term number.

(\frac{a_5}{a_2}=r^3=\frac{324}{12}=27), so (r=3). In exams a gap of (3) positions gives (r^3).

From \(729\left(\frac{1}{3}\right)^{n-1}=1\), \(3^{6-(n-1)}=1\), so \(n=7\). In exams count decreasing powers carefully.

For a geometric progression, the seventh term is \(a_7=ar^6\). Hence, \(192=3r^6\), so \(r^6=64=2^6\). Since \(r\) is stated to be positive, \(r=2\). If \(r=3\), then \(3\times3^6\) is not 192. Exam tip: in \(a_n=ar^{n-1}\), the exponent of \(r\) is always \(n-1\).

Here, the first term is \(a=4\), 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, \(S_7=\frac{4(3^7-1)}{3-1}=\frac{4(2187-1)}{2}=4372\). Hence, 4372 is correct. An answer such as 4374 can result from a small error while evaluating \(3^7-1\) or dividing. Exam tip: write down \(a\), \(r\), and \(n\) separately before substituting in the formula.

(\frac{384}{48}=8=r^3), so (r=2) and (a_1=\frac{48}{2^2}=12). In exams find (r) first and then the first term.

The fifth term is (x\cdot2^4=16x=80), so (x=5). In exams put the algebraic first term in (ar^{n-1}) too.

Here, \(S_4\) is the sum of the first four terms: \(2+8+32+128=170\). Therefore, the correct answer is 170. Option 168 is incorrect because the sum of all four given terms is 170. Exam tip: when only a few terms are given, direct addition is often the quickest and safest method.