Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, ...
Hint. Identify a and r=(second term)/(first term), then apply the general term formula.
a=5/2, r=(5/4)/(5/2)=1/2. an=a.r^(n-1)=(5/2)(1/2)^(n-1)=5/2^n. a20=5/2^20.
✦ Working through each part gives: a20=5/2^20; an=5/2^n.
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Hint. a12 is 4 terms beyond a8, so a12=a8.r^4 without needing the first term at all.
a12=a8.r^(12-8)=192(2)^4=192(16)=3072.
✦ Working through each part gives: 3072.
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q^2=ps.
Hint. Write each term using the general-term formula, then compare q^2 with the product ps directly.
p=ar^4, q=ar^7, s=ar^10. q^2=(ar^7)^2=a^2r^14. ps=(ar^4)(ar^10)=a^2r^14. Since both equal a^2r^14, q^2=ps.
✦ Working through each part gives: proved: q^2=ps, since both sides equal a^2.r^14.
The 4th term of a G.P. is the square of its second term, and the first term is -3. Determine its 7th term.
Hint. Write the condition a4=(a2)^2 in terms of a and r, and solve for r using the given value of a.
a4=ar^3, a2=ar. Given ar^3=(ar)^2=a^2r^2, so dividing by ar^2 (nonzero): r=a. Since a=-3, r=-3. a7=ar^6=(-3)(-3)^6=(-3)(729)=-2187.
✦ Working through each part gives: -2187.
Which term of the following sequences: (a) 2, 2sqrt2, 4, ... is 128? (b) sqrt3, 3, 3sqrt3, ... is 729? (c) 1/3, 1/9, 1/27, ... is 1/19683?
Hint. For each part, find a and r, set the general term equal to the target value, and solve for n by matching powers of the same base.
(a) a=2, r=2sqrt2/2=sqrt2. an=2(sqrt2)^(n-1)=128=2^7, so (sqrt2)^(n-1)=2^6, i.e. 2^((n-1)/2)=2^6, giving (n-1)/2=6, n=13. (b) a=sqrt3, r=3/sqrt3=sqrt3. an=(sqrt3)^n=3^(n/2)=729=3^6, so n/2=6, n=12. (c) a=1/3, r=1/3. an=(1/3)^n=1/19683=1/3^9, so n=9.
✦ Working through each part gives: (a) the 13th term. (b) the 12th term. (c) the 9th term.
For what values of x are the numbers -2/7, x, -7/2 in G.P.?
Hint. The middle term squared equals the product of the outer two terms.
x^2=(-2/7)(-7/2)=1, so x=1 or x=-1.
✦ Working through each part gives: x = 1 or x = -1.
Find the sum to 20 terms of the G.P. 0.15, 0.015, 0.0015, ...
Hint. Identify a and r, note r<1, and apply the finite sum formula with n=20.
a=0.15, r=0.1. S20=a(1-r^20)/(1-r)=0.15(1-0.1^20)/0.9=(1/6)(1-10^-20).
✦ Working through each part gives: s20 = (1/6)(1 - 10^-20).
Find the sum to n terms of the G.P. sqrt7, sqrt21, 3sqrt7, ...
Hint. Confirm the common ratio is sqrt3, then apply the finite sum formula.
a=sqrt7, r=sqrt21/sqrt7=sqrt3 (check: 3sqrt7/sqrt21=3sqrt7/(sqrt7.sqrt3)=3/sqrt3=sqrt3, consistent). Sn=sqrt7((sqrt3)^n-1)/(sqrt3-1).
✦ Working through each part gives: sn = sqrt7[(sqrt3)^n - 1] / (sqrt3 - 1).
Find the sum to n terms of the sequence 1, -a, a^2, -a^3, ... (if a is not equal to -1).
Hint. This is a G.P. with first term 1 and common ratio -a; apply the finite sum formula.
First term 1, r=-a. Sn=[1-(-a)^n]/[1-(-a)]=[1-(-a)^n]/(1+a).
✦ Working through each part gives: sn = [1 - (-a)^n] / (1+a).
Find the sum to n terms of the sequence x^3, x^5, x^7, ... (if x is not equal to plus or minus 1).
Hint. This is a G.P. with first term x^3 and common ratio x^2; apply the finite sum formula.
a=x^3, r=x^2. Sn=x^3(x^(2n)-1)/(x^2-1).
✦ Working through each part gives: sn = x^3(x^(2n) - 1) / (x^2 - 1).
Evaluate the sum from k=1 to 11 of (2+3^k).
Hint. Split the sum into a constant part and a geometric part, then evaluate each separately.
Sum = sum(2, k=1 to 11) + sum(3^k, k=1 to 11) = 22 + 3(3^11-1)/(3-1) = 22 + 3(177147-1)/2 = 22 + 3(177146)/2 = 22+265719=265741.
✦ Working through each part gives: 265741.
The sum of the first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Hint. Write the three terms as a/r, a, ar so the product condition immediately gives a, then solve the sum condition as a quadratic in r.
Product: (a/r)(a)(ar)=a^3=1, so a=1. Sum: 1/r+1+r=39/10, so 1/r+r=29/10, giving 10r^2-29r+10=0. Discriminant=841-400=441=21^2. r=(29+-21)/20=5/2 or 2/5.
✦ Working through each part gives: common ratio 5/2 (terms 2/5, 1, 5/2) or equivalently 2/5 (same terms in reverse order).
How many terms of the G.P. 3, 3^2, 3^3, ... are needed to give the sum 120?
Hint. Apply the finite sum formula with a=r=3 and solve for n.
Sn=3(3^n-1)/(3-1)=120, so 3^n-1=80, 3^n=81=3^4, n=4.
✦ Working through each part gives: 4 terms.
The sum of the first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio, and the sum to n terms.
Hint. Write both sums using the common factor (1+r+r^2), then divide one by the other to isolate r^3.
a(1+r+r^2)=16 and ar^3(1+r+r^2)=128. Dividing: r^3=128/16=8, so r=2. Then a(1+2+4)=16, so 7a=16, a=16/7. Sn=(16/7)(2^n-1)/(2-1)=(16/7)(2^n-1).
✦ Working through each part gives: a=16/7, r=2, Sn=(16/7)(2^n-1).
Given a G.P. with a=729 and 7th term 64, determine S7.
Hint. Find r from a7=ar^6, then apply the finite sum formula.
729.r^6=64, so r^6=64/729=(2/3)^6, giving r=2/3. S7=729(1-(2/3)^7)/(1-2/3)=729x3x(1-128/2187)=2187x(2059/2187)=2059.
✦ Working through each part gives: 2059.
Find a G.P. for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.
Hint. The second condition gives r directly (r^2=4); use it in the first condition to find a.
a5=4.a3 means ar^4=4ar^2, so r^2=4, r=+-2. If r=2: a(1+2)=-4, a=-4/3. If r=-2: a(1-2)=-4, a=4.
✦ Working through each part gives: either -4/3, -8/3, -16/3, ... (r=2) or 4, -8, 16, -32, ... (r=-2).
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively, prove that x, y, z are in G.P.
Hint. Express each term with the general-term formula, then compare the ratios y/x and z/y.
x=ar^3, y=ar^9, z=ar^15. y/x=r^6 and z/y=r^6. Since the two ratios are equal, x, y, z are themselves in G.P.
✦ Working through each part gives: proved: y/x = z/y = r^6, so x, y, z are in G.P.
Find the sum to n terms of the sequence 8, 88, 888, 8888, ...
Hint. Factor out 8, then relate the resulting 1,11,111,... pattern to a genuine G.P. of powers of 10.
Sn=8[1+11+111+...n terms]=(8/9)[9+99+999+...]=(8/9)[(10+10^2+...+10^n)-n]=(8/9)[10(10^n-1)/9-n]=80(10^n-1)/81-8n/9.
✦ Working through each part gives: sn = 80(10^n - 1)/81 - 8n/9.
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.
Hint. The products themselves form a new G.P.; identify its first term and common ratio, then sum all 5 terms.
Products: 2x128=256, 4x32=128, 8x8=64, 16x2=32, 32x(1/2)=16. This is a G.P.: a=256, r=1/2, n=5. Sum=256(1-(1/2)^5)/(1-1/2)=256x2x(31/32)=512x31/32=496.
✦ Working through each part gives: 496.
Show that the products of the corresponding terms of the sequences a, ar, ar^2, ..., ar^(n-1) and A, AR, AR^2, ..., AR^(n-1) form a G.P., and find the common ratio.
Hint. Write out the k-th product term and the (k+1)-th product term, then compute their ratio.
The k-th product term is ar^(k-1).AR^(k-1)=aA(rR)^(k-1). The ratio of consecutive product terms is [aA(rR)^k]/[aA(rR)^(k-1)]=rR, a constant, so the products form a G.P. with common ratio rR.
✦ Working through each part gives: the products form a G.P. with common ratio rR (first term aA).
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the fourth by 18.
Hint. Write the four terms as a, ar, ar^2, ar^3, translate both conditions into equations, then divide one by the other to eliminate a.
ar^2-a=9, i.e. a(r^2-1)=9. ar-ar^3=18, i.e. -ar(r^2-1)=18, i.e. ar(r^2-1)=-18. Dividing: r=-18/9=-2. Then a(4-1)=9, so a=3.
✦ Working through each part gives: 3, -6, 12, -24.
If the pth, qth and rth terms of a G.P. are a, b and c, respectively, prove that a^(q-r).b^(r-p).c^(p-q)=1.
Hint. Express a, b, c using the general term formula with a common first term and ratio, then show both the exponents of the first term and of the ratio sum to zero.
a=AR^(p-1), b=AR^(q-1), c=AR^(r-1) for some first term A and ratio R. The exponent of A in a^(q-r)b^(r-p)c^(p-q) is (q-r)+(r-p)+(p-q)=0. The exponent of R is (p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q), which expands and simplifies to 0 as well. So the whole expression equals A^0.R^0=1.
✦ Working through each part gives: proved: a^(q-r).b^(r-p).c^(p-q) = 1.
If the first and the nth term of a G.P. are a and b, respectively, and P is the product of the n terms, prove that P^2=(ab)^n.
Hint. Write P as the product of all n terms, collecting the powers of r using the sum 0+1+2+...+(n-1), then compare P^2 with (ab)^n.
P=a.ar.ar^2...ar^(n-1)=a^n.r^(n(n-1)/2). So P^2=a^(2n).r^(n(n-1)). Also (ab)^n=(a.ar^(n-1))^n=(a^2r^(n-1))^n=a^(2n)r^(n(n-1)), matching P^2 exactly.
✦ Working through each part gives: proved: P^2 = (ab)^n.
Show that the ratio of the sum of the first n terms of a G.P. to the sum of the terms from the (n+1)th to the (2n)th term is 1/r^n.
Hint. Write the second sum as S(2n)-Sn, factor out r^n from it, and compare to the first-n-terms sum.
Sfirst=a(r^n-1)/(r-1). The next-n-terms sum equals S(2n)-Sn=a(r^(2n)-1)/(r-1)-a(r^n-1)/(r-1)=a.r^n(r^n-1)/(r-1)=r^n.Sfirst. Ratio=Sfirst/(r^n.Sfirst)=1/r^n.
✦ Working through each part gives: proved: the ratio equals 1/r^n.
If a, b, c and d are in G.P., show that (a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2.
Hint. Substitute b=ar, c=ar^2, d=ar^3 into both sides and simplify, using the common factor (1+r^2+r^4) that appears on both sides.
LHS=(a^2+a^2r^2+a^2r^4)(a^2r^2+a^2r^4+a^2r^6)=a^2(1+r^2+r^4).a^2r^2(1+r^2+r^4)=a^4r^2(1+r^2+r^4)^2. RHS=(a^2r+a^2r^3+a^2r^5)^2=[a^2r(1+r^2+r^4)]^2=a^4r^2(1+r^2+r^4)^2. LHS=RHS.
✦ Working through each part gives: proved: both sides equal a^4.r^2.(1+r^2+r^4)^2.
Insert two numbers between 3 and 81 so that the resulting sequence is a G.P.
Hint. The four terms 3, G1, G2, 81 form a G.P. with 81 as the 4th term; find r first.
81=3.r^3, so r^3=27, r=3. G1=3(3)=9, G2=3(3)^2=27.
✦ Working through each part gives: 9 and 27 (giving the G.P. 3, 9, 27, 81).
Find the value of n so that (a^(n+1)+b^(n+1))/(a^n+b^n) may be the geometric mean between a and b.
Hint. Set the expression equal to sqrt(ab), cross-multiply, and factor to isolate a common term of (sqrt(a)-sqrt(b)).
Setting (a^(n+1)+b^(n+1))/(a^n+b^n)=sqrt(ab) and cross-multiplying gives a^(n+1)+b^(n+1)=a^(n+1/2)b^(1/2)+a^(1/2)b^(n+1/2). Rearranging: a^(n+1/2)(a^(1/2)-b^(1/2))=b^(n+1/2)(a^(1/2)-b^(1/2)). For a not equal to b, dividing both sides by (a^(1/2)-b^(1/2)) gives a^(n+1/2)=b^(n+1/2), which forces n+1/2=0.
✦ Working through each part gives: n = -1/2.
The sum of two numbers is 6 times their geometric mean, show that the numbers are in the ratio (3+2sqrt2):(3-2sqrt2).
Hint. Set a=(3+2sqrt2)m and b=(3-2sqrt2)m for a scale factor m, then verify a+b=6.sqrt(ab) directly.
Let a=(3+2sqrt2)m, b=(3-2sqrt2)m. Then a+b=6m, and ab=[(3)^2-(2sqrt2)^2]m^2=(9-8)m^2=m^2, so sqrt(ab)=m. Then 6.sqrt(ab)=6m=a+b, confirming the condition holds exactly for this ratio.
✦ Working through each part gives: proved: with a:b = (3+2sqrt2):(3-2sqrt2), a+b = 6.sqrt(ab) holds exactly.
If A and G are the A.M. and G.M., respectively, between two positive numbers, prove that the numbers are A +- sqrt((A+G)(A-G)).
Hint. Use (a-b)^2=(a+b)^2-4ab rewritten in terms of A and G, then combine with a+b=2A to isolate a and b.
a+b=2A and ab=G^2. So (a-b)^2=(a+b)^2-4ab=4A^2-4G^2=4(A+G)(A-G), giving a-b=+-2.sqrt((A+G)(A-G)). Combined with a+b=2A: a=[ (a+b)+(a-b) ]/2=A +- sqrt((A+G)(A-G)), and b is the same expression with the sign flipped.
✦ Working through each part gives: the numbers are A + sqrt((A+G)(A-G)) and A - sqrt((A+G)(A-G)).
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present originally, how many will be present at the end of the 2nd hour, the 4th hour, and the nth hour?
Hint. Model the count as a G.P. with first term 30 and common ratio 2, where the count at the end of hour n is the (n+1)-th term of the sequence starting from the original count.
Count at end of hour n = 30x2^n. End of 2nd hour: 30x2^2=120. End of 4th hour: 30x2^4=480. End of nth hour: 30x2^n.
✦ Working through each part gives: 120 at the end of the 2nd hour, 480 at the end of the 4th hour, and 30x2^n at the end of the nth hour.
What will Rs 500 amount to in 10 years after being deposited in a bank paying annual compound interest at 10% per year?
Hint. Compound growth is a G.P. with first term 500 and common ratio 1.1; the amount after 10 years is the 10th multiplication, i.e. 500 times r^10.
Amount = 500(1.1)^10. Numerically, (1.1)^10 is approximately 2.5937, so the amount is approximately 500x2.5937 which is about Rs 1296.87.
✦ Working through each part gives: 500(1.1)^10, approximately Rs 1296.87.
If the A.M. and G.M. of the roots of a quadratic equation are 8 and 5, respectively, obtain the quadratic equation.
Hint. Use the A.M. and G.M. to recover the sum and product of the roots, then build the quadratic from those.
If the roots are alpha, beta: (alpha+beta)/2=8, so alpha+beta=16. sqrt(alpha.beta)=5, so alpha.beta=25. The quadratic is x^2-(sum)x+(product)=0.
✦ Working through each part gives: x^2 - 16x + 25 = 0.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh108.pdf) — Exercise 8.1 (14 questions on general sequences), Exercise 8.2 (32 questions on Geometric Progression — the book's only numbered exercise on G.P.; no 'Exercise 8.3' exists), plus the chapter's Miscellaneous Exercise (18 questions), 64 questions total. Confirmed against the CBSE curriculum PDF that Arithmetic Progression is not re-taught in this chapter (it is Class 10 content; only Arithmetic Mean reappears here) and that the sum to infinity of a G.P. is genuinely CBSE-summative syllabus content despite the current book never deriving it anywhere, including its own Summary — that formula and its recurring-decimal application are taught in the chapter .md but are not sourced from an actual NCERT exercise question, so no corresponding entry appears in this solutions file. Several garbled stacked-fraction and surd expressions (Exercise 8.1 Q4/Q5/Q6/Q8/Q10; Exercise 8.2 Q1/Q5/Q6/Q11) were cross-checked via 300dpi page renders. Miscellaneous Exercise Q16 (simple interest) was deliberately solved as an Arithmetic Progression, not a Geometric one, and flagged as a commonError — a recurring trap since it appears inside a G.P.-focused exercise. Questions are referenced from the NCERT textbook for identification.
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