Miscellaneous Exercise — Sequences and Series

1. 8.M.14 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   If f is a function satisfying f(x+y)=f(x).f(y) for all x,y in N, such that f(1)=3 and the sum from x=1 to n of f(x) equals 120, find the value of n.

   Hint. The functional equation forces f(x)=3^x, turning the given sum into a G.P. sum with a=r=3.

   f(x+y)=f(x)f(y) with f(1)=3 forces f(x)=3^x (since f(2)=f(1)f(1)=9=3^2, f(3)=f(2)f(1)=27=3^3, and so on). So the sum is a G.P. with a=3, r=3: 3(3^n-1)/(3-1)=120, giving 3^n-1=80, 3^n=81=3^4, n=4.

   ✦ Working through each part gives: n = 4.

2. 8.M.24 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   The sum of some terms of a G.P. is 315, whose first term and common ratio are 5 and 2, respectively. Find the last term and the number of terms.

   Hint. Apply the finite sum formula with a=5, r=2 to solve for n, then compute the last term directly.

   Sn=5(2^n-1)/(2-1)=5(2^n-1)=315, so 2^n-1=63, 2^n=64=2^6, n=6. Last term=a.r^(n-1)=5(2)^5=5(32)=160.

   ✦ Working through each part gives: 6 terms, with last term 160.

3. 8.M.34 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of the G.P.

   Hint. Write both terms using a=1, form a quadratic in r^2, and solve.

   a=1, so a3+a5=r^2+r^4=90. Let y=r^2: y^2+y-90=0, giving y=9 or y=-10. Since y=r^2 must be non-negative, y=9, so r^2=9, r=+-3.

   ✦ Working through each part gives: r = 3 or r = -3.

4. 8.M.45 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

   Hint. Write the GP terms as a/r, a, ar; use the sum condition and the AP condition on the shifted values to form two equations, then solve.

   Let the terms be a/r, a, ar with a/r+a+ar=56. After subtracting 1,7,21: (a/r-1),(a-7),(ar-21) are in AP, so 2(a-7)=(a/r-1)+(ar-21). Simplifying gives 2a+8=a(1/r+r), and from the sum equation a(1/r+r)=56-a, so 2a+8=56-a, giving a=16. Then 1/r+r=(56-16)/16=5/2, giving 2r^2-5r+2=0, so r=2 or r=1/2.

   ✦ Working through each part gives: 8, 16, 32 (subtracting 1,7,21 gives the AP 7,9,11).

5. 8.M.55 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places, find its common ratio.

   Hint. Write the total sum and the odd-position sum (itself a G.P. with ratio r^2) using the finite sum formula, then divide one by the other.

   For 2m terms, S=a(r^(2m)-1)/(r-1). The odd-position terms a, ar^2, ar^4,...,ar^(2m-2) form a G.P. with ratio r^2: S_odd=a(r^(2m)-1)/(r^2-1). Given S=5.S_odd: 1/(r-1)=5/[(r-1)(r+1)], so r+1=5, r=4.

   ✦ Working through each part gives: r = 4.

6. 8.M.65 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   If (a+bx)/(a-bx)=(b+cx)/(b-cx)=(c+dx)/(c-dx), with x not equal to 0, show that a, b, c, d are in G.P.

   Hint. Cross-multiply each pair of equal ratios separately to get two clean relations, b^2=ac and c^2=bd, then combine them.

   From the first equality: (a+bx)(b-cx)=(a-bx)(b+cx), which simplifies to 2b^2x=2acx, so b^2=ac (since x is not 0). From the second equality similarly: c^2=bd. Since b^2=ac gives b/a=c/b, and c^2=bd gives c/b=d/c, all three ratios b/a, c/b, d/c are equal, so a,b,c,d are in G.P.

   ✦ Working through each part gives: proved: b^2=ac and c^2=bd together give b/a=c/b=d/c, so a,b,c,d are in G.P.

7. 8.M.75 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   Let S be the sum, P the product, and R the sum of reciprocals of n terms in a G.P. Prove that P^2.R^n=S^n.

   Hint. Express P and R in terms of a, r, n using previously derived identities, then substitute both into P^2.R^n and simplify.

   S=a(r^n-1)/(r-1). P=a^n.r^(n(n-1)/2) (product of n terms). R=sum of reciprocals=(1/a)(r^n-1)/(r-1)/r^(n-1)=S/(a^2.r^(n-1)) (derived by summing the reciprocal G.P.). Then P^2.R^n=a^(2n)r^(n(n-1)).[S/(a^2r^(n-1))]^n=a^(2n)r^(n(n-1)).S^n/(a^(2n)r^(n(n-1)))=S^n.

   ✦ Working through each part gives: proved: P^2.R^n = S^n.

8. 8.M.84 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   If a, b, c, d are in G.P., prove that (a^n+b^n), (b^n+c^n), (c^n+d^n) are in G.P.

   Hint. Substitute b=ar, c=ar^2, d=ar^3, factor out the common (1+r^n) from all three expressions, and compare consecutive ratios.

   a^n+b^n=a^n(1+r^n). b^n+c^n=a^n.r^n(1+r^n). c^n+d^n=a^n.r^(2n)(1+r^n). The ratio of the 2nd to the 1st is r^n, and the ratio of the 3rd to the 2nd is also r^n, so the three quantities are in G.P.

   ✦ Working through each part gives: proved: all three quantities share the common ratio r^n.

9. 8.M.95 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

   If a and b are the roots of x^2-3x+p=0, and c, d are the roots of x^2-12x+q=0, where a, b, c, d form a G.P., prove that (q+p):(q-p)=17:15.

   Hint. Use Vieta's formulas to get a+b and c+d, and ab=p, cd=q, then use the G.P. structure to relate p and q.

   a+b=3, ab=p, c+d=12, cd=q. With b=ar, c=ar^2, d=ar^3: a(1+r)=3 and ar^2(1+r)=12; dividing gives r^2=4. p=ab=a^2r, q=cd=a^2r^5, so q/p=r^4=16 (true for r=+-2). So q=16p, giving (q+p)/(q-p)=(17p)/(15p)=17/15.

   ✦ Working through each part gives: proved: (q+p):(q-p) = 17:15.

10. 8.M.105 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    The ratio of the A.M. and G.M. of two positive numbers a and b is m:n. Show that a:b = (m+sqrt(m^2-n^2)) : (m-sqrt(m^2-n^2)).

    Hint. Let k=a/b, express the AM/GM ratio purely in terms of k, and solve the resulting quadratic in sqrt(k).

    Let k=a/b. AM/GM=(a+b)/(2sqrt(ab))=(k+1)/(2sqrt(k))=m/n. Let u=sqrt(k): n(u^2+1)=2mu, so nu^2-2mu+n=0, giving u=[m+-sqrt(m^2-n^2)]/n. Taking the root with u>1 (since a>=b makes k>=1): k=u^2=[m+sqrt(m^2-n^2)]^2/n^2. Since n^2=(m+sqrt(m^2-n^2))(m-sqrt(m^2-n^2)), this simplifies to k=(m+sqrt(m^2-n^2))/(m-sqrt(m^2-n^2)).

    ✦ Working through each part gives: proved: a:b = (m+sqrt(m^2-n^2)) : (m-sqrt(m^2-n^2)).

11. 8.M.116 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    Find the sum of the following series up to n terms: (i) 5+55+555+... (ii) 0.6+0.66+0.666+...

    Hint. For each part, factor out the leading digit, then relate the resulting repunit-style pattern to powers of 10, splitting the sum into a constant part and a geometric part.

    (i) Sn=5[1+11+111+...n terms]=(5/9)[9+99+999+...]=(5/9)[(10+10^2+...+10^n)-n]=(5/9)[10(10^n-1)/9-n]=50(10^n-1)/81-5n/9. (ii) Sn=6[0.1+0.11+0.111+...]=(6/9)[(1-0.1)+(1-0.01)+...+(1-0.1^n)]=(2/3)[n-(0.1+0.01+...+0.1^n)]. The bracketed geometric sum is 0.1(1-0.1^n)/(1-0.1)=(1/9)(1-10^-n). So Sn=(2n/3)-(2/27)(1-10^-n).

    ✦ Working through each part gives: (i) Sn = 50(10^n - 1)/81 - 5n/9. (ii) Sn = (2n/3) - (2/27)(1 - 10^-n).

12. 8.M.124 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    Find the 20th term of the series 2x4 + 4x6 + 6x8 + ... + n terms.

    Hint. Write the k-th term as a product of two linear expressions in k, then substitute k=20.

    The k-th term is (2k)(2k+2)=4k(k+1). T20=4(20)(21)=4(420)=1680.

    ✦ Working through each part gives: 1680.

13. 8.M.135 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

    Hint. The unpaid balance drops by Rs 500 each year, forming an AP from 6000 down to 500; sum that AP to get total interest, then add it to the cash and principal already paid.

    Balance=12000-6000=6000, paid off in 12 instalments of Rs 500 principal. Unpaid balances before each payment: 6000,5500,...,500 (12 terms, AP). Total interest=12% of their sum=0.12x(12/2)(6000+500)=0.12x6x6500=0.12x39000=4680. Total cost=cash(6000)+principal instalments(6000)+interest(4680)=16680.

    ✦ Working through each part gives: rs 16680.

14. 8.M.145 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

    Hint. The unpaid balance drops by Rs 1000 each year, forming an AP from 18000 down to 1000; sum that AP to get total interest, then add it to the cash and principal already paid.

    Balance=22000-4000=18000, paid off in 18 instalments of Rs 1000 principal. Unpaid balances before each payment: 18000,17000,...,1000 (18 terms, AP). Total interest=10% of their sum=0.10x(18/2)(18000+1000)=0.10x9x19000=0.10x171000=17100. Total cost=cash(4000)+principal instalments(18000)+interest(17100)=39100.

    ✦ Working through each part gives: rs 39100.

15. 8.M.155 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    A person writes a letter to four of his friends, asking each to copy it and mail it to four different persons with the same instruction. Assuming the chain is not broken and it costs 50 paise to mail one letter, find the amount spent on postage when the 8th set of letters is mailed.

    Hint. The number of letters in each successive set forms a G.P. with first term 4 and ratio 4; sum the first 8 terms of that G.P., then multiply by the cost per letter.

    Set k has 4^k letters, a G.P. with a=4, r=4. Total letters through set 8 = 4(4^8-1)/(4-1)=4(65536-1)/3=4(65535)/3=87380. Cost=87380x0.50=43690.

    ✦ Working through each part gives: rs 43690.

16. 8.M.164 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    A man deposited Rs 10000 in a bank at 5% simple interest annually. Find the amount in the 15th year since he deposited the amount, and the total amount after 20 years.

    Hint. This is simple interest, so the amounts form an Arithmetic Progression, not a G.P. -- the amount after k completed years is 10000 plus 500 times k.

    Amount after k years=10000+500k (5% of 10000 is 500 added each year). Amount in the 15th year (after 14 completed years)=10000+500(14)=10000+7000=17000. Amount after 20 years=10000+500(20)=10000+10000=20000.

    ✦ Working through each part gives: rs 17000 in the 15th year; Rs 20000 after 20 years.

    Where students slip. This question sits inside the G.P. chapter but describes simple interest, which grows by a constant amount each year -- an Arithmetic Progression. Applying a G.P. formula here is the standard trap.

17. 8.M.174 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

    Hint. Depreciating by 20% each year means the value forms a G.P. with common ratio 0.8; apply the general term formula with n=5.

    Value forms a G.P.: a=15625, r=0.8=4/5. Value after 5 years=a.r^5=15625x(4/5)^5. Since 15625=5^6 and (4/5)^5=4^5/5^5=1024/3125: value=5^6x1024/5^5=5x1024=5120.

    ✦ Working through each part gives: rs 5120.

18. 8.M.185 marksNCERT Class 11 Mathematics, Sequences and Series, Reprint 2026-27

    150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day, 4 more on the third day, and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

    Hint. Let n be the originally planned number of days, so the total work is 150n worker-days; the actual work is the sum of an AP of workers over (n+8) days, starting at 150 and dropping by 4 each day, and these two totals must be equal.

    Total work=150n (planned). Actual work=sum of AP with first term 150, common difference -4, over (n+8) terms = [(n+8)/2][300-4(n+7)]. Setting this equal to 150n and simplifying leads to n^2+15n-544=0, which factors via the quadratic formula to n=(-15+sqrt(2401))/2=(-15+49)/2=17 (rejecting the negative root). So the work took n+8=25 days.

    ✦ Working through each part gives: 25 days.

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.