Exercise 8.1
Write the first five terms of the sequence whose nth term is an = n(n + 2).
Answer
a1 = 1×3 = 3, a2 = 2×4 = 8, a3 = 3×5 = 15, a4 = 4×6 = 24, a5 = 5×7 = 35.
First five terms: 3, 8, 15, 24, 35.
Write the first five terms of the sequence whose nth term is an = n/(n+1).
Answer
a1 = 1/2, a2 = 2/3, a3 = 3/4, a4 = 4/5, a5 = 5/6.
Write the first five terms of the sequence whose nth term is an = 2n.
Answer
2, 4, 8, 16, 32.
Write the first five terms of the sequence whose nth term is an = (2n − 3)/6.
Answer
a1 = −1/6, a2 = 1/6, a3 = 1/2, a4 = 5/6, a5 = 7/6.
Write the first five terms: an = (−1)n−1 5n+1.
Answer
25, −125, 625, −3125, 15625.
Write the first five terms: an = n(n2+5)/4.
Answer
a1 = 1×6/4 = 3/2, a2 = 2×9/4 = 9/2, a3 = 21/2, a4 = 21, a5 = 75/2.
Find a17 and a24 for an = 4n − 3.
Answer
a17 = 4(17) − 3 = 65. a24 = 4(24) − 3 = 93.
Find a7 for an = n2/2n.
Answer
a7 = 49/128.
Find a9 for an = (−1)n−1 n3.
Answer
a9 = (−1)8 × 729 = 729.
Find a20 for an = n(n−2)/(n+3).
Answer
a20 = 20×18/23 = 360/23.
a1 = 3, an = 3an−1 + 2 for n > 1. Write first five terms and series.
Answer
3, 11, 35, 107, 323. Series: 3 + 11 + 35 + 107 + 323 + …
a1 = −1, an = an−1/n, n ≥ 2. Write first five terms and series.
Answer
−1, −1/2, −1/6, −1/24, −1/120. Series: −1 + (−1/2) + (−1/6) + (−1/24) + (−1/120) + …
a1 = a2 = 2, an = an−1 − 1, n > 2. Write first five terms and series.
Answer
2, 2, 1, 0, −1. Series: 2 + 2 + 1 + 0 + (−1) + …
Fibonacci sequence: a1 = a2 = 1, an = an−1 + an−2, n > 2. Find an+1/an for n = 1, 2, 3, 4, 5.
Answer
Sequence: 1, 1, 2, 3, 5, 8. Ratios: n=1: 1/1=1; n=2: 2/1=2; n=3: 3/2; n=4: 5/3; n=5: 8/5.
Exercise 8.2
Find 20th and nth terms of G.P. 5/2, 5/4, 5/8, …
Answer
a = 5/2, r = 1/2. an = (5/2)(1/2)n−1 = 5/2n. a20 = 5/220.
Find 12th term of G.P. whose 8th term is 192 and common ratio is 2.
Answer
a×27 = 192
⇒ a = 3/2. a12 = (3/2)×211 = 3072.
5th, 8th and 11th terms of a G.P. are p, q, s. Show q2 = ps.
Answer
p = ar4, q = ar7, s = ar10. q2 = a2r14 = ar4×ar10 = ps. (Proved)
4th term of G.P. equals square of 2nd term, first term is −3. Find 7th term.
Answer
ar3 = (ar)2
⇒ r = a = −3. a7 = (−3)(−3)6 = −2187.
Q5. Which term of the sequences
(a) 2, 2√2, 4, … is 128?
Answer
a = 2, r = √2. 2(√2)n−1 = 128 = 27. 2(n+1)/2 = 27
⇒ n = 13.
(b) √3, 3, 3√3, … is 729?
Answer
a = √3, r = √3. (√3)n = 729 = (√3)12
⇒ n = 12.
(c) 1/3, 1/9, 1/27, … is 1/19683?
Answer
(1/3)n = 3−9 → n = 9.
For what values of x, −2/7, x, −7/2 are in G.P.?
Answer
x2 = (−2/7)(−7/2) = 1 → x = ±1.
Sum to 20 terms: 0.15, 0.015, 0.0015, …
Answer
a = 0.15, r = 0.1. S20 = (0.15/0.9)(1 − 10−20) = (1/6)(1 − 10−20).
Sum to n terms: √7, √21, 3√7, …
Answer
a = √7, r = √3. Sn = √7(√3n − 1)/(√3 − 1) = (√7/2)(√3+1)(√3n − 1).
Sum to n terms: 1, −a, a2, −a3, … (a ≠ −1).
Answer
First term = 1, ratio = −a. Sn = (1−(−a)n)/(1+a).
Sum to n terms: x3, x5, x7, … (x ≠ ±1).
Answer
a = x3, r = x2. Sn = x3(x2n − 1)/(x2 − 1).
Evaluate Σk=111 (2 + 3k).
Answer
= 2×11 + 3(311−1)/(3−1) = 22 + (3/2)(177147−1) = 22 + 265719 = 265741.
Sum of first three terms of G.P. is 39/10 and product is 1. Find common ratio and terms.
Answer
Let a/r, a, ar. a3 = 1
⇒ a = 1. 1/r+1+r = 39/10
⇒ 10r2−29r+10 = 0
⇒ r = 5/2 or 2/5. Terms: 2/5, 1, 5/2.
How many terms of G.P. 3, 32, 33, … are needed to give sum 120?
Answer
3(3n−1)/2 = 120 → 3n = 81 → n = 4.
Sum of first three terms of G.P. is 16, sum of next three is 128. Find first term, ratio and Sn.
Answer
S3=16, S6−S3=128
⇒ r3=8
⇒ r=2. a(7)=16
⇒ a=16/7. Sn=(16/7)(2n−1).
a = 729, 7th term = 64. Find S7.
Answer
729r6=64
⇒ r=2/3. S7=729(1−(2/3)7)/(1/3) = 2187×(1−128/2187) = 2059.
Sum of first two terms = −4, fifth term = 4 × third term. Find G.P.
Answer
r2=4
⇒ r=±2. For r=2: a=−4/3; G.P.: −4/3, −8/3, −16/3, … For r=−2: a=4; G.P.: 4, −8, 16, −32, …
4th, 10th and 16th terms are x, y, z. Prove x, y, z are in G.P.
Answer
x=ar3, y=ar9, z=ar15. y2=a2r18=xz. (Proved)
Sum to n terms: 8, 88, 888, 8888, …
Answer
Sn = (8/9)[(10+102+…+10n) − n] = (8/9)[10(10n−1)/9 − n] = 80(10n−1)/81 − 8n/9.
Find the sum of products of corresponding terms of 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.
Answer
Products: 256, 128, 64, 32, 16 (G.P., a=256, r=1/2). Sum = 256(1−(1/2)5)/(1/2) = 496.
Products of corresponding terms of a, ar, …arn−1 and A, AR, …ARn−1 form G.P. Find common ratio.
Answer
Products: aA, aArR, aA(rR)2, … Common ratio = rR. (Proved)
Four numbers in G.P.: third term exceeds first by 9, second exceeds fourth by 18. Find them.
Answer
Let a, ar, ar2, ar3. ar2−a=9 and ar−ar3=18. Solving: r=−2, a=3. Numbers: 3, −6, 12, −24.
pth, qth, rth terms of G.P. are a, b, c. Prove aq−rbr−pcp−q = 1.
Answer
a=ARp−1, b=ARq−1, c=ARr−1. The product = A0×R0 = 1 (exponents of A and R each sum to 0). (Proved)
First term = a, nth term = b, product of n terms = P. Prove P2 = (ab)n.
Answer
P = anrn(n−1)/2, b = arn−1. P2 = a2nrn(n−1) = (a·arn−1)n = (ab)n. (Proved)
Ratio of sum of first n terms to sum of (n+1)th to (2n)th term is 1/rn. Prove it.
Answer
Sn = a(rn−1)/(r−1). Sum of next n terms = Sn×rn. Ratio = 1/rn. (Proved)
a, b, c, d in G.P. Show (a2+b2+c2)(b2+c2+d2) = (ab+bc+cd)2.
Answer
With b=ar, c=ar2, d=ar3: LHS = a4r2(1+r2+r4)2 = RHS = (a2r+a2r3+a2r5)2. (Proved)
Insert two numbers between 3 and 81 so that the sequence is G.P.
Answer
r3 = 81/3 = 27
⇒ r = 3. Numbers: 9 and 27.
Find n so that (an+1+bn+1)/(an+bn) is G.M. of a and b.
Answer
Setting equal to √(ab) and simplifying gives n = −1/2.
Sum of two numbers is 6 times their G.M. Show numbers are in ratio (3+2√2):(3−2√2).
Answer
a+b=6√(ab). (a+b)2=36ab, (a−b)2=32ab
⇒ (a−b)=4√2×√(ab). Then a:b = (3+2√2):(3−2√2). (Proved)
If A and G are A.M. and G.M. between two positive numbers, prove the numbers are A±√(A2−G2).
Answer
a+b=2A, ab=G2. (a−b)2=4A2−4G2
⇒ a−b=±2√(A2−G2). Combined with a+b=2A: a,b=A±√(A2−G2). (Proved)
Bacteria doubles every hour, 30 initially. Find count at end of 2nd, 4th and nth hour.
Answer
Count at end of t hours = 30×2t. End 2nd: 120. End 4th: 480. End nth: 30×2n.
Rs 500 deposited at 10% compound interest. Find amount after 10 years.
Answer
A = 500×(1.1)10 = 500×2.5937 ≈ Rs 1296.87.
A.M. of roots = 8 and G.M. = 5. Find the quadratic equation.
Answer
α+β = 16, αβ = 25. Equation: x2−16x+25 = 0.
Miscellaneous Exercise on Chapter 8
f(x+y) = f(x)f(y), f(1) = 3, Σf(x) = 120 (x from 1 to n). Find n.
Answer
f(x) = 3x. Σ3x = 3(3n−1)/2 = 120
⇒ 3n = 81
⇒ n = 4.
Sum of G.P. is 315, first term = 5, common ratio = 2. Find last term and number of terms.
Answer
5(2n−1) = 315
⇒ 2n = 64
⇒ n = 6. Last term = 5×25 = 160.
First term of G.P. is 1. Sum of 3rd and 5th terms is 90. Find common ratio.
Answer
r2+r4 = 90
⇒ r4+r2−90=0
⇒ (r2−9)(r2+10)=0
⇒ r = ±3.
Three numbers in G.P. sum to 56. Subtracting 1, 7, 21 gives A.P. Find the numbers.
Answer
a(1+r+r2)=56 and a(r−1)2=8. Solving: r=2, a=8 or r=1/2, a=32. Numbers: 8, 16, 32.
G.P. has even terms. Sum of all = 5 × sum at odd places. Find common ratio.
Answer
Let 2n terms, ratio r. Sall/Sodd = r+1 = 5
⇒ r = 4.
(a+bx)/(a−bx) = (b+cx)/(b−cx) = (c+dx)/(c−dx), x≠0. Show a, b, c, d are in G.P.
Answer
By componendo-dividendo: a/b = b/c
⇒ b2=ac, similarly c2=bd. Hence a, b, c, d are in G.P. (Proved)
S=sum, P=product, R=sum of reciprocals of n terms in G.P. Prove P2Rn=Sn.
Answer
S = a(rn−1)/(r−1). P = anrn(n−1)/2. R = (rn−1)/(arn−1(r−1)). P2Rn = an(rn−1)n/(r−1)n = Sn. (Proved)
a, b, c, d in G.P. Prove (an+bn), (bn+cn), (cn+dn) are in G.P.
Answer
With b=ar, c=ar2, d=ar3: (bn+cn)2 = a2nr2n(1+rn)2 = (an+bn)(cn+dn). (Proved)
a, b are roots of x2−3x+p=0; c, d are roots of x2−12x+q=0; a, b, c, d in G.P. Prove (q+p):(q−p) = 17:15.
Answer
a+b=3, ab=p; c+d=12, cd=q. G.P.: a, ar, ar2, ar3. a(1+r)=3, ar2(1+r)=12
⇒ r2=4, r=2, a=1. p=2, q=32. (q+p):(q−p)=34:30=17:15. (Proved)
A.M./G.M. = m:n for two positive numbers. Show a:b = (m+√(m2−n2)):(m−√(m2−n2)).
Answer
(a+b)/2√(ab) = m/n. Using (√a+√b)/(√a−√b) = (m+n)/(m−n) by componendo-dividendo, then applying again yields the required ratio. (Proved)
(i). Sum to n terms: 5 + 55 + 555 + …
Answer
Sn = (5/9)[10(10n−1)/9 − n] = 5(10n+1−10−9n)/81.
(ii). Sum to n terms: 0.6 + 0.66 + 0.666 + …
Answer
Sn = (2/3)[n − (1/9)(1−10−n)].
Find 20th term of 2×4 + 4×6 + 6×8 + …
Answer
nth term = 2n(2n+2) = 4n2+4n. 20th term = 4(400)+80 = 1680.
Farmer pays Rs 500 per year plus 12% interest on unpaid amount (balance Rs 6000). Total cost?
Answer
12 instalments. Interest = 720+660+…+60 (A.P.) = 6×780 = Rs 4680. Total cost = 6000+6000+4680 = Rs 16680.
Shamshad Ali: balance Rs 18000, Rs 1000 per year plus 10% interest on unpaid. Total cost?
Answer
18 instalments. Total interest = (18/2)(1800+100) = 9×1900 = Rs 17100. Total cost = 4000+18000+17100 = Rs 39100.
Letters mailed to 4 each, 50 paise each. Amount spent when 8th set mailed.
Answer
8th set has 48 = 65536 letters. Cost = 65536×0.50 = Rs 32768.
Rs 10000 at 5% simple interest. Amount in 15th year; total after 20 years.
Answer
Amount in nth year = 10000 + (n−1)×500. 15th year: 10000+7000 = Rs 17000. Total after 20 years: 10000+20×500 = Rs 20000.
Machine costs Rs 15625 depreciates 20% annually. Value after 5 years?
Answer
Value = 15625×(0.8)5 = 15625×0.32768 = Rs 5120.
150 workers; 4 drop out each day from 2nd day. Took 8 more days. Find original days planned.
Answer
Let n = original days. Setting total work equal: (n+8)(136−2n) = 150n
⇒ n2+15n−544=0
⇒ n=17. Work completed in 25 days.
Important Extra Questions
If first term of G.P. is 2 and S6 = 9×S3, find common ratio.
Answer
r3+1 = 9 → r3 = 8 → r = 2.
Three numbers in G.P.: sum = 14, product = 64. Find them.
Answer
a3 = 64
⇒ a = 4. 4/r+4+4r = 14
⇒ r = 2 or 1/2. Numbers: 2, 4, 8.
Sum of infinite G.P. = 3 and sum of squares = 3. Find G.P.
Answer
a/(1−r) = 3 and a2/(1−r2) = 3
⇒ a/(1+r)=1
⇒ r = 1/2, a = 3/2. G.P.: 3/2, 3/4, 3/8, …
Find sum to n terms: 1 + (1+2) + (1+2+3) + …
Answer
nth term = n(n+1)/2. Sn = (1/2)[Σn2+Σn] = n(n+1)(n+2)/6.
a, b, c in G.P. and ax = by = cz. Prove 1/x, 1/y, 1/z are in A.P.
Answer
b2=ac
⇒ k2/y=k1/x+1/z
⇒ 2/y = 1/x+1/z
⇒ 1/x, 1/y, 1/z in A.P. (Proved)