Miscellaneous Exercise answers: Sequences and Series

Open NCERT p. 147Matches NCERT’s answer

1. Put x = y = 1: f(2) = f(1)f(1) = 3×3 = 9.
2. In general, f(k) = f(1)^k = 3^k for every natural number k, since adding 1 to the input each time multiplies the output by f(1).
3. So f(1)+f(2)+…+f(n) is the G.P. sum 3+3²+…+3ⁿ = 3(3ⁿ−1)/(3−1) = (3ⁿ⁺¹−3)/2.
4. Setting this equal to 120: (3ⁿ⁺¹−3)/2 = 120, so 3ⁿ⁺¹ = 243 = 3⁵.
5. So n+1 = 5, giving n = 4.

Answern = 4

Watch this explained “Backwards: the run length is the unknown”, 8:45 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 147Matches NCERT’s answer

1. First term a = 5, common ratio r = 2, sum Sₙ = 315.
2. Sₙ = a(rⁿ−1)/(r−1) = 5(2ⁿ−1) = 315, so 2ⁿ−1 = 63, i.e. 2ⁿ = 64 = 2⁶.
3. So the number of terms n = 6.
4. The last term is a×r⁽ⁿ⁻¹⁾ = 5×2⁵ = 160.

AnswerLast term = 160, number of terms = 6

Watch this explained “Backwards: the run length is the unknown”, 8:45 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 147Matches NCERT’s answer

1. First term a = 1. Third term = a r² = r², fifth term = a r⁴ = r⁴.
2. Given r² + r⁴ = 90. Let y = r²: y² + y − 90 = 0.
3. Solving, y = 9 or y = −10. Since y = r² cannot be negative, y = 9.
4. So r² = 9, giving r = 3 or r = −3.

AnswerCommon ratio = 3 or −3

Watch this explained “The count shows in the exponent”, 1:55 into Reaching any term without walking through the earlier ones

Open NCERT p. 147Matches NCERT’s answer

1. Let the three numbers in G.P. be a/r, a, ar. Their sum is 56, so a/r + a + ar = 56, which gives a/r + ar = 56 − a.
2. Subtracting 1, 7, 21 gives a/r − 1, a − 7, ar − 21, which are in A.P. So twice the middle equals the sum of the outer two: 2(a − 7) = (a/r − 1) + (ar − 21), which gives a/r + ar = 2a + 8.
3. Equate the two expressions for a/r + ar: 56 − a = 2a + 8, so 3a = 48 and a = 16.
4. Then 16/r + 16r = 40. Multiply by r and divide by 8: 2r² − 5r + 2 = 0, so (2r − 1)(r − 2) = 0, giving r = 2 or r = 1/2.
5. r = 2 gives 8, 16, 32. Subtracting 1, 7, 21 gives 7, 9, 11, an A.P. with common difference 2.
6. r = 1/2 gives 32, 16, 8. Subtracting 1, 7, 21 gives 31, 9, −13, also an A.P. (common difference −22). So both orders work, and the numbers are the same.

AnswerThe numbers are 8, 16, 32 (they also work in the order 32, 16, 8)

Watch this explained “A total and a product together”, 11:21 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 147Matches NCERT’s answer

1. Let the G.P. have 2n terms: a, ar, ar², …, ar⁽²ⁿ⁻¹⁾.
2. Total sum = a(r²ⁿ−1)/(r−1).
3. The odd-placed terms a, ar², ar⁴, …, ar⁽²ⁿ⁻²⁾ themselves form a G.P. with ratio r², and their sum is a(r²ⁿ−1)/(r²−1).
4. Given total = 5 × (odd-place sum): a(r²ⁿ−1)/(r−1) = 5a(r²ⁿ−1)/(r²−1).
5. Cancel the common factor and use r²−1 = (r−1)(r+1): 1/(r−1) = 5/[(r−1)(r+1)], so r+1 = 5.
6. So r = 4.

AnswerCommon ratio = 4

Watch the lesson Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 148One way to think about it

1. Take the first two ratios: (a+bx)/(a−bx) = (b+cx)/(b−cx). Cross-multiply: (a+bx)(b−cx) = (b+cx)(a−bx).
2. Expand the left side: ab − acx + b²x − bcx². Expand the right side: ab − b²x + acx − bcx².
3. Subtract the right side from the left: 2b²x − 2acx = 0, that is 2x(b² − ac) = 0. Since x ≠ 0, b² = ac, so b/a = c/b.
4. The same working on the second and third ratios (with b, c, d in place of a, b, c) gives c² = bd, so c/b = d/c.
5. So b/a = c/b = d/c: each term divided by the one before it gives the same number. That is what it means for a, b, c, d to be in G.P.

In shorta, b, c, d are in G.P., because b² = ac and c² = bd give b/a = c/b = d/c.

Watch this explained “The definition, both clauses”, 4:38 into A constant ratio between neighbours is the entire definition

Open NCERT p. 148One way to think about it

1. Let the G.P. be a, ar, ar², …, arⁿ⁻¹. First take r ≠ 1 (the case r = 1 is at the end).
2. Sum: S = a(rⁿ − 1)/(r − 1).
3. Product: P = a × ar × ar² × … × arⁿ⁻¹ = aⁿ r^{0+1+…+(n−1)} = aⁿ r^n(n−1)/2, so P² = a²ⁿ rⁿ⁽ⁿ⁻¹⁾.
4. The reciprocals 1/a, 1/(ar), …, 1/(arⁿ⁻¹) form a G.P. with first term 1/a and ratio 1/r, so R = (1/a)(1 − 1/rⁿ)/(1 − 1/r). Multiplying top and bottom by rⁿ gives R = (rⁿ − 1)/[a rⁿ⁻¹(r − 1)].
5. So Rⁿ = (rⁿ − 1)ⁿ / [aⁿ rⁿ⁽ⁿ⁻¹⁾ (r − 1)ⁿ].
6. Multiply: P²Rⁿ = a²ⁿ rⁿ⁽ⁿ⁻¹⁾ × (rⁿ − 1)ⁿ / [aⁿ rⁿ⁽ⁿ⁻¹⁾ (r − 1)ⁿ] = aⁿ(rⁿ − 1)ⁿ/(r − 1)ⁿ = [a(rⁿ − 1)/(r − 1)]ⁿ = Sⁿ.
7. If r = 1, every term is a: S = na, P = aⁿ and R = n/a, so P²Rⁿ = a²ⁿ × nⁿ/aⁿ = nⁿaⁿ = (na)ⁿ = Sⁿ.

In shortP²Rⁿ = Sⁿ

Watch the lesson Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 148One way to think about it

1. Prove: (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.
2. Since a, b, c, d are in G.P. with common ratio r: b = ar, c = ar², d = ar³.
3. aⁿ + bⁿ = aⁿ + aⁿrⁿ = aⁿ(1 + rⁿ).
4. bⁿ + cⁿ = aⁿrⁿ + aⁿr²ⁿ = aⁿrⁿ(1 + rⁿ).
5. cⁿ + dⁿ = aⁿr²ⁿ + aⁿr³ⁿ = aⁿr²ⁿ(1 + rⁿ).
6. So (bⁿ + cⁿ)/(aⁿ + bⁿ) = rⁿ and (cⁿ + dⁿ)/(bⁿ + cⁿ) = rⁿ. (Dividing needs 1 + rⁿ ≠ 0; if 1 + rⁿ = 0 all three expressions would be 0, and the question takes for granted that this does not happen.)
7. Both ratios equal rⁿ, so (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

In short(aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P. with common ratio rⁿ.

Watch this explained “The definition, both clauses”, 4:38 into A constant ratio between neighbours is the entire definition

Open NCERT p. 148One way to think about it

1. a, b are roots of x²−3x+p=0, so a+b=3 and ab=p. c, d are roots of x²−12x+q=0, so c+d=12 and cd=q.
2. Since a, b, c, d are in G.P. with common ratio r: b=ar, c=ar², d=ar³.
3. From a+b=3: a(1+r)=3. From c+d=12: ar²(1+r)=12. Dividing the second by the first gives r²=4.
4. p=ab=a²r and q=cd=a²r⁵=a²r·r⁴, so q+p = a²r(r⁴+1) and q−p = a²r(r⁴−1).
5. (q+p)/(q−p) = (r⁴+1)/(r⁴−1). Since r²=4, r⁴=16, so this is (16+1)/(16−1) = 17/15.
6. So (q+p) : (q−p) = 17 : 15.

In short(q+p) : (q−p) = 17 : 15

Watch this explained “Dividing leaves the gap”, 8:01 into Reaching any term without walking through the earlier ones

Open NCERT p. 148One way to think about it

1. A.M. = (a+b)/2, G.M. = √(ab). Given A.M. : G.M. = m : n, write A.M. = mk and G.M. = nk for some k.
2. The square of the gap between a and b equals the square of their total minus four times their product: (a−b)² = (a+b)² − 4ab = 4A.M.² − 4G.M.².
3. So a−b = 2√(A.M.² − G.M.²) = 2k√(m²−n²) (taking a>b), and a+b = 2A.M. = 2mk.
4. Adding and subtracting these: a = k(m+√(m²−n²)), b = k(m−√(m²−n²)).
5. Dividing, the factor k cancels: a : b = (m+√(m²−n²)) : (m−√(m²−n²)).

In shorta : b = (m+√(m²−n²)) : (m−√(m²−n²))

Watch this explained “Running it backwards”, 6:40 into Why one of the two means can never overtake the other

Open NCERT p. 148Matches NCERT’s answer

(i) 5 + 55 + 555 + …

1. Every term is 5 times a repunit: 5, 55, 555, … = 5×1, 5×11, 5×111, …
2. Write each repunit as (10^k−1)/9, so the sum becomes (5/9)×[(10−1)+(10²−1)+…+(10ⁿ−1)].
3. This splits into (5/9)×[(10+10²+…+10ⁿ) − n], and the bracket's first part is a G.P. sum: 10(10ⁿ−1)/9.
4. Simplifying the whole expression gives 5(10ⁿ⁺¹−9n−10)/81.

Answer5(10ⁿ⁺¹ − 9n − 10)/81

(ii) .6 +.66 +.666+…

1. Each term is 2/3 times (1 − 10⁻ᵏ): 0.6 = (2/3)(1−0.1), 0.66 = (2/3)(1−0.01), and so on.
2. Summing n such terms: (2/3)×[n − (0.1+0.01+…+10⁻ⁿ)].
3. The bracket's second part is a G.P. sum: (1−10⁻ⁿ)/9.
4. This gives (2n)/3 − (2/27)(1−10⁻ⁿ), which simplifies to 2(9n−1+10⁻ⁿ)/27.

Answer2(9n − 1 + 10⁻ⁿ)/27

Watch this explained “Check first, convert second”, 9:56 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 148Matches NCERT’s answer

1. The kth term of the series is (2k)×(2k+2), since the two factors climb by 2 each time: 2,4,6,… and 4,6,8,…
2. For the 20th term, put k=20: (2×20)×(2×20+2) = 40×42.
3. 40×42 = 1680.

Answer1680

Watch this explained “A rule read off the terms must be tested”, 4:08 into A rule that turns a position number into a term

Open NCERT p. 148Matches NCERT’s answer

1. Balance to be paid in instalments = 12000 − 6000 = ₹6000.
2. With instalments of Rs 500 each, the number of instalments = 6000/500 = 12.
3. Interest is 12% on whatever is still unpaid just before each instalment: on 6000, 5500, 5000, …, 500 — an A.P. of 12 terms.
4. Total interest = 12% of (6000+5500+…+500) = 12% of 39000 = ₹4680.
5. Total cost = cash paid + instalment amounts + interest = 6000 + 6000 + 4680 = ₹16680.

Answer₹16680

Open NCERT p. 148Matches NCERT’s answer

1. Balance to be paid in instalments = 22000 − 4000 = ₹18000.
2. With instalments of Rs 1000 each, the number of instalments = 18000/1000 = 18.
3. Interest is 10% on the unpaid amount just before each instalment: on 18000, 17000, …, 1000 — an A.P. of 18 terms summing to 171000.
4. Total interest = 10% of 171000 = ₹17100.
5. Total cost = 4000 + 18000 + 17100 = ₹39100.

Answer₹39100

Open NCERT p. 148Matches NCERT’s answer

1. Set 1 has 4 letters, set 2 has 4×4=16, set 3 has 4³=64, …, set k has 4^k letters.
2. Total letters mailed through the 8th set = 4+4²+…+4⁸ = 4(4⁸−1)/(4−1) = 87380.
3. Postage = 50 paise = ₹0.5 per letter, so total postage = 87380×0.5 = ₹43690.

Answer₹43690

Watch this explained “Forwards: total a stated run”, 7:42 into Subtracting a scaled copy of the total to collapse it to two terms

Open NCERT p. 148Checked by computerReads two ways: both answers shown

1. Simple interest each year = 5% of ₹10000 = ₹500. It is the same every year, so the amounts form an AP with common difference 500.
2. 'The amount in the 15th year' can mean the amount during the 15th year (before that year's interest is added) or at the end of it. NCERT's answer key uses the first, so it leads.
3. During the 15th year: in year 1 the amount is 10000, so a = 10000, d = 500, and a₁₅ = 10000 + 14 × 500 = ₹17000.
4. At the end of the 15th year: 15 years of interest have been added: 10000 + 15 × 500 = ₹17500.
5. After 20 years, 20 years of interest have been added: 10000 + 20 × 500 = ₹20000.

AnswerRead as the amount during the 15th year (NCERT's answer key): ₹17000. Read as the amount at the end of the 15th year: ₹17500. Total amount after 20 years: ₹20000.

Open NCERT p. 148Matches NCERT’s answer

1. Depreciating by 20% each year means each year's value is 80% of the year before — a G.P. with first term ₹15625 and ratio 4/5.
2. Value after 5 years = 15625 × (4/5)⁵.
3. 15625 × 1024/3125 = ₹5120.

Answer₹5120

Watch this explained “The rule, and its off-by-one”, 3:31 into Reaching any term without walking through the earlier ones

Open NCERT p. 148Matches NCERT’s answer

1. Let the job be planned to finish in n days with 150 workers throughout: total work = 150n worker-days.
2. Workers actually present each day form an A.P.: 150, 146, 142, …, dropping by 4 daily, over n+8 days (8 more than planned).
3. Total work actually done = sum of this A.P. over (n+8) days = (n+8)×150 − 2(n+7)(n+8).
4. Setting this equal to 150n and simplifying gives (n+7)(n+8) = 600, i.e. n²+15n−544 = 0.
5. Solving, n = 17 (the negative root is rejected).
6. So the work actually took n+8 = 25 days.

Answer25 days