Extra Practice Questions
Arithmetic Progressions (R.D. Sharma Level)
1. If the ratio of the sum of the first n terms of two APs is (7n + 1) : (4n + 27), find the ratio of their 9th terms.
Let the two APs have first terms a₁, a₂ and common differences d₁, d₂.
Given: Sn of AP₁Sn of AP₂ = (n/2)[2a₁+(n-1)d₁](n/2)[2a₂+(n-1)d₂] = 7n+14n+27
This simplifies to 2a₁+(n-1)d₁2a₂+(n-1)d₂ = a₁ + ((n-1)/2)d₁a₂ + ((n-1)/2)d₂ = 7n+14n+27.
We need the ratio of the 9th terms: a₁ + 8d₁a₂ + 8d₂. Comparing this with the above form, we need (n-1)/2 = 8, which means n-1=16, so n=17.
Substitute n=17 into the given ratio: 7(17)+14(17)+27 = 119+168+27 = 12095 = 2419.
Answer: The ratio of their 9th terms is 24:19.
2. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.
Let the three parts in AP be (a-d), a, (a+d).
Sum: (a-d) + a + (a+d) = 3a = 207 => a = 69.
The parts are (69-d), 69, (69+d). The two smaller parts are (69-d) and 69.
Product: (69-d) × 69 = 4623 => 69-d = 4623/69 = 67 => d = 2.
The parts are (69-2), 69, (69+2), which are 67, 69, and 71.
Answer: The three parts are 67, 69, and 71.
3. The sum of the first n terms of an AP is given by Sₙ = 3n² + 5n. Find the 25th term of this AP.
The nth term (aₙ) is the difference between the sum of n terms and the sum of (n-1) terms: aₙ = Sₙ – Sₙ₋₁.
aₙ = (3n² + 5n) – [3(n-1)² + 5(n-1)]
= (3n² + 5n) – [3(n²-2n+1) + 5n-5]
= 3n² + 5n – [3n² – 6n + 3 + 5n – 5]
= 3n² + 5n – (3n² – n – 2) = 6n + 2.
Now find the 25th term: a₂₅ = 6(25) + 2 = 150 + 2 = 152.
Answer: The 25th term is 152.
4. Find the sum of all integers between 100 and 200 that are divisible by 9.
The first integer greater than 100 divisible by 9 is 108. The last integer less than 200 divisible by 9 is 198.
This forms an AP: 108, 117, …, 198. Here, a=108, d=9, aₙ=198.
Find n: 198 = 108 + (n-1)9 => 90 = (n-1)9 => n-1=10 => n=11.
Sum S₁₁ = n/2 (a + l) = 11/2 (108 + 198) = 11/2 (306) = 11 × 153 = 1683.
Answer: The sum is 1683.
… (16 more questions follow)
5. If m times the mth term of an AP is equal to n times its nth term (m ≠ n), show that the (m+n)th term of the AP is zero.
Given: m × aₘ = n × aₙ
m[a + (m-1)d] = n[a + (n-1)d]
ma + m(m-1)d = na + n(n-1)d
ma – na = [n(n-1) – m(m-1)]d
a(m-n) = [n² – n – m² + m]d
a(m-n) = [-(m²-n²) + (m-n)]d
a(m-n) = [-(m-n)(m+n) + (m-n)]d
Divide by (m-n): a = [-(m+n) + 1]d = [1 – m – n]d.
We need to show aₘ₊ₙ = 0.
aₘ₊ₙ = a + (m+n-1)d
= [1 – m – n]d + (m+n-1)d = [- (m+n-1)]d + (m+n-1)d = 0.
(Proved)
6. The angles of a quadrilateral are in AP. If the greatest angle is double the least, find all the four angles.
Let the angles be a-3d, a-d, a+d, a+3d. Sum of angles = 360°.
(a-3d) + (a-d) + (a+d) + (a+3d) = 4a = 360° => a = 90°.
Greatest angle = a+3d. Least angle = a-3d.
Given: a+3d = 2(a-3d) => a+3d = 2a-6d => a = 9d.
Since a=90°, 90 = 9d => d = 10°.
The angles are: 90-3(10)=60°, 90-10=80°, 90+10=100°, 90+3(10)=120°.
Answer: 60°, 80°, 100°, 120°.
7. If the sum of first p terms of an AP is q and the sum of first q terms is p, then show that the sum of first (p+q) terms is -(p+q).
Given: Sₚ = q => p/2[2a+(p-1)d] = q => 2a+(p-1)d = 2q/p —(1)
Given: Sq = p => q/2[2a+(q-1)d] = p => 2a+(q-1)d = 2p/q —(2)
Subtract (2) from (1): d(p-q) = 2q/p – 2p/q = -2(p²-q²)/pq. So, d = -2(p+q)/pq.
Substitute d in (1): 2a = 2q/p – (p-1)[-2(p+q)/pq].
a = q/p + (p-1)(p+q)/pq = [q² + p² + pq – p – q] / pq.
Sp+q = (p+q)/2 [2a + (p+q-1)d]
= (p+q)/2 [ 2(q²+p²+pq-p-q)/pq – 2(p+q-1)(p+q)/pq ]
= (p+q)/pq [ q²+p²+pq-p-q – (p²+q²+2pq-p-q) ]
= (p+q)/pq [ -pq ] = -(p+q).
(Proved)
8. Insert 5 numbers between 8 and 26 such that the resulting sequence is an AP.
We need to insert 5 arithmetic means. The sequence will be 8, A₁, A₂, A₃, A₄, A₅, 26.
This is an AP with a = 8 and the 7th term a₇ = 26.
a₇ = a + 6d => 26 = 8 + 6d => 18 = 6d => d = 3.
The numbers are:
A₁ = a+d = 8+3=11
A₂ = a+2d = 8+6=14
A₃ = a+3d = 8+9=17
A₄ = a+4d = 8+12=20
A₅ = a+5d = 8+15=23.
Answer: The numbers are 11, 14, 17, 20, 23.
9. Find the sum of all two-digit numbers which when divided by 3, leave a remainder of 1.
The first such number is 10. The next is 13. The last such number less than 100 is 97.
The AP is 10, 13, 16, …, 97. Here a=10, d=3, aₙ=97.
Find n: 97 = 10 + (n-1)3 => 87 = (n-1)3 => n-1 = 29 => n=30.
Sum S₃₀ = 30/2 (10 + 97) = 15(107) = 1605.
Answer: The sum is 1605.
10. In an AP, the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference.
Given a=2, l=29, Sₙ=155.
Sₙ = n/2 (a+l) => 155 = n/2 (2+29) = n/2 (31) => 310 = 31n => n=10.
The last term is the 10th term. a₁₀ = 29.
a₁₀ = a + 9d => 29 = 2 + 9d => 27 = 9d => d=3.
Answer: The common difference is 3.
11. If Sₙ, the sum of first n terms of an AP, is given by Sₙ = 5n² + 3n, then find its nth term and common difference.
aₙ = Sₙ – Sₙ₋₁ = (5n²+3n) – [5(n-1)²+3(n-1)]
= 5n²+3n – [5(n²-2n+1)+3n-3] = 5n²+3n – (5n²-10n+5+3n-3) = 5n²+3n – (5n²-7n+2)
aₙ = 10n – 2.
To find d, we can find a₁ and a₂.
a₁ = 10(1)-2 = 8.
a₂ = 10(2)-2 = 18.
d = a₂ – a₁ = 18 – 8 = 10.
Answer: nth term aₙ = 10n-2, common difference d = 10.
12. Which term of the AP 121, 117, 113, … is its first negative term?
Here, a=121, d = -4. We need to find the smallest n for which aₙ < 0.
a + (n-1)d < 0
121 + (n-1)(-4) < 0
121 – 4n + 4 < 0
125 < 4n => n > 125/4 => n > 31.25.
Since n must be an integer, the smallest integer value for n is 32.
Answer: The 32nd term is the first negative term.
13. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
Sum of houses preceding x = Sx-1. Sum of houses following x = S₄₉ – Sx.
Given Sx-1 = S₄₉ – Sx => Sx-1 + Sx = S₄₉.
Using Sₙ = n(n+1)/2:
(x-1)x/2 + x(x+1)/2 = 49(50)/2.
x(x-1+x+1)/2 = 49 × 25 => x(2x)/2 = 1225 => x² = 1225.
x = √1225 = 35.
Answer: The value of x is 35.
14. The sum of the first n terms of an AP is 3n² + 6n. Find the nth term and the 15th term of this AP.
Sₙ = 3n² + 6n.
aₙ = Sₙ – Sₙ₋₁ = (3n²+6n) – [3(n-1)²+6(n-1)]
= 3n²+6n – [3(n²-2n+1)+6n-6] = 3n²+6n – [3n²-6n+3+6n-6] = 3n²+6n – (3n²-3) = 6n+3.
nth term aₙ = 6n + 3.
15th term a₁₅ = 6(15) + 3 = 90 + 3 = 93.
Answer: aₙ = 6n+3, a₁₅ = 93.
15. If the pth term of an AP is 1/q and the qth term is 1/p, prove that the sum of the first pq terms is 1/2 (pq + 1).
aₚ = a+(p-1)d = 1/q —(1)
aq = a+(q-1)d = 1/p —(2)
Subtract (2) from (1): d(p-q) = 1/q – 1/p = (p-q)/pq. So, d = 1/pq.
Substitute d in (1): a + (p-1)/pq = 1/q => a = 1/q – (p-1)/pq = [p-(p-1)]/pq = 1/pq.
So, a=d=1/pq.
Spq = pq/2 [2a + (pq-1)d] = pq/2 [2/pq + (pq-1)/pq]
= pq/2 [ (2+pq-1)/pq ] = 1/2 (pq+1).
(Proved)
16. A man arranges to pay off a debt of ₹3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first installment.
Total debt S₄₀ = 3600. n=40. Let first installment be ‘a’.
S₄₀ = 40/2 [2a + 39d] = 3600 => 20[2a+39d]=3600 => 2a+39d=180 —(1)
Debt paid in 30 installments = 2/3 of total debt = (2/3)*3600 = 2400.
S₃₀ = 30/2 [2a + 29d] = 2400 => 15[2a+29d]=2400 => 2a+29d=160 —(2)
Subtract (2) from (1): 10d = 20 => d=2.
Substitute d in (2): 2a + 29(2) = 160 => 2a + 58 = 160 => 2a = 102 => a=51.
Answer: The first installment was ₹51.
17. Find the sum of all three-digit numbers which leave a remainder of 2 when divided by 5.
The numbers end in 2 or 7. First number is 102. Last number is 997.
The AP is 102, 107, 112, …, 997. a=102, d=5, l=997.
997 = 102 + (n-1)5 => 895 = (n-1)5 => n-1=179 => n=180.
S₁₈₀ = 180/2 (102 + 997) = 90(1099) = 98910.
Answer: The sum is 98910.
18. The ratio of the 11th term to the 18th term of an AP is 2:3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first 5 terms to the sum of the first 21 terms.
Given a₁₁/a₁₈ = (a+10d)/(a+17d) = 2/3 => 3a+30d=2a+34d => a=4d.
Ratio of 5th to 21st term: (a+4d)/(a+20d) = (4d+4d)/(4d+20d) = 8d/24d = 1/3.
Ratio of sums S₅/S₂₁ = [5/2(2a+4d)] / [21/2(2a+20d)] = [5(a+2d)] / [21(a+10d)]
= [5(4d+2d)] / [21(4d+10d)] = [5(6d)]/[21(14d)] = 30d/294d = 5/49.
Answer: Ratio of terms is 1:3, ratio of sums is 5:49.
19. If S₁, S₂, S₃ are the sum of n, 2n, 3n terms respectively of an AP, prove that S₃ = 3(S₂ – S₁).
S₁ = n/2[2a+(n-1)d]
S₂ = 2n/2[2a+(2n-1)d] = n[2a+(2n-1)d]
S₃ = 3n/2[2a+(3n-1)d]
Consider RHS: 3(S₂ – S₁) = 3[ n(2a+(2n-1)d) – n/2(2a+(n-1)d) ]
= 3n/2 [ 2(2a+(2n-1)d) – (2a+(n-1)d) ]
= 3n/2 [ 4a+4nd-2d – 2a-nd+d ]
= 3n/2 [ 2a + 3nd – d ] = 3n/2 [ 2a + (3n-1)d ] = S₃ = LHS.
(Proved)
20. A spiral is made of 20 semicircles, with radii r₁, r₂, … such that r₂ = r₁ + 1, r₃ = r₂ + 1, and so on, starting with r₁=1 cm. Find the total length of the spiral. (Leave answer in terms of π).
The radii form an AP: 1, 2, 3, …, 20.
The length of each semicircle is πr.
The total length is the sum of the lengths of the 20 semicircles: πr₁ + πr₂ + … + πr₂₀.
Total Length = π(r₁ + r₂ + … + r₂₀) = π × (Sum of the AP of radii).
The sum of radii S₂₀ = 20/2(1 + 20) = 10(21) = 210.
Total Length = π × 210 = 210π cm.
Answer: The total length of the spiral is 210π cm.
