NCERT Mathematics · Class 10 · English · 2026-27
Arithmetic Progressions — Exercise 5.3
Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.
Choose a question
- Question 1
- Question 2
- Question 3
- Question 4
- Question 5
- Question 6
- Question 7
- Question 8
- Question 9
- Question 10
- Question 11
- Question 12
- Question 13
- Question 14
- Question 15
- Question 16
- Question 17
- Question 18
- Question 19
- Question 20
Question 1
Calculate the four finite AP sums in Question 1.
Step-by-step solution
\[S_n=\frac n2[2a+(n-1)d]\]
(i)
a=2,d=5,n=10: S₁₀=5(4+45)=245.
(ii)
a=−37,d=4,n=12: S₁₂=6(−74+44)=−180.
(iii)
a=0.6,d=1.1,n=100: S₁₀₀=50(1.2+108.9)=5505.
(iv)
1/15,1/12,1/10,… has d=1/60. S₁₁=(11/2)[2/15+10/60]=(11/2)(3/10)=33/20.
Resource ID: CB-10-M-CH5-EX5.3-Q1
Question 2
Find the sums of the APs ending at 84, 10 and −230.
Step-by-step solution
(i)
7,10½,14,…,84: a=7,d=7/2. From 84=7+(n−1)7/2, n=23. S=23(7+84)/2=2093/2=1046.5.
(ii)
34,32,30,…,10: n=1+(10−34)/(−2)=13. S=13(34+10)/2=286.
(iii)
−5,−8,−11,…,−230: n=1+(−230+5)/(−3)=76. S=76(−5−230)/2=−8930.
Resource ID: CB-10-M-CH5-EX5.3-Q2
Question 3
Find the unknown quantities in the ten AP cases.
Step-by-step solution
\[a_n=a+(n-1)d,\qquad S_n=\frac n2(a+a_n)\]
(i)
50=5+3(n−1) gives n=16; S₁₆=16(5+50)/2=440.
(ii)
35=7+12d gives d=7/3; S₁₃=13(7+35)/2=273.
(iii)
37=a+11×3 gives a=4; S₁₂=12(4+37)/2=246.
(iv)
a+2d=15 and 125=5(2a+9d). Thus 2a+4d=30 and 2a+9d=25; d=−1, a=17 and a₁₀=8.
(v)
75=(9/2)(2a+40) gives 2a=−70/3, so a=−35/3; a₉=a+40=85/3.
(vi)
90=(n/2)[4+8(n−1)] gives 2n²−n−45=0=(2n+9)(n−5). Reject negative n, so n=5; a₅=34.
(vii)
210=n(8+62)/2=35n gives n=6. Then 62=8+5d, so d=54/5.
(viii)
a=4−2(n−1)=6−2n. Then −14=n(a+4)/2=n(5−n), so (n−7)(n+2)=0. The valid n=7 gives a=−8.
(ix)
192=4(6+7d) gives d=6.
(x)
144=9(a+28)/2 gives a+28=32, so a=4.
Resource ID: CB-10-M-CH5-EX5.3-Q3
Question 4
How many terms of 9,17,25,… sum to 636?
Step-by-step solution
636=(n/2)[18+8(n−1)]=n(4n+5). Hence 4n²+5n−636=0=(n−12)(4n+53). Only the positive integer root is valid.
Resource ID: CB-10-M-CH5-EX5.3-Q4
Question 5
An AP starts at 5, ends at 45 and sums to 400. Find n and d.
Step-by-step solution
400=n(5+45)/2=25n gives n=16. Then 45=5+15d.
Resource ID: CB-10-M-CH5-EX5.3-Q5
Question 6
An AP starts at 17, ends at 350 and has d=9. Find its count and sum.
Step-by-step solution
350=17+9(n−1) gives n=38. S=38(17+350)/2=19×367.
Resource ID: CB-10-M-CH5-EX5.3-Q6
Question 7
Find S₂₂ when d=7 and a₂₂=149.
Step-by-step solution
149=a+21×7 gives a=2. S₂₂=22(2+149)/2.
Resource ID: CB-10-M-CH5-EX5.3-Q7
Question 8
Find S₅₁ when the second and third terms are 14 and 18.
Step-by-step solution
d=18−14=4; a=14−4=10. Then S₅₁=(51/2)[20+50×4]=51×110.
Resource ID: CB-10-M-CH5-EX5.3-Q8
Question 9
Given S₇=49 and S₁₇=289, find Sₙ.
Step-by-step solution
49=(7/2)(2a+6d) gives a+3d=7. Similarly 289=(17/2)(2a+16d) gives a+8d=17. Thus 5d=10, d=2, a=1. Substitute into the sum formula: Sₙ=(n/2)[2+2(n−1)].
Resource ID: CB-10-M-CH5-EX5.3-Q9
Question 10
Show that aₙ=3+4n and aₙ=9−5n define APs, and find S₁₅.
Step-by-step solution
(i)
aₙ₊₁−aₙ=[3+4(n+1)]−(3+4n)=4, constant. The first term is 7. S₁₅=(15/2)(14+14×4)=525.
(ii)
aₙ₊₁−aₙ=[9−5(n+1)]−(9−5n)=−5, constant. The first term is 4. S₁₅=(15/2)(8−70)=−465.
Resource ID: CB-10-M-CH5-EX5.3-Q10
Question 11
If Sₙ=4n−n², find the first two sums and the requested terms.
Step-by-step solution
S₁=3, S₂=8−4=4. Therefore a₁=3 and a₂=S₂−S₁=1. In general aₙ=Sₙ−Sₙ₋₁=(4n−n²)−[4(n−1)−(n−1)²]=5−2n. This also gives a₁ correctly.
Resource ID: CB-10-M-CH5-EX5.3-Q11
Question 12
Sum the first 40 positive integers divisible by 6.
Step-by-step solution
These are 6,12,…,240, an AP with 40 terms. S=40(6+240)/2.
Resource ID: CB-10-M-CH5-EX5.3-Q12
Question 13
Sum the first fifteen multiples of 8.
Step-by-step solution
The positive multiples are 8,16,…,120. S=15(8+120)/2.
Resource ID: CB-10-M-CH5-EX5.3-Q13
Question 14
Sum all odd numbers between 0 and 50.
Step-by-step solution
The AP is 1,3,…,49, with n=1+(49−1)/2=25. S=25(1+49)/2.
Resource ID: CB-10-M-CH5-EX5.3-Q14
Question 15
Daily delay penalties begin at ₹200 and increase ₹50 per day. Find the penalty for 30 days.
Step-by-step solution
The daily penalties form an AP: a=200,d=50. Total S₃₀=(30/2)[400+29×50]=15×1850.
Resource ID: CB-10-M-CH5-EX5.3-Q15
Question 16
Divide ₹700 into seven prizes, each ₹20 below the preceding prize.
Step-by-step solution
Let the largest prize be a. Then d=−20 and 700=(7/2)(2a−120). Thus 2a−120=200, giving a=160. Subtract ₹20 successively.
Resource ID: CB-10-M-CH5-EX5.3-Q16
Question 17
Three sections in every class from I to XII plant as many trees per section as their class number. Find the total.
Step-by-step solution
Each class k contributes 3k trees. Total=3(1+2+⋯+12)=3×12×13/2.
Resource ID: CB-10-M-CH5-EX5.3-Q17
Question 18
Find the total arc length of thirteen semicircles with radii 0.5,1,1.5,… cm, using π=22/7.
Step-by-step solution
A semicircular arc has length πr, without a diameter segment. The radii sum to (13/2)(0.5+6.5)=45.5 cm. Hence total length=(22/7)×45.5.
Resource ID: CB-10-M-CH5-EX5.3-Q18
Question 19
Stack 200 logs with 20 in the bottom row, then 19,18,… . Find the number of rows and the top-row count.
Step-by-step solution
The row counts form an AP with a=20,d=−1. Set 200=(n/2)[40−(n−1)], so n²−41n+400=0=(n−16)(n−25). Although both solve the sum equation, n=25 would give a negative top count 21−25. Thus n=16 and top count=20−15.
Resource ID: CB-10-M-CH5-EX5.3-Q19
Question 20
Find the total running distance for ten potato pickups, starting 5 m away with potatoes 3 m apart.
Step-by-step solution
The one-way distances are 5,8,11,…,32 m. Each potato requires an outward and a return trip. Total=2×[10(5+32)/2].
Resource ID: CB-10-M-CH5-EX5.3-Q20