NCERT Mathematics · Class 10 · English · 2026-27
Arithmetic Progressions — Exercise 5.2
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
Complete the five rows of the AP table.
Step-by-step solution
Use aₙ=a+(n−1)d and solve for the missing entry.
(i)
a₈=7+7×3=28.
(ii)
0=−18+9d gives d=2.
(iii)
−5=a+17(−3) gives a=46.
(iv)
3.6=−18.9+(n−1)2.5 gives n−1=9, so n=10.
(v)
With d=0 every term is 3.5; a₁₀₅=3.5.
Resource ID: CB-10-M-CH5-EX5.2-Q1
Question 2
Choose and justify the correct 30th and 11th terms.
Step-by-step solution
(i)
For 10,7,4,…, a=10,d=−3. a₃₀=10+29(−3)=−77. Choice C.
(ii)
For −3,−1/2,2,…, d=5/2. a₁₁=−3+10×5/2=22. Choice B.
Resource ID: CB-10-M-CH5-EX5.2-Q2
Question 3
Fill the missing entries in the five APs shown in boxes.
Step-by-step solution
For equally spaced terms, d=(last−first)/(number of gaps).
(i)
2,□,26 has two gaps: d=(26−2)/2=12. Completed: 2,14,26.
(ii)
□,13,□,3: 13 to 3 spans two gaps, so d=−5. Completed: 18,13,8,3.
(iii)
5,□,□,9½: three gaps give d=(9.5−5)/3=1.5. Completed: 5,6½,8,9½.
(iv)
−4,□,□,□,□,6: five gaps give d=2. Completed: −4,−2,0,2,4,6.
(v)
□,38,□,□,□,−22: four gaps from 38 to −22 give d=−15. Completed: 53,38,23,8,−7,−22.
Resource ID: CB-10-M-CH5-EX5.2-Q3
Question 4
Locate 78 in 3,8,13,18,… .
Step-by-step solution
Here a=3,d=5. Set 78=3+5(n−1). Then 75=5(n−1), so n−1=15.
Resource ID: CB-10-M-CH5-EX5.2-Q4
Question 5
Count terms in each finite AP ending at 205 and −47.
Step-by-step solution
For 7,13,19,…,205: 205=7+6(n−1), hence n=34. For 18,15½,13,…,−47: −47=18−2.5(n−1), hence n−1=26.
Resource ID: CB-10-M-CH5-EX5.2-Q5
Question 6
Does −150 occur in 11,8,5,2,…?
Step-by-step solution
Set −150=11−3(n−1). This gives 3n=164 and n=164/3. A term index must be a positive integer.
Resource ID: CB-10-M-CH5-EX5.2-Q6
Question 7
Find the 31st term when a₁₁=38 and a₁₆=73.
Step-by-step solution
Subtract a+10d=38 from a+15d=73: 5d=35, so d=7. Then a=38−70=−32 and a₃₁=−32+30×7.
Resource ID: CB-10-M-CH5-EX5.2-Q7
Question 8
An AP has 50 terms, a₃=12 and a₅₀=106. Find a₂₉.
Step-by-step solution
Subtract a+2d=12 from a+49d=106: 47d=94, so d=2. Thus a=8 and a₂₉=8+28×2.
Resource ID: CB-10-M-CH5-EX5.2-Q8
Question 9
Find the zero term when a₃=4 and a₉=−8.
Step-by-step solution
The difference gives 6d=−12, so d=−2. From a+2d=4, a=8. Set 0=8−2(n−1), giving n−1=4.
Resource ID: CB-10-M-CH5-EX5.2-Q9
Question 10
Find d if the 17th term exceeds the 10th by 7.
Step-by-step solution
a₁₇−a₁₀=(a+16d)−(a+9d)=7d. Thus 7d=7.
Resource ID: CB-10-M-CH5-EX5.2-Q10
Question 11
Which term of 3,15,27,39,… exceeds its 54th term by 132?
Step-by-step solution
Here d=12. The difference aₙ−a₅₄=(n−54)12=132, so n−54=11.
Resource ID: CB-10-M-CH5-EX5.2-Q11
Question 12
Two APs share d and their 100th terms differ by 100. Compare their 1000th terms.
Step-by-step solution
If first terms are a and b, the nth-term difference is [a+(n−1)d]−[b+(n−1)d]=a−b. It is independent of n.
Resource ID: CB-10-M-CH5-EX5.2-Q12
Question 13
Count the three-digit multiples of 7.
Step-by-step solution
The first is 105 and last 994. Set 994=105+7(n−1): n−1=889/7=127.
Resource ID: CB-10-M-CH5-EX5.2-Q13
Question 14
Count multiples of 4 strictly between 10 and 250.
Step-by-step solution
They are 12,16,…,248. Thus 248=12+4(n−1), giving n−1=59.
Resource ID: CB-10-M-CH5-EX5.2-Q14
Question 15
When do the nth terms of 63,65,67,… and 3,10,17,… agree?
Step-by-step solution
Their nth terms are 63+2(n−1) and 3+7(n−1). Equating gives 60=5(n−1), so n=13. Each term then equals 87.
Resource ID: CB-10-M-CH5-EX5.2-Q15
Question 16
Find an AP with third term 16 and a₇−a₅=12.
Step-by-step solution
The term difference is 2d=12, so d=6. Then a+2×6=16 gives a=4.
Resource ID: CB-10-M-CH5-EX5.2-Q16
Question 17
Find the 20th term from the end of 3,8,13,…,253.
Step-by-step solution
Reverse the AP: first term 253, difference −5. Its 20th term is 253+19(−5).
Resource ID: CB-10-M-CH5-EX5.2-Q17
Question 18
Find the first three terms if a₄+a₈=24 and a₆+a₁₀=44.
Step-by-step solution
The equations are 2a+10d=24 and 2a+14d=44. Subtract to get 4d=20, hence d=5. Then 2a=24−50=−26, so a=−13.
Resource ID: CB-10-M-CH5-EX5.2-Q18
Question 19
An annual salary starts at ₹5000 in 1995 and rises ₹200 yearly. When does it reach ₹7000?
Step-by-step solution
7000=5000+200(n−1) gives n−1=10. The first year is 1995, so add ten calendar years.
Resource ID: CB-10-M-CH5-EX5.2-Q19
Question 20
Weekly savings start at ₹5 and rise ₹1.75 each week. Locate ₹20.75.
Step-by-step solution
20.75=5+1.75(n−1). Thus n−1=15.75/1.75=9.
Resource ID: CB-10-M-CH5-EX5.2-Q20