NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions (Exercise 5.2)
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Q1. Fill in the blanks in the following table, given that a is the first term, d the common difference and aₙ the nth term of the A.P.
Solutions:
(i)
Given: a = 7, d = 3, n = 8
We know, aₙ = a + (n − 1)d
⇒ a₈ = 7 + (8 − 1)×3 = 7 + 21 = 28(ii)
Given: a = −18, aₙ = 0, n = 10
aₙ = a + (n − 1)d
0 = −18 + (10 − 1)d
18 = 9d ⇒ d = 2
Hence, d = 2(iii)
Given: d = −3, n = 18, aₙ = −5
aₙ = a + (n − 1)d
−5 = a + 17(−3) ⇒ −5 = a − 51 ⇒ a = 46
Hence, a = 46(iv)
Given: a = −18.9, d = 2.5, aₙ = 3.6
aₙ = a + (n − 1)d
3.6 = −18.9 + (n − 1)×2.5
22.5 = (n − 1)×2.5 ⇒ n − 1 = 9 ⇒ n = 10(v)
Given: a = 3.5, d = 0, n = 105
aₙ = a + (n − 1)d
aₙ = 3.5 + (104)×0 = 3.5 -
Q2. Choose the correct choice in the following and justify:
(i) 30th term of A.P. 10, 7, 4, …
a = 10, d = −3
aₙ = a + (n − 1)d
a₃₀ = 10 + (29)(−3) = 10 − 87 = −77
Hence, correct answer is Option (C) −77.(ii) 11th term of A.P. −3, −1/2, 2, …
a = −3, d = (−1/2) − (−3) = 5/2
a₁₁ = −3 + (10)(5/2) = −3 + 25 = 22
Hence, correct answer is Option (B) 22. -
Q3. In the following A.P.s, find the missing terms:
(i) 2, __, 26
a = 2, a₃ = 26
26 = 2 + 2d ⇒ d = 12
a₂ = 2 + 12 = 14(ii) __, 13, __, 3
From a₂ = 13 and a₄ = 3
13 = a + d …(i)
3 = a + 3d …(ii)
Subtract: −10 = 2d ⇒ d = −5
From (i): 13 = a − 5 ⇒ a = 18
a₃ = 18 + 2(−5) = 8
Hence, missing terms: 18, 8(iii) 5, __, __, 19/2
a = 5, a₄ = 19/2
19/2 = 5 + 3d ⇒ d = 3/2
a₂ = 5 + 3/2 = 13/2, a₃ = 5 + 3 = 8
Hence, missing terms: 13/2, 8(iv) −4, __, __, __, __, 6
a = −4, a₆ = 6
6 = −4 + 5d ⇒ d = 2
a₂ = −2, a₃ = 0, a₄ = 2, a₅ = 4
Hence, missing terms: −2, 0, 2, 4(v) __, 38, __, __, __, −22
a₂ = 38, a₆ = −22
38 = a + d, −22 = a + 5d
Subtract: −60 = 4d ⇒ d = −15
a = 38 − (−15) = 53
a₃ = 23, a₄ = 8, a₅ = −7
Hence, missing terms: 53, 23, 8, −7 -
Q4. Which term of the A.P. 3, 8, 13, 18, … is 78?
a = 3, d = 5, aₙ = 78
78 = 3 + (n − 1)×5
75 = 5(n − 1) ⇒ n = 16
Hence, 16th term is 78. -
Q5. Find the number of terms in the A.P. 7, 13, 19, …, 205
a = 7, d = 6, aₙ = 205
205 = 7 + (n − 1)×6
198 = 6(n − 1) ⇒ n = 34
Hence, total 34 terms. -
Q6. Check whether −150 is a term of the A.P. 11, 8, 5, 2, …
a = 11, d = −3, aₙ = −150
−150 = 11 + (n − 1)(−3)
−150 = 14 − 3n ⇒ n = 54.6 (not integer)
Hence, −150 is not a term of the A.P. -
Q7. Find the 31st term of an A.P. whose 11th term is 38 and 16th term is 73.
a₁₁ = 38, a₁₆ = 73
38 = a + 10d, 73 = a + 15d
Subtract ⇒ 35 = 5d ⇒ d = 7
a = −32
a₃₁ = −32 + 30×7 = 178 -
Q8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
a₃ = 12 = a + 2d
a₅₀ = 106 = a + 49d
Subtract ⇒ 94 = 47d ⇒ d = 2
a = 8
a₂₉ = 8 + 28×2 = 64 -
Q9. If 3rd and 9th terms of an A.P. are 4 and −8 respectively, which term is 0?
a₃ = 4 = a + 2d
a₉ = −8 = a + 8d
Subtract ⇒ −12 = 6d ⇒ d = −2
a = 8
0 = 8 + (n − 1)(−2)
n = 5
Hence, 5th term = 0 -
Q10. If 17th term exceeds 10th term by 7, find the common difference.
a₁₇ − a₁₀ = 7
(a + 16d) − (a + 9d) = 7 ⇒ 7d = 7 ⇒ d = 1 -
Q11. Which term of the A.P. 3, 15, 27, 39,… will be 132 more than its 54th term?
a = 3, d = 12
a₅₄ = 3 + 53×12 = 639
aₙ = 639 + 132 = 771
771 = 3 + (n − 1)×12 ⇒ n = 65
Hence, 65th term is 132 more than 54th term. -
Q12. Two A.P.s have same common difference. Difference between their 100th terms is 100. Find the difference between 1000th terms.
(a₁ + 99d) − (a₂ + 99d) = 100 ⇒ a₁ − a₂ = 100
(a₁ + 999d) − (a₂ + 999d) = a₁ − a₂ = 100
Hence, difference = 100. -
Q13. How many three-digit numbers are divisible by 7?
First term, a = 105; last term, aₙ = 994; d = 7
aₙ = a + (n − 1)d ⇒ 994 = 105 + (n − 1)×7
n = 128
Hence, 128 numbers are divisible by 7. -
Q14. How many multiples of 4 lie between 10 and 250?
a = 12, d = 4, aₙ = 248
248 = 12 + (n − 1)×4 ⇒ n = 60
Hence, 60 multiples of 4. -
Q15. For what value of n are nth terms of A.P.s 63, 65, 67,… and 3, 10, 17,… equal?
First A.P.: a = 63, d = 2 ⇒ aₙ = 61 + 2n
Second A.P.: a = 3, d = 7 ⇒ aₙ = 7n − 4
Equating: 61 + 2n = 7n − 4 ⇒ n = 13
Hence, 13th term of both A.P.s are equal. -
Q16. Determine the A.P. whose third term is 16 and 7th term exceeds 5th term by 12.
a₃ = 16 ⇒ a + 2d = 16
a₇ − a₅ = 12 ⇒ 2d = 12 ⇒ d = 6
a = 4
Hence, A.P. is 4, 10, 16, 22, … -
Q17. Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Reverse A.P.: 253, 248, 243, … (d = −5)
a₂₀ = 253 + (19)(−5) = 158
Hence, 20th term from last = 158. -
Q18. The sum of 4th and 8th terms of an A.P. is 24 and sum of 6th and 10th terms is 44. Find first three terms.
a₄ + a₈ = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 …(i)
a₆ + a₁₀ = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22 …(ii)
Subtract ⇒ 2d = 10 ⇒ d = 5
a = −13
Hence, A.P. = −13, −8, −3, … -
Q19. Subba Rao started work in 1995 at ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
a = 5000, d = 200, aₙ = 7000
7000 = 5000 + (n − 1)×200
n = 11 ⇒ 1995 + 10 = 2005. -
Q20. Ramkali saved ₹5 in first week and increased by ₹1.75 weekly. In which week did she save ₹20.75?
a = 5, d = 1.75, aₙ = 20.75
20.75 = 5 + (n − 1)×1.75
n = 10
Hence, 10th week.