CBSE Class 10 Mathematics Arithmetic Progression Assignment Set D

Very Short Answer Type Questions

Question. If the numbers x − 2, 4x − 1 and 5x + 2 are in A.P. Find the value of x.
Answer: ⸪ x − 2, 4x − 1 and 5x + 2 are in A.P.
∴ (4x − 1) − (x − 2) = (5x + 2) − (4x − 1)
⇒ 3x + 1 = x + 3
⇒ 2x = 2 ⇒ x = 1

Question. Which term of the A.P. 4, 9, 14, ..... is 109?
Answer: Let 109 is the nth term,
∴ Using T_n = a + (n − 1) d, we have:
109 = 4 + (n − 1) 5 [ä a = 4 and d = 9 − 4 = 5]
⇒ n − 1 = 109 - 4/5 = 105/5 = 21
⇒ n = 21 + 1 = 22
Thus, the 22nd term is 109.

Question. If a, (a − 2) and 3a are in A.P. then what is the value of a?
Answer: ⸪ a, (a − 2) and 3a are in A.P.
∴ (a − 2) − a = 3a − (a − 2)
⇒ a − 2 − a = 3a − a + 2
⇒ − 2 = 2a + 2
⇒ 2a = − 2 − 2 = − 4
⇒ a = −4/2 = - 2
Thus, the required value of a is − 2.

Question. How many terms are there in the A.P.?
7, 10, 13, ....., 151
Answer: Here, a = 7, d = 10 − 7 = 3
Let there are n-terms.
∴ Tn = a + (n − 1) d
⇒ T51 = 7 + (n − 1) × 3
⇒ 151 - 7/3 = n − 1
⇒ 144/3 = n − 1 ⇒ n = 48 + 1 = 49
i.e., n = 49

Question. Which term of the A.P. 72, 63, 54, ..... is 0?
Answer: Here, a = 72
d = 63 − 72 = − 9
Let nth term of this A.P. be 0
∴ T_n = a + (n − 1) d
⇒ 72 + (n − 1) × (− 9) = 0
⇒ (n − 1) = − 72/−9 = 8
⇒ n = 8 + 1 = 9
Thus the 9th term of the A.P. is 0.

Question. The first term of an A.P. is 6 and its common difference is − 2. Find its 18th term.
Answer: Using T_n = a + (n − 1) d, we have:
T₁₈ = 6 + (18 − 1) × (− 2)
= 6 + 17 × (− 2)
= 6 − 34 = − 28
Thus, the 18th term is − 28.

Question. The 4th term of an A.P. is 14 and its 12th term 70. What is its first term?
Answer: Let the first term = a
If ‘d’ is the common difference,
Then T₄ = a + 3d = 14 ...(1)
And T₁₂ = a + 11d = 70 ...(2)
Subtracting (1) from (2),
a + 11d − a − 3d = 70 − 14
⇒ 8d = 56 ⇒ d = 56/8 = 7
∴ From (1), a + 3 (7) = 14
⇒ a + 21 = 14
⇒ a = 14 − 21 = (− 7)
Thus, the first term is − 7.

Question. Which term of A.P. 5, 2, − 1, − 4 ..... is − 40?
Answer: Here, a = 5
d = 2 − 5 = − 3
Let nth term be − 40
∴ Tn = a + (n − 1) d
⇒ − 40 = 5 + (n − 1) × (− 3)
⇒ n − 1 = − 40/−5/−3 = −45/−3 = 15
⇒ n = 15 + 1 = 16
i.e., The 16th term of the A.P. is − 40.

Question. What is the sum of all the natural numbers from 1 to 100?
Answer: We have:
1, 2, 3, 4, ....., 100 are in an A.P. such that
a = 1 and l = 100
∴ S_n = n/2 [a + l]
⇒ S₁₀₀ = 100/2 [1 + 100] = 50 × 101 = 5050.

Question. For an A.P., the 8th term is 17 and the 14th term is 29. Find its common difference.
Answer: Let the common difference = d and first term = a
∴ T₈ = a + 7d = 17 ...(1)
T₁₄ = a + 13d = 29 ...(2)
Subtracting (1) from (2), we have:
a + 13d − a − 7d = 29 − 17
⇒ 6d = 12
⇒ d = 12/6 = 2
∴ The required common difference = 2.

Question. If the first and last terms of an A.P. are 10 and − 10. How many terms are there? Given that d = − 1.
Answer: Let the required number of terms is n and
1st term a = 10
nth term T_n = − 10
Let common difference be d then using,
T_n = a + (n − 1) d, we have:
− 10 = 10 + (n − 1) × (− 1)
⇒ − 10 = 10 − n + 1
⇒ − n + 1 = − 10 − 10 = − 20
⇒ − n = − 20 − 1 = − 21
⇒ n = 21

Question. The nth term of an A.P. is (3n − 2) find its first term.
Answer: ⸪ T_n = 3n − 2
∴ T₁ = 3 (1) − 2 = 3 − 2 = 1
⇒ First term = 1

Question. The nth term of an A.P. is (2n − 3) find the common difference.
Answer: Here, Tn = 2n − 3
∴ T₁ = 2 (1) − 3 = − 1
T₂ = 2 (2) − 3 = 1
∴ d = T₂ − T1 = 1 − (− 1) = 2
Thus the common difference is 2.

Question. If the nth term of an A.P. is (7n − 5). Find its 100th term.
Answer: Here, T_n = 7n − 5
∴ T₁ = 7 (1) − 5 = 2
T₂ = 7 (2) − 5 = 9
∴ a = 2
and d = T₂ − T₁
= 9 − 2 = 7
Now T100 = 2 + (100 − 1) 7 [using Tn = a + (n − 1) d]
= 2 + 99 × 7
= 2 + 693 = 695.

Question. Find the sum of first 12 terms of the A.P. 5, 8, 11, 14, ...... .
Answer: Here, a = 5
d = 8 − 5 = 3
n = 12
Using S_n = n/2 [2 (a) + (n − 1) d]
we have: S₁₂ = 12/2 [2 (5) + (12 − 1) × 3]
= 6 [10 + 33]
= 6 × 43 = 258

Question. Write the common difference of an A.P. whose nth term is 3n + 5.
Answer: T_n = 3n + 5
∴ T₁ = 3 (1) + 5 = 8
T₂ = 3 (2) + 5 = 11
⇒ d = T2 − T1
= 11 − 8 = 3
Thus, the common difference = 3.

Question. Write the value of x for which x + 2, 2x, 2x + 3 are three consecutive terms of an A.P.
Answer: Here, T₁ = x + 2
T₂ = 2x
T₃ = 2x + 3
For an A.P., we have:
∴ 2x − (x + 2) = 2x + 3 − 2x
⇒ 2x − x − 2 = 2x + 3 − 2x
⇒ x − 2 = 3
⇒ x = 3 + 2 = 5
Thus, x = 5

Question. What is the common difference of an A.P. whose nth term is 3 + 5n? 
Answer: ⸪ T_n = 3 + 5n
∴ T₁ = 3 + 5 (1) = 8
And T₂ = 3 + 5 (2) = 13
⸪ d = T₂ − T1
∴ d = 13 − 8 = 5
Thus, common difference = 5.

Question. For what value of k, are the numbers x, (2x + k) and (3x + 6) three consecutive terms of an A.P.?
Answer: Here, T₁ = x, T₂ = (2x + k) and T₃ = (3x + 6)
For an A.P., we have
T₂ − T₁ = T3 − T₂
i.e., 2x + k − x = 3x + 6 − (2x + k)
⇒ x + k = 3x + 6 − 2x − k
⇒ x + k = x + 6 − k
⇒ k + k = x + 6 − x
⇒ 2k = 6
⇒ k = 6/2 = 3

Question. If 4/5 , a, 2 are three consecutive terms of an A.P., then find the value of a? 
Answer: Here, T₁ = 4/5
T₂ = a
T₃ = 2
⸪ For an A.P.,
T₂ − T₁ = T₃ − T₂
∴ a − 4/5 = 2 − a
⇒ a + a = 2 + 4/5
⇒ 2a = 14/5
⇒ a = 14/5 x 1/2 = 7/5
Thus, a = 7/5

Question. For what value of p are 2p − 1, 7 and 3p three consecutive terms of an A.P.?
Answer: Here, T₁ = 2p − 1
T₂ = 7
T₃ = 3p
⸪ For an A.P., we have:
T₂ − T₁ = T₃ − T₂
⇒ 7 − (2p − 1) = 3p − 7
⇒ 7 − 2p + 1 = 3p − 7
⇒ − 2p − 3p = − 7 − 1 − 7
⇒ − 5p = − 15
⇒ p = -15/-5 = 3
Thus, p = 3

Question. For what value of p are 2p + 1, 13 and 5p − 3 three consecutive terms of an A.P.?
Answer: Here, T₁ = 2p + 1
T₂ = 13
T₃ = 5p − 3
For an A.P., we have:
T₂ − T₁ = T₃ − T₂
⇒ 13 − (2p + 1) = 5p − 3 − 13
⇒ 13 − 2p − 1 = 5p − 16
⇒ − 2p + 12 = 5p − 16
⇒ − 2p − 5p = − 16 − 12 = − 28
⇒ − 7p = − 28
⇒ p = −28/−7 = 28/7 = 4
∴ p = 4

Question. The nth term of an A.P. is 7 − 4n. Find its common difference.
Answer: ⸪ T_n = 7 − 4n
∴ T₁ = 7 − 4 (1) = 3
T₂ = 7 − 4 (2) = − 1
∴ d = T2 − T1
= (− 1) − 3 = − 4
Thus, common difference = − 4

Question. The nth term of an A.P. is 6n + 2. Find the common difference.
Answer: Here, T_n = 6_n + 2
∴ T₁ = 6 (1) + 2 = 8
T₂ = 6 (2) + 2 = 14
⇒ d = T₂ − T₁ = 14 − 8 = 6
∴ Common difference = 6.

Question. Write the next term of the A.P. √8 , √18 , √32 , ..... 
Answer: Here, T₁ = √8 = √4 × 2 = 2√2
T₂ = √18 = √9 × 2 = 3√2
T₃ = √32 = √16 × 2 = 4√2
∴ a = 2√2
Now, d = T₂ − T₁
= 3√2 − 2√2 = √2(3 − 2) = √2
∴ T₄ = a + 3d
= 2√2 + 3 (√2)
= 2√2 + 3√2
= √2 (2 + 3) = 5√2 or 50
Thus, the next term of the A.P. is 5√2 or 50 .

Question. The first term of an A.P. is p and its common difference is q. Find the 10th term.
Answer: Here, a = p and d = q
⸪ T_n = a + (n − 1) d
∴ T₁₀ = p + (10 − 1) q
= p + 9q
Thus, the 10th term is p + 9q.

Question. Find the next term of the A.P. 2 , 8 , 18 .....
Answer: Here, T1 = 2 ⇒ a = 2
T2 = 8 = 2 2
T3 = 18 = 3 2
Now, d = T2 − T1
= 2 2 − 2
= 2
Now, using Tn = a + (n − 1) × d, we have
T4 = a + 3d
= 2 + 3 e 2j
= 2 1 + 3 = 4 2
= 16 × 2 = 32
Thus, the next term = 32 .

Question. Which term of the A.P.: 21, 18, 15, ..... is zero?
Answer: Here, a = 21
d = 18 − 21 = − 3
Since T_n = a + (n − 1) d
⇒ 0 = 21 + (n − 1) × (− 3)
⇒ − 3 (n − 1) = − 21
⇒ (n − 1) = −21/−3 = 7
⇒ n = 7 + 1 = 8
Thus, the 8th term of this A.P. will be 0.

Question. Which term of the A.P.: 14, 11, 8, ..... is − 1? 
Answer: Here, a = 14
d = 11 − 14 = − 3
Let the nth term be (− 1)
∴ Using Tn = a + (n − 1) d, we get
− 1 = 11 + (n − 1) × (− 3)
⇒ − 1 − 14 = − 3 (n − 1)
⇒ − 15 = − 3 (n − 1)
∴ n − 1 = −15/−3 = 5
⇒ n = 5 + 1 = 6
Thus, −1 is the 6th term of the A.P.

Short Answer Type Questions

Question. If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term.
Answer: Let ‘a’ be the first term and ‘d’ be the common difference.
Now, Using Tn = a + (n − 1) d, we have
T₉ = a + 8d ⇒ a + 8d = 0 ...                        (1) [⸪ T₉ = 0 Given]
T₁₉ = a + 18d = (a + 8d) + 10d = (0) + 10d = 10d                         ...(2) [⸪ a + 8d = 0]
T₂₉ = a + 28d
= (a + 8d) + 20d
= 0 + 20d = 20d                                 [⸪ a + 8d = 0]
= 2 × (10d) = 2 (T₁₉)                         [⸪ T₁₉ = 10d]
⇒ T₂₉ = 2 (T19)
Thus, the 29th term of the A.P. is double of its 19th term.

Question. If T_n = 3 + 4n then find the A.P. and hence find the sum of its first 15 terms.
Answer: Let the first term be ‘a’ and the common difference be ‘d’.
ä T_n = a + (n − 1) d
∴ T₁ = a + (1 − 1) d = a + 0 × d = a
T₂ = a + (2 − 1) d = a + d
But it is given that
T_n = 3 + 4n
∴ T₁ = 3 + 4 (1) = 7
⇒ First term, a = 7
Also, T₂ = a + d = 3 + 4 (2) = 11
∴ d = T₂ − T₁ = 11 − 7 = 4
Now, using S_n = n/2 [2a + (n − 1) d], we get
S₁₅ = 15/2 [2 (7) + (15 − 1) × 4]
= 15/2 [14 + 14 × 4]
= 15/2 [70]
= 15 × 35 = 525
Thus, the sum of first 15 terms = 525.

Question. Which term of the A.P.: 3, 15, 27, 39, ..... will be 120 more than its 53rd term?
Answer: The given A.P. is:
3, 15, 27, 39, .....
∴ a = 3
d = 15 − 3 = 12
∴ Using, T_n = a + (n − 1) d, we have:
T₅₃ = 3 + (53 − 1) × 12
= 3 + (52 × 12)
= 3 + 624 = 627
Now, T₅₃ + 120 = 627 + 120 = 747.
Let the required term be T_n
∴ T_n = 747
or a + (n − 1) d = 747
∴ 3 + (n − 1) × 12 = 747
⇒ (n − 1) × 12 = 747 − 3 = 744
⇒ n − 1 = 744/12 = 62
⇒ n = 62 + 1 = 63
Thus, the 63rd term of the given A.P. is 120 more than its 53rd term.

Question. Find the 31st term of an A.P. whose 10th term is 31 and the 15th term is 66.
Answer: Let the first term is ‘a’ and the common difference is ‘d’.
Using Tn = a + (n − 1) d, we have:
T₁₀ = a + 9d
⇒ 31 = a + 9d ...(1)
Also T₁₅ = a + 14d
⇒ 66 = a + 14d ...(2)
Subtracting (1) from (2), we have:
a + 14d − a − 9d = 66 − 31
⇒ 5d = 35
⇒ d = 35/5 = 7
∴ From (1), a + 9d = 31
⇒ a + 9 (7) = 31
⇒ a + 63 = 31
⇒ a = 31 − 63
⇒ a = − 32
Now, T₃₁ = a + 30d
= − 32 + 30 (7)
= − 32 + 210 = 178
Thus, the 31st term of the given A.P. is 178.

Question. If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Hence find the sum of the first 15 terms of the A.P.
Answer: Let the 1st term = a
And the common difference = d
∴ Using T_n = a + (n − 1) d
∴ T₈ = a + 7d
⇒ 37 = a + 7d ...(1)
Also T₁₅ = a + 14d
And T₁₂ = a + 11d
According to the question,
T₁₅ = T₁₂ + 15
⇒ a + 14d = a + 11d + 15
⇒ a − a + 14d − 11d = 15
⇒ 3d = 15 ⇒ d = 15/3 = 5
From (1), we have:
a + 7 (5) = 37
⇒ a + 35 = 37
⇒ a = 37 − 35 = 2
Since an, A.P. is given by :
a, a + d, a + 2d, a + 3d, ....
∴ The required A.P. is given by 2, 2 + 5, 2 + 2(5),... 2, 7, 12, ...
Now, using Sn = n/2 [2a + (n − 1) d]
∴ S15 = 15/2 [2 (2) + 14 × 5]
= 15/2 [4 + 70]
= 15/2 × 74 = 15 × 37 = 555.

Question. The 5th and 15th terms of an A.P. are 13 and − 17 respectively. Find the sum of first 21 terms of the A.P.
Answer: Let ‘a’ be the first term and ‘d’ be the common difference.
∴ Using T_n = a + (n − 1) d, we have:
T₁₅ = a + 14d = − 17 ...(1)
T₅ = a + 4d = 13 ...(2)
Subtracting (2) from (1), we have:
(T₁₅ − T₅) = − 17 − 13 = − 30
⇒ a + 14d − a − 4d = − 30
⇒ 10d = − 30 ⇒ d = − 3
Substituting d = − 3 in (2), we get
a + 4d = 13
⇒ a + 4 (− 3) = 13
⇒ a + (− 12) = 13
⇒ a = 13 + 12 = 25
Now using S_n = n/2 [2a + (n − 1) d] we have:
S₂₁ = 21/2 [2 (25) + (21 − 1) × (− 3)]
= 21/2 [50 + (− 60)]
= 21/2 × − 10
= 21 × (− 5) = −105
Thus, the sum of first fifteen terms = − 105.

Question. The 1st and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9 how many terms are there in the A.P.? What is their sum?
Answer: Here, first term, a = 17
Last term T_n = 350 = l
⸪ Common difference (d) = 9.
∴ Using T_n = a + (n − 1) d, we have:
350 = 17 + (n − 1) × 9
⇒ n − 1 = 350 - 17 /9
= 333/9 = 37
⇒ n = 37 + 1 = 38
Thus, there are 38 terms.
Now, using, S_n = n/2 [a + l], we have
S₃₈ = 38/2 [17 + 350]
= 19 [367] = 6973
Thus, the required sum of 38 terms = 6973.

Question. The first and last term of an A.P. are 4 and 81 respectively. If the common difference is 7, how many terms are there in the A.P. and what is their sum?
Answer: Here, first term = 4 ⇒ a = 4 and d = 7.
Last term, l = 81 ⇒ T_n = 81
⸪ T_n = a + (n − 1) d
∴ 81 = 4 + (n − 1) × 7
⇒ 81 − 4 = (n − 1) × 7
⇒ 77 = (n − 1) × 7 ⇒ n = 77/7 + 1 = 11 + 1 = 12
⇒ There are 12 terms.
Now, using
S_n = n/2 (a + l)
⇒ S₁₂ = 12/2 (4 + 81)
⇒ S₁₂ = 6 × 85 = 510
∴ The sum of 12 terms of the A.P. is 510.