CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set B

VERY SHORT ANSWER TYPE QUESTIONS :

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
∴ T_n = 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. 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 × 1/2 = 7/5
Thus, a = 7/5

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

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. The nth term of an A.P. is (2n − 3) find the common difference.
Answer : Here, T_n = 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. 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 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 T_n = a + (n − 1) d]
= 2 + 99 × 7
= 2 + 693 = 695.

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

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. 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 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 
T₁₄ = a + 13d = 29 
Subtracting (1) from (2), we have:
a + 13d − a − 7d = 29 − 17
⇒ 6d = 12
⇒ d = 12/6 = 2
∴ The required common difference = 2.

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. 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                                                        [∵ 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.
∴ T_n = 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 
And T₁₂ = a + 11d = 70 
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. 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. 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. 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. 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. 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 T_n = 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.

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. 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. 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 value of the middlemost term (s) of the AP : –11, –7, –3, ...49.
Answer : a = –11, a_n = 49 and d = (–7) – (–11) = 4
∴ a_n = a + (n – 1)d
⇒ 49 = –11 + (n – 1) × 4 ⇒ n = 16
Since, n is an even number
∴ There will be two middle terms, which are:
16/2 th and {16/2+1} th
or 8th and 9th
Now, a₈ = a + (8 – 1)d
= –11 + 7 × 4 = 17
a₉ = a + (9 – 1)d
= –11 + 8 × 4 = 21
Thus, the values of the two middlemost terms are : 17 and 21.

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 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.

SHORT ANSWER TYPE QUESTIONS :

Question. In an A.P., the first term is 8, nth term is 33 and sum of first n terms is 123. Find n and d, the common difference.
Answer : Here,
First term T₁ = 8 ⇒ a = 8
nth term T_n = 33 = l
∵ S_n = 123 [Given]
∴ Using, S_n = n/2 [a + l], we have 
S_n = n/2 [8 + 33]
⇒ 123 = n/2 × 41
⇒ n = 123 × 2/41 = 6
Now, T₆ = 33
⇒ a + 5d = 33
⇒ 8 + 5d = 33
⇒ 5d = 33 − 8 = 25
⇒ d = 25/5 = 5
Thus, n = 6 and d = 5.

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

Question. Find the sum of all three digit numbers which are divisible by 7. 
Answer : The three digit numbers which are divisible by 7 are:
105, 112, 119, ....., 994.
It is an A.P. such that
a = 105
d = 112 − 105 = 7
T_n = 994 = l
∴ T_n = a + (n − 1) × d
∴ 994 = 105 + (n − 1) × 7
⇒ n − 1 = 994 − 105/7 = 889/7 = 127
⇒ n = 127 + 1 = 128
Now, using S_n = n/2 [a + l]
We have S₁₂₈ = 128/2 [105 + 994]
= 64 [1099]
= 70336
Thus, the required sum = 70336.

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 [∵ T₉ = 0 Given]
T₁₉ = a + 18d = (a + 8d) + 10d = (0) + 10d = 10d [∵ 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. In an A.P., the first term is 22, nth term is − 11 and sum of first n terms is 66. Find n and d, the common difference. 
Answer : We have

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.

Question. Find the sum of all the three digit numbers which are divisible by 9. 
Answer : All the three digit numbers divisible by 9 are:
117, 126, ....., 999 and they form an A.P.
Here, a = 108
d = 117 − 108 = 9
T_n = 999 = l
Now, using T_n = a + (n − 1) d, we have
999 = 108 + (n − 1) (9)
⇒ 999 − 108 = (n − 1) × 9
⇒ 891 = (n − 1) × 9
⇒ n − 1 = 891/9 = 99
⇒ n = 99 + 1 = 100
Now, the sum of n term of an A.P. is given
S_n = n/2 [a + l]
∴ S₁₀₀ = 100/2 [108 + 999]
= 50 [1107]
= 55350
Thus, the required sum is 55350.

Question. The angles of a quadrilateral are in A.P. whose common difference is 15°. Find the angles.
Answer : Let one of the angles = a
∴ The angles are in an A.P.
∴ The angles are:
a°, (a + d)°, (a + 2d)° and (a + 3d)°
∴ d = 15

Question. Find the middle term of the A.P. 10, 7, 4, ....., − 62. 
Answer : Here, a = 10
d = 7 − 10 = − 3
T_n = (− 62)
∴ Using T_n = a + (n − 1) d, we have
− 62 = 10 + (n − 1) × (− 3)
⇒ n − 1 = − 62 − 10/−3 = − 72/− 3 = 24
⇒ n = 24 + 1 = 25
⇒ Number of terms = 25
∴ Middle term = {n + 1/2} th term
= 25 + 1/2 th term
= 13th term
Now T₁₃ = 10 + 12d
= 10 + 12 (− 3)
= 10 − 36 = − 26
Thus, the middle term = − 26.

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 
T₅ = a + 4d = 13 
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 angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle. 
Answer : Let a, b, c are the angles of the triangle, such that
c = 2a 
Since a, b, c are in A.P.
Then b = a + c/2
From (1) and (2), we get

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’.
∵ 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. Find the 10th term from the end of the A.P.:
8, 10, 12, ....., 126 
Answer : Here, a = 8
d = 10 − 8 = 2
T_n = 126
Using T_n = a + (n − 1) d
⇒ 126 = 8 + (n − 1) × 2
⇒ n − 1 = 126 − 8/2 = 59
⇒ n = 59 + 1 = 60
∴ l = 60
Now 10th term from the end is given by
l − (10 − 1) = 60 − 9 = 51
Now, T₅₁ = a + 50d
= 8 + 50 × 2
= 8 + 100 = 108
Thus, the 10th term from the end is 108.

Question. The sum of first six terms of an AP is 42. The ratio of 10th term to its 30th term is 1 : 3. Calculate the first term and 13th term of A.P.
Answer : S₆ = 6/2 {2a+(6 − 1)d} = 42
∴ 6a + 15d = 42 
Also, (a1₀) : (a₃₀) = 1 : 3
or a+9d/a+29d = 1/3
⇒ 3(a + 9d) = a + 29d
⇒ 3a + 27d = a + 27d
⇒ 2a = 2d
⇒ a = d
From (1) 6d + 15d = 42 ⇒ d = 2
From (2) a = d ⇒ d = 2
Now, a₁₃ = a + 12d
= 2 + 12 × 2 = 26

Question. Which term of the A.P. 4, 12, 20, 28, ..... will be 120 more than its 21st term?
Answer : Here, a = 4
d = 12 − 4 = 8
Using T_n = a + (n − 1) d
∴ T₂₁ = 4 + (21 − 1) × 8
= 4 + 20 × 8 = 164
∵ The required nth term = T₂₁ + 120
∴ nth term = 164 + 120 = 284
∴ 284 = a + (n − 1) d
⇒ 284 = 4 + (n − 1) × 8
⇒ 284 − 4 = (n − 1) × 8
⇒ n − 1 = 280/8 = 35
⇒ n = 35 + 1 = 36
Thus, the required term is the 36th term of the A.P.

Question. If S_nSn, the sum of first n terms of an A.P. is given by
S_n = 5n² + 3n
Then find the nth term. 
Answer : ∴ S_n = 5n² + 3n
∴ S_{n− 1} = 5 (n − 1)² + 3 (n − 1)
= 5 (n² − 2n + 1) + 3 (n − 1)
= 5n² − 10n + 5 + 3n − 3
= 5n² − 7n + 2
Now, nth term = S_n − S_{n− 1}
∴ The required nth term
= [5n² + 3n] − [5n²− 7n + 2]
= 10n − 2.

Question. If S_n the sum of n terms of an A.P. is given by S_n = 3n² − 4n, find the nth term.
Answer : We have:
S_n − 1 = 3 (n − 1)² − 4 (n − 1)
= 3 (n² − 2n + 1) − 4n + 4
= 3n² − 6n + 3 − 4n + 4
= 3n² − 10n + 7
∴ nth term = S_n − S_n − 1
= 3n² − 4n − [3n² − 10n + 7]
= 3n² − 4n − 3n² + 10n − 7
= 6n − 7.

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. 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
∴ Tn = 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. The sum of 4th and 8th terms of an A.P. is 24, and the sum of 6th and 10th terms is 44. Find the A.P. 
Answer : Let, the first term = a
Common difference be = d

Question. Which term of the A.P. 3, 15, 27, 39, ..... will be 120 more than its 21st term?
Answer : Let the 1st term is ‘a’ and common difference = d
∴ a = 3 and d = 15 − 3 = 12
Now, using T_n = a + (n − 1) d
∴ T₂₁ = 3 + (21 − 1) × 12
= 3 + 20 × 12
= 3 + 240 = 243
Let the required term be the nth term.
∵ nth term = 120 + 21st term
= 120 + 243 = 363
Now T_n = a + (n − 1) d
⇒ 363 = 3 + (n − 1) × 12
⇒ 363 − 3 = (n − 1) × 12
⇒ n − 1 = 360/12 = 30
⇒ n = 30 + 1 = 31
Thus the required term is the 31st term of the A.P.

Question. If m times the mth term of an A.P. is equal to n times the nth term, find the (m + n)th term of the A.P. 
Answer : Let the first term (T₁) = a and the common difference be ‘d’.

Question. Find the sum of all the three digit numbers which are divisible by 11. 
Answer : All the three digit numbers divisible by 11 are 110, 121, 132, ....., 990.
Here, a = 110
d = 121 − 110 = 11
T_n = 990
∴ Using T_n = a + (n − 1) d, we have
990 = 110 + (n − 1) × 11
⇒ n − 1 = 990 − 110/11 = 80
⇒ n = 80 + 1 = 81
Now, using S_n = n/2 [a + l], we have
S₈₁ = 81/2 [110 + 990]
= 81/2 [1100]
= 81 × 550 = 44550
Thus, the required sum = 44550.

Question. The sum of 5th and 9th terms of an A.P. is 72 and the sum of 7th and 12th term of 97. Find the A.P. 
Answer : Let ‘a’ be the 1st term and ‘d’ be the common difference of the A.P.
Now, using T_n = a + (n − 1) d, we have

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 T_n = a + (n − 1) d, we have:
T₁₀ = a + 9d
⇒ 31 = a + 9d 
Also T₁₅ = a + 14d
⇒ 66 = a + 14d 
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. The sum of n terms of an A.P. is 3n² + 5n. Find the A.P. Hence, find its 16th term.
Answer : We have,
S_n = 3n² + 5n
∴ S₁ = 3 (1)² + 5 (1)
= 3 + 5 = 8
⇒ T₁ = 8 ⇒ a = 8
S₂ = 3 (2)² + 5 (2)
= 12 + 10 = 22
⇒ T₂ = 22 − 8 = 14
Now d = T₂ − T₁ = 14 − 8 = 6
∵ An A.P. is given by,
a, (a + d), (a + 2d), .....
∴ The required A.P. is:
8, (8 + 6), [8 + 2 (6)], .....
⇒ 8, 14, 20, .....
Now, using T_n = a + (n − 1) d, we hve
T₁₆ = a + 15d
= 8 + 15 × 6 = 98
Thus, the 16th term of the A.P. is 98.

Question. In an A.P. the sum of its first ten terms is –150 and the sum of its next term is –550. Find the A.P.
Answer : Let the first term = a
And the common difference = d
∴ S₁₀ = 10/2 [2a+ (10 − 1)d] = –150

Question. The sum of n terms of an A.P. is 5n² − 3n. Find the A.P. Hence find its 10th term.
Answer : We have:
S_n = 5n² − 3n
∴ S₁ = 5 (1)² − 3 (1) = 2
⇒ First term T₁ = (a) = 2
S₂ = 5 (2)² − 3 (2) = 20 − 6 = 14
⇒ Second term T₂ = 14 − 2 = 12
Now the common difference = T₂ − T₁
⇒ d = 12 − 2 = 10
∵ An A.P. is given by
a, (a + d), (a + 2d) .....
∴ The required A.P. is:
2, (2 + 10), [2 + 2 (10)], .....
⇒ 2, 12, 22, .....
Now, using T_n = a + (n − 1) d, we have
T₁₀ = 2 + (10 − 1) × 10
= 2 + 9 × 10
= 2 + 90 = 92.

Extra based Questions :

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

Question. The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
Answer : Let us consider,
The first term (T₁) = Fare for the first km = ₹ 15 since, the taxi fare beyond the first km is ₹ 8 for each additional km. ⇒ T₁ = 15
∴ Fare for 2 km = ₹ 15 + 1 × ₹ 8 ⇒ T₂ = a + 8              [where a = 15]
Fare for 3 km = ₹ 15 + 2 × ₹ 8 ⇒ T₃ = a + 16
Fare for 4 km = ₹ 15 + 3 × ₹ 8 ⇒ T₄ = a + 24
Fare for 5 km = ₹ 15 + 4 × ₹ 8 ⇒ T₅ = a + 32
Fare for n km = ₹ 15 + (n − 1) 8 ⇒ T_n = a + (n − 1) 8
We see that above terms form an A.P.

Question. The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Answer : Let the amount of air in the cylinder = x
∴ Air removed in 1st stroke = 1/4 x

Question. The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
Answer : Here, The cost of digging for 1st metre = ₹ 150
The cost of digging for first 2 metres = ₹ 150 + ₹ 50 = ₹ 200
The cost of digging for first 3 metres = ₹ 150 + (₹ 50) × 2 = ₹ 250
The cost of digging for first 4 metres = ₹ 150 + (₹ 50) × 3 = ₹ 300
∴ The terms are: 150, 200, 250, 300, ...
Since, 200 − 150 = 50
And 250 − 200 = 50
⇒ (200 − 150) = (250 − 200)
∴ The above terms form an A.P.

Question. The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.
Answer :

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

Question. a = 10, d = 10
Answer : T_n = a + (n − 1) d
∴ For a = 10 and d = 10, we have:
T₁ = 10 + (1 − 1) × 10 = 10 + 0 = 10
T₂ = 10 + (2 − 1) × 10 = 10 + 10 = 20
T₃ = 10 + (3 − 1) × 10 = 10 + 20 = 30
T₄ = 10 + (4 − 1) × 10 = 10 + 30 = 40
Thus, the first four terms of A.P. are:
10, 20, 30, 40.

Question. a = − 2, d = 0
Answer : T_n = a + (n − 1) d
∴ For a = − 2 and d = 0, we have:
T₁ = −2 + (1 − 1) × 0 = −2 + 0 = −2
T₂ = −2 + (2 − 1) × 0 = −2 + 0 = −2
T₃ = −2 + (3 − 1) × 0 = −2 + 0 = −2
T₄ = −2 + (4 − 1) × 0 = −2 + 0 = −2
∴ The first four terms are:
− 2, − 2, − 2, − 2.

Question. a = 4, d = − 3
Answer : T_n = a + (n − 1) d
∴ For a = 4 and d = − 3, we have:
T₁ = 4 + (1 − 1) × (− 3) = 4 + 0 = 4
T₂ = 4 + (2 − 1) × (− 3) = 4 + (− 3) = 1
T₃ = 4 + (3 − 1) × (− 3) = 4 + (− 6) = − 2
T₄ = 4 + (4 − 1) × (− 3) = 4 + (− 9) = − 5
Thus, the first four terms are:
4, 1, − 2, − 5.

Question. a = − 1, d = 1/2
Answer : T_n = a + (n − 1) d
For a = − 1 and d = 1/2 , we get
T₁ = − 1 + (1 − 1) × 1/2 = − 1 + 0 = − 1
T₂ = − 1 + (2 − 1) × 1/2 = − 1 + 1/2 = − 1/2
T₃ = − 1 + (3 − 1) × 1/2 = − 1 + 1 = 0
T₄ = − 1 + (4 − 1) × 1/2 = − 1 + 3/2 = 1/2
∴ The first four terms are:
− 1, − 1/2, 0, 1/2.

Question. a = − 1.25, d = − 0.25 
Answer : T_n = a + (n − 1) d
∴ For a = − 1.25 and d = − 0.25, we get
T₁ = − 1.25 + (1 − 1) × (− 0.25) = − 1.25 + 0 = − 1.25
T₂ = − 1.25 + (2 − 1) × (− 0.25) = − 1.25 + (− 0.25) = − 1.50
T₃ = − 1.25 + (3 − 1) × (− 0.25) = − 1.25 + (− 0.50) = − 1.75
T₄ = − 1.25 + (4 − 1) × (− 0.25) = − 1.25 + (− 0.75) = − 2.0
Thus, the four terms are:
− 1.25, − 1.50, − 1.75, − 2.0

For the following APs, write the first term and the common difference:

Question. 3, 1, − 1, − 3, ...
Answer : We have: 3, 1, − 1, − 3, .....
⇒ T₁ = 3 ⇒ a = 3
T₂ = 1
T₃ = − 1
T₄ = − 3
∴ T₂ − T₁ = 1 − 3 = − 2
T₄ − T₃ = − 3 − (− 1) = − 3 + 2 = − 2 } ⇒ d = − 2
Thus, a = 3 and d = − 2

Question. − 5, − 1, 3, 7, ...
Answer : We have: − 5, − 1, 3, 7, .....
⇒ T₁ = − 5 }   ⇒ a = − 5
T₂ = − 1     }  ⇒ d = T₂ − T₁ = − 1 − (− 5) = − 1 + 5 = 4
T₃ = 3        }  ⇒ d = − 1 + 5 = 4
T₄ = 7        }      T₄ − T₃ = 7 − 3 = 4 } ⇒ d = 4
Thus, a = − 5 and d = 4

Question. 1/3, 5/3, 9/3, 13/3, ...
Answer : 1/3, 5/3, 9/3, 13/3, .....
⇒ T₁ = 1/3 ⇒ a = 1/3
T₂ = 5/3 ⇒ d = T₂ − T₁ = 5/3 − 1/3 = 4/3
T₃ = 9/13  } ⇒ d = T₄ − T₃ = 13/3 − 9/3 = 4/3
T₄ = 13/3  }
Thus, a = 1/3 and d = 4/3

Question. 0.6, 1.7, 2.8, 3.9, ... 
Answer : We have: 0.6, 1.7, 2.8, 3.9, .....
⇒ T₁ = 0.6 ⇒ a = 0.6
T₂ = 1.7 ⇒ d = T₂ − T₁ = 1.7 − 0.6 = 1.1
T₃ = 2.8
T₄ = 3.9 ⇒ d = T₄ − T₃ = 3.9 − 2.8 = 1.1
Thus, a = 0.6 and d = 1.1

Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

Question. 2, 4, 8, 16, ...
Answer : We have: 2, 4, 8, 16, .....
T₁ = 2 }
T₂ = 4 }   ⇒ T₂ − T₁ = 4 − 2 = 2 
T₃ = 8         }
T₄ = 16       } ⇒ T₄ − T₃ = 16 − 8 = 8
Since 2 ≠ 8  }
∴ T₂ − T₁ ≠ T₄ − T₃
∴ The given numbers do not form an A.P.

Question. 2, 5/2, 3, 7/2, ...
Answer : We have: 2, 5/2, 3, 7/2, .....
∴ T₁ = 2, T₂ = 5/2, T₃ = 3, T₄ = 7/2
T₂ − T₁ = 5/2 −2 = 1/2
T₃ − T₂ = 3 − 5/2 = 1/2
T₄ − T₃ = 7/2 −3 = 1/2
∴ T₂ − T₁ = T₃ − T₂ = T₄ − T₃ = 1/2 ⇒ d = 1/2
∴ The given numbers form an A.P.
∴ T₅ = T₄ + 1/2 = 7/2 + 1/2 = 4
T₆ = T₅ + 1/2 = 4+1/2 = 9/2
T₇ = T₆ + 1/2 = 9/2+1/2 = 5
Thus, d = 1/2 and T₅ = 4, T₆ = 9/2 and T₇ = 5

Question. − 1.2, − 3.2, − 5.2, − 7.2, ...
Answer : We have: − 1.2, − 3.2, − 5.2, − 7.2, .....
∴ T₁ = −1.2, T₂ = −3.2, T₃ = −5.2, T₄ = −7.2
T₂ − T₁ = −3.2 + 1.2 = −2
T₃ − T₂ = −5.2 + 3.2 = −2
T₄ − T₃ = −7.2 + 5.2 = −2
∴ T₂ − T₁ = T₃ − T₂ = T₄ − T₃ = −2 ⇒ d = −2
∴ The given numbers form an A.P.
Such that d = − 2.
Now, T₅ = T₄ + (− 2) = − 7.2 + (− 2) = − 9.2
T₆ = T₅ + (− 2) = − 9.2 + (− 2) = − 11.2
T₇ = T₆ + (− 2) = − 11.2 + (− 2) = − 13.2
Thus, d = − 2 and T₅ = − 9.2, T₆ = − 11.2 and T₇ = − 13.2

Question. − 10, − 6, − 2, 2, ...
Answer : We have: − 10, − 6, − 2, 2, .....
∴ T₁ = −10, T₂ = −6, T₃ = −2, T₄ = 2
T₂ − T₁ = −6 + 10 = 4
T₃ − T₂ = −2 + 6 = 4
T₄ − T₃ = 2 + 2 = 4
∴ T₂ − T₁ = T₃ − T₂ = T₄ − T₃ = 4 ⇒ d = 4
∴ The given numbers form an A.P.
Now, T₅ = T₄ + 4 = 2 + 4 = 6
T₆ = T₅ + 4 = 6 + 4 = 10
T₇ = T₆ + 4 = 10 + 4 = 14
Thus, d = 4 and T₅ = 6, T₆ = 10, T₇ = 14

Question. 3, 3 + √2 , 3 + 2√2 , 3 + 3√2 , ...
Answer : We have:

Question. 0.2, 0.22, 0.222, 0.2222, ...
Answer : We have: 0.2, 0.22, 0.222, 0.2222, .....
∴ T₁ = 0.2 }
T₂ − 0.22  }     ⇒ T₂ − T₁ = 0.2 = 0.02
T₃ = 0.222    } ⇒ T₄ − T₃ = 0.2222 − 0.222 = 0.0002.
T₄ = 0.2222  }
Since,
T₂ − T₁ ≠ T₄ − T₃ 
∴ The given numbers do not form an A.P.

Question. 0, − 4, − 8, − 12, ...
Answer : We have: 0, − 4, − 8, − 12, .....
∴ T₁ = 0, T₂ = −4, T₃ = −8, T₄ = −12
T₂ − T₁= −4 − 0 = −4
T₃ − T₂ = −8 + 4 = −4
T₄ − T₃ = −12 + 8 = −4
∴ T₂ − T₁ = T₃ − T₂ = T₄ − T₃ = −4 ⇒ d = −4
∴ The given numbers form an A.P.
Now, T₅ = T₄ + (− 4) = − 12 + (− 4) = − 16
T₆ = T₅ + (− 4) = − 16 + (− 4) = − 20
T₇ = T₆ + (− 4) = − 20 + (− 4) = − 24
Thus, d = − 4 and T₅ = − 16, T₆ = − 20, T₇ = − 24

Question. −1/2, −1/2, −1/2, −1/2, ...
Answer : We have:

Question. 1, 3, 9, 27, ...
Answer : We have: 1, 3, 9, 27, .....
Here, T₁ = 1 }  ⇒ T₂ − T₁ = 3 − 1 = 2
T₂ = 3          }
T₃ = 9      }     ⇒ T₄ − T₃ = 27 − 9 = 18
T₄ = 27    }
∴ T₂ − T₁ ≠ T₄ − T₃ 
∴ The given numbers do not form an A.P.

Question. a, 2a, 3a, 4a, ...
Answer : We have: a, 2a, 3a, 4a, .....
∴ T₁ = a, T₂ = 2a, T₃ = 3a, T₄ = 4a
T₂ − T₁ = 2a − a = a
T₃ − T₂ = 3a − 2a = a
T₄ − T₃ = 4a − 3a = a
∴ T₂ − T₁ = T₃ − T₂ = T₄ − T₃ = a ⇒ d = a
∴ The numbers form an A.P.
Now, T₅ = T₄ + a = 4a + a = 5a
T₆ = T₅ + a = 5a + a = 6a
T₇ = T₆ + a = 6a + a = 7a
Thus, d = a and T₅ = 5a, T₆ = 6a, T₇ = 7a

Question. a, a², a³, a⁴, ...
Answer :We have: a, a², a³, a⁴, .....
∴ T₁ = a }    ⇒ T₂ − T₁ = a² − a = a [a − 1]
T₂ = a²   }
T₃ = a³      } ⇒ T₄ − T₃ = a⁴ − a³ = a³ [a − 1]
T₄ = a⁴      }
Since,
T₂ − T₁ ≠ T₄ − T₃ ∴ The given terms are not in A.P.

Question. √2 , √8 , √18 , √32 , ...
Answer : We have: √2 , √8 , √18 , √32 , .....
∴ T₁ = √2 , T₂ = 8√ , T₃ = √18 , T₄ = √32
T₂ − T₁ = √8 − √2 = 2√2 − √2 = 2√
T₃ − T₂ = √18 − √8 = 3√2 − 2√2 = √2
T₄ − T₃ = √32 − √18 = 4√2 − 3√2 = √2
∴ T₂ − T₁ = T₃ − T₂ = T₄ − T₃ = √2 ⇒ d = √2
∴ The given numbers form an A.P.
Now, T₅ = 4√2 + √2 = 5√2 = √50
T₆ = 5√2 + √2 = 6√2 = √72
T₇ = 6√2 + √2 = 7√2 = √98
Thus, d = √2 and T₅ = √50 , T₆ = √72 , T₇ = √98

Question. √3 , √6 , √9 , √12 , ...
Answer : We have: √3 , √6 , √9 , √12 , .....
∴ T₁ = √3 }      ⇒ T₂ − T₁ = √6 − √3 = √3(√2 − 1)
T₁ = √6   } 
and T₃ = √9  } ⇒ T₄ − T₃ = √12 − √9 = 2√3 − 3 = √3(2 − √3)
T₄ = √12      } 
∴ T₂ − T₁ ≠ T₄ − T₃ 
⇒ The given terms do not form an A.P.

Question. 1², 3², 5², 7², ...
Answer : We have: 1², 3², 5², 7², .....
∴ T₁ = 1² = 1 } T  ⇒ T₂ − T₁ = 9 − 1 = 8
T₂ = 3² = 9   }
T₃ = 5² = 25    }   ⇒ T₄ − T₃ = 49 − 25 = 24
T₄ = 7² = 49    }
∴ T₂ − T₁ ≠ T₄ − T₃ 
∴ The given terms do not form an A.P.

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.:

                 a  d  n a_n
Question. 7  3  8  …
Answer : a_n = a + (n − 1) d
⇒ a₈ = 7 + (8 − 1) 3
= 7 + 7 × 3
= 7 + 21
⇒ a₈ = 28

                       a  d    n  a_n
Question. − 18  …  10  0
Answer : a_n = a + (n − 1) d
⇒ a₁₀ = − 18 + (10 − 1) d
⇒ 0 = − 18 + 9d
⇒ 9d = 18 ⇒ d = 18/9 = 2
∴ d = 2

                  a     d    n    a_n
Question. …  − 3  18  − 5
Answer : a_n = a + (n − 1) d
⇒ − 5 = a + (18 − 1) × (− 3)
⇒ − 5 = a + 17 × (− 3)
⇒ − 5 = a − 51
⇒ a = − 5 + 51 = 46
Thus, a = 46

                       a       d   n  a_n
Question. − 18.9  2.5  …  3.6
Answer : a_n = a + (n − 1) d
⇒ 3.6 = − 18.9 + (n − 1) × 2.5
⇒ (n − 1) × 2.5 = 3.6 + 18.9
⇒ (n − 1) × 2.5 = 22.5
⇒ n − 1 = 22.5/2 5 = 9
⇒ n = 9 + 1 = 10
Thus, n = 10

                  a   d    n    a_n
Question. 3.5  0  105  …
Answer : a_n = a + (n − 1) d
⇒ a_n = 3.5 + (105 − 1) × 0
⇒ a_n = 3.5 + 104 × 0
⇒ a_n = 3.5 + 0 = 3.5
Thus, a_n = 3.5

Question. Which term of the A.P.: 3, 8, 13, 18, ..., is 78?
Answer : Let the nth term = 78
Here, a = 3, ⇒ T₁ = 3 and T₂ = 8
∴ d = T₂ − T₁ = 8 − 3 = 5
Now, T_n = a + (n − 1) d
⇒ 78 = 3 + (n − 1) × 5
⇒ 78 − 3 = (n − 1) × 5
⇒ 75 = (n − 1) × 5
⇒ (n − 1) = 75 ÷ 5 = 15
⇒ n = 15 + 1 = 16
Thus, 78 is the 16th term of the given A.P.

Find the number of terms in each of the following A.Ps. :

Question. 7, 13, 19, ..., 205
Answer : Here, a = 7
d = 13 − 7 = 6
Let the number of terms be n
∴ T_n = 205
Now, T_n = a + (n − 1) × d
⇒ 7 + (n − 1) × 6 = 205
⇒ (n − 1) × 6 = 205 − 7 = 198
⇒ n − 1 = 198/6 = 33
∴ n = 33 + 1 = 34
Thus, the required number of terms is 34.

Question. 18, 15 1/2, 13, ..., − 47
Answer : Here, a = 18
d = 15 1/2 − 18 = − 2 1/2
Let the nth term = − 47

Question. Check whether − 150 is a term of the A.P.: 11, 8, 5, 2 ...
Answer : For the given A.P., we have 
a = 11
d = 8 − 11 = − 3
Let − 150 is the nth term of the given A.P.
∴ T_n = a + (n − 1) d
⇒ − 150 = 11 + (n − 1) × (− 3)
⇒ − 150 − 11 = (n − 1) × (− 3)
⇒ − 161 = (n − 1) × (− 3)
⇒ n − 1 = − 161/− 3 = 161/3
⇒ n = 161/3 + 1 = 164/3 = 54 2/3
But n should be a positive integer.
Thus, − 150 is not a term of the given A.P.

Question. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Answer : Here, T₃₁ = ?

Question. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer : Here, n = 50
T₃ = 12
T_n = 106 ⇒ T₅₀ = 106
If first term = a and the common difference = d
∴ T₃ = a + 2d = 12
T₅₀ = a + 49d = 106 
⇒ T₅₀ − T₃ ⇒ a + 49d − (a + 2d) = 106 − 12
⇒ 47d = 94
⇒ d = 94/47 = 2
From (1), we have
a + 2d = 12 ⇒ a + 2 (2) = 12
⇒ a = 12 − 4 = 8
Now, T₂₉ = a + (29 − 1) d
= 8 + (28) × 2
= 8 + 56 = 64
Thus, the 29th term is 64.

Question. If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively, which term of this A.P. is zero ?
Answer : Here, T₃ = 4 and T₉ = − 8
∴ Using T_n = a + (n − 1) d
⇒ T₃ = a + 2d = 4 
T₉ = a + 8d = − 8 
Subtracting (1) from (2) we get
(a + 8d) − (a + 2d) = − 8 − 4
⇒ 6d = − 12
⇒ d = −12/6 = 2
Now, from (1), we have:
a + 2d = 4
⇒ a + 2 (− 2) = −4
⇒ a − 4 = 4
⇒ a = 4 + 4 = 8
Let the nth term of the A.P. be 0.
∴ T_n = a + (n − 1) d = 0
⇒ 8 + (n − 1) × (− 2) = 0
⇒ (n − 1) × − 2 = − 8
⇒ n − 1 = −8/−2 = 4
⇒ n = 4 + 1 = 5
Thus, the 5th term of the A.P. is 0. 

Question. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Answer : Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Now, using T_n = a + (n − 1) d
T₁₇ = a + 16d
T₁₀ = a + 9d
According to the condition,
T_n + 7 = T₁₇
⇒ (a + 9d) + 7 = a + 16d
⇒ a + 9d − a − 16d = − 7
⇒ − 7d = − 7 ⇒ d = 1
Thus, the common difference is 1.

Question. Which term of the A.P.: 3, 15, 27, 39, ... will be 132 more than its 54th term?
Answer : Here, a = 3
d = 15 − 3 = 12
Using T_n = a + (n − 1) d, we get
T₅₄ = a + 53d
= 3 + 53 × 12
= 3 + 636 = 639
Let a_n be 132 more than its 54th term.
∴ a_n = T₅₄ + 132
⇒ a_n = 639 + 132 = 771
Now a_n = a + (n − 1) d = 771
⇒ 3 + (n − 1) × 12 = 771
⇒ (n − 1) × 12 = 771 − 3 = 768
⇒ (n − 1) = 768/12 = 64
⇒ n = 64 + 1 = 65
Thus, 132 more than 54th term is the 65th term.

Question. Two A.Ps. have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Answer : Let for the 1st A.P., the first term = a
∴ T₁₀₀ = a + 99d
And for the 2nd A.P., the first term = a′
∴ T′₁₀₀ = a′ + 99d
According to the condition, we have:
T₁₀₀ − T′₁₀₀ = 100
⇒ a + 99d − (a′ + 99d) = 100
⇒ a − a′ = 100
Let, T′₁₀₀₀ − T′₁₀₀₀ = x
∴ a + 999d − (a′ + 999d) = x
⇒ a − a′ = x ⇒ x = 100
∴ The difference between the 1000th terms is 100.

Question. How many three-digit numbers are divisible by 7?
Answer : The first three digit number divisible by 7 is 105.
The last such three digit number is 994.
∴ The A.P. is 105, 112, 119, ....., 994
Here, a = 105 and d = 7
Let n be the required number of terms.
∴ T_n = a + (n − 1) d
⇒ 994 = 105 + (n − 1) × 7
⇒ (n − 1) × 7 = 994 − 105 = 889
⇒ (n − 1) = 889/7 = 127
⇒ n = 127 + 1 = 128
Thus, 128 numbers of 3-digit are divisible by 7.

Question. How many multiples of 4 lie between 10 and 250?
Answer : ∵ The first multiple of 4 beyond 10 is 12.
The multiple of 4 just below 250 is 248.
∴ The A.P. is given by:
12, 16, 20, ....., 248
Here, a = 12 and d = 4
Let the number of terms = n
∴ Using T_n = a + (n − 1) d, we get
∴ T_n = 12 + (n − 1) × 4
⇒ 248 = 12 + (n − 1) × 4
⇒ (n − 1) × 4 = 248 − 12 = 236
⇒ n − 1 = 236/4 = 59
⇒ n = 59 + 1 = 60
Thus, the required number of terms = 60.

Question. For what value of n, are the nth terms of two A.Ps.: 63, 65, 67, ... and 3, 10, 17, ... equal?
Answer : For the 1st A.P.
∵ a = 63 and d = 65 − 63 = 2
∴ T_n = a + (n − 1) d
⇒ T_n = 63 + (n − 1) × 2
For the 2nd A.P.
∵ a = 3 and d = 10 − 3 = 7
∴ T_n = a + (n − 1) d
⇒ T_n = 3 + (n − 1) × 7
Now, according to the condition,
3 + (n − 1) × 7 = 63 + (n − 1) × 2
⇒ (n − 1) × 7 − (n − 1) × 2 = 63 − 3
⇒ 7n − 7 − 2n + 2 = 60
⇒ 5n − 5 = 60
⇒ 5n = 60 + 5 = 65
⇒ n = 65/5 = 13
Thus, the 13th terms of the two given A.Ps. are equal.

Question. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer : Let the first term = a and the common difference = d.
∴ Using Tn = a + (n − 1) d, we have:
T₃ = a + 2d
⇒ a + 2d = 16 
And T₇ = a + 6d, T₅ = a + 4d
According to the condition,
T₇ − T₅ = 12
⇒ (a + 6d) − (a + 4d) = 12
⇒ a + 6d − a − 4d = 12
⇒ 2d = 12
⇒ d = 12/2 = 6 
Now, from (1) and (2), we have:
a + 2 (6) = 16
⇒ a + 12 = 16
⇒ a = 16 − 12 = 4
∴ The required A.P. is
4, [4 + 6], [4 + 2 (6)], [4 + 3 (6)], .....
or 4, 10, 16, 22, .....

Question. Find the 20th term from the last term of the A.P.: 3, 8, 13, ..., 253.
Answer : We have, the last term l = 253
Here, d = 8 − 3 = 5
Since, the nth term before the last term is given by l − (n − 1) d,
∴ We have
20th term from the end = l − (20 − 1) × 5
= 253 − 19 × 5
= 253 − 95 = 158

Question. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44.
Find the first three terms of the A.P.
Answer : Let the first term = a
And the common difference = d
∴ Using T_n = a + (n − 1) d,
T₄ + T₈ = 24
⇒ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24

Question. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Answer : Here, a = ₹ 5000 and d = ₹ 200
Say, in the nth year he gets ₹ 7000.
∴ Using T_n = a + (n − 1) d, we get
7000 = 5000 + (n − 1) × 200
⇒ (n − 1) × 200 = 7000 − 5000 = 2000
⇒ n − 1 = 2000/200 = 10
⇒ n = 10 + 1 = 11
Thus, his income becomes ₹ 7000 in 11 years.

Find the sum of the following A.Ps. :

Question. 2, 7, 12, ..., to 10 terms.
Answer : Here, a = 2
d = 7 − 2 = 5
n = 10
Since, Sn = n/2 [2a + (n − 1) d]
∴ S₁₀ = 10/2 [2 × 2 + (10 − 1) × 5]
⇒ S₁₀ = 5 [4 + 9 × 5]
⇒ S₁₀ = 5 [49] = 245
Thus, the sum of first 10 terms is 245.

Question. − 37, − 33, − 29, ..., to 12 terms.
Answer : We have:
a = − 37
d = − 33 − (− 37) = 4
n = 12
∴ S_n = n/2 [2a + (n − 1) d]
⇒ S₁₂ = 12/2 [2 (− 37) + (12 − 1) × 4]
= 6 [− 74 + 11 × 4]
= 6 [− 74 + 44]
= 6 × [− 30] = − 180
Thus, sum of first 12 terms = −180.

Question. 0.6, 1.7, 2.8, ..., to 100 terms.
Answer : Here, a = 0.6
d = 1.7 − 0.6 = 1.1
n = 100
∴ S_n = n/2 [2a + (n − 1) d]
S₁₀₀ = 100/2 [2 (0.6) + (100 − 1) × 1.1]
= 50 [1.2 + 99 × 1.1]
= 50 [1.2 + 108.9]
= 50 [110.1]
= 5505
Thus, the required sum of first 100 terms is 5505.

Question. 1/15, 112, 110, ..., to 11 terms.
Answer : Here, a = 1/15

Find the sums given below:

Question. 7 + 10 1/2 + 14+ ... + 84
Answer : Here, a = 7
d = 10 1/2 −7 = 3 1/2 = 7/2
l = 84
Let n be the number of terms
∴ T_n = a + (n − 1) d
⇒ 84 = 7 + (n − 1) × 7/2
⇒ (n − 1) × 7/2 = 84 − 7 = 77
⇒ n − 1 = 77 × 2/7  = 22
⇒ n = 22 + 1 = 23
Now, S_n = n/2 (a + l)
⇒ S₂₃ = 23/2 (7 + 84)
= 23/2 × 91 = 2093/2 = 1046 1/2
Thus, the required sum = 1046 1/2.

Question. 34 + 32 + 30 + ... + 10
Answer : Here, a = 34
d = 32 − 34 = − 2
l = 10
Let the number of terms be n

Question. − 5 + (− 8) + (− 11) + ... + (− 230)
Answer : Here, a = − 5
d = − 8 − (− 5) = − 3
l = − 230
Let n be the number of terms.
∴ T_n = − 230
⇒ − 230 = − 5 + (n − 1) × (− 3)
⇒ (n − 1) × (− 3) = − 230 + 5 = − 225
⇒ n − 1 = −225/−3 = 75
⇒ n = 75 + 1 = 76
Now, S76 = 76/2 [(− 5) + (− 230)]
= 38 × (− 235) 
= − 8930
∴ The required sum = − 8930.

Question. given a = 5, d = 3, a_n = 50, find n and S_n.
Answer : Here, a = 5, d = 3 and an = 50 = l
∵ a_n = a + (n − 1) d
∴ 50 = 5 + (n − 1) × 3
⇒ 50 − 5 = (n − 1) × 3
⇒ (n − 1) × 3 = 45
⇒ (n − 1) = 45/3 = 15
⇒ n = 15 + 1 = 16
Now S_n = n/2 (a + l)
= 16/2 (5 + 50)
= 8 (55) = 440
Thus, n = 16 and S_n = 440

Question. given d = 5, S₉ = 75, find a and a₉.
Answer : Here, d = 5, S₉ = 75
Let the first term of the A.P. is ‘a’.

Question. given a = 2, d = 8, S_n = 90, find n and a_n.
Answer : Here, a = 2, d = 8 and S_n = 90
∵ S_n = n/2 [2a + (n − 1) d]

Question. given a = 7, a₁₃ = 35, find d and S₁₃.
Answer : Here, a = 7 and a₁₃ = 35 = l
∴ a_n = a + (n − 1) d
⇒ 35 = 7 + (13 − 1) d
⇒ 35 − 7 = 12d
⇒ 28 = 12d
⇒ d = 28/12 = 7/3
Now, using
S_n = n/2 (a + l)
S₁₃ = 13/2 (7 + 35)
= 13/2 × 42
= 13 × 21 = 273
S_n = 273 and d = 7/3

Question. given a₁₂ = 37, d = 3, find a and S₁₂.
Answer : Here, a₁₂ = 37 = l and d = 3
Let the first term of the A.P. be ‘a’.
Now a₁₂ = a + (12 − 1) d
⇒ 37 = a + 11d
⇒ 37 = a + 11 × 3
⇒ 37 = a + 33
⇒ a = 37 − 33 = 4
Now, S_n =n/2 (a + l)
⇒ S₁₂ = 12/2 (4 + 37)
⇒ S₁₂ = 6 × (41) = 246
Thus, a = 4 and S₁₂ = 246

Question. given a = 8, a_n = 62, S_n = 210, find n and d.
Answer : Here, a = 8, a_n = 62 = l and S_n = 210
Let the common difference = d

Question. given a_n = 4, d = 2, S_n = − 14, find n and a.
Answer : Here, a_n = 4, d = 2 and S_n = − 14
Let the first term be ‘a’.
∵ a_n = 4
∴ a + (n − 1) 2 = 4

Question. given a = 3, n = 8, S = 192, find d.
Answer : Here, a = 3, n = 8 and S_n = 192
Let the common difference = d.
∵ S_n = n/2 [2a + (n − 1) d]
∴ 192 = 8/2 [2 (3) + (8 − 1) d]
⇒ 192 = 4 [6 + 7d]
⇒ 192 = 24 + 28d
⇒ 28d = 192 − 24 = 168
⇒ d = 168/28 = 6
Thus, d = 6.

Question. given l = 28, S = 144, and there are total 9 terms. Find a.
Answer : Here, l = 28 and S₉ = 144
Let the first term be ‘a’.
Then S_n = n/2 (a + l)
⇒ S₉ = 9/2 (a + 28)
⇒ 144 = 9/2 (a + 28)
⇒ a + 28 = 144 × 2/9 = 16 × 2 = 32
⇒ a = 32 − 28 = 4
Thus, a = 4.

Question. How many terms of the A.P.: 9, 17, 25, ... must be taken to give a sum of 636?
Answer : Here, a = 9
d = 17 − 9 = 8
S_n = 636
∵ S_n = n/2 [2a + (n − 1) d] = 636
∴ n/2 [(2 × 9) + (n − 1) × 8] = 636
⇒ n [18 + (n − 1) × 8] = 1272
⇒ n (8n + 10) = 1272
⇒ 8n² + 10n − 1272 = 0
⇒ 4n² + 5n − 636 = 0
⇒ 4n² + 53n − 48n − 636 = 0
⇒ n (4n + 53) − 12 (4n + 53) = 0
⇒ (n − 12) (4n + 53) = 0 ⇒ n = 12 and n = − 53/4
Rejecting n = − 53/4 , we have n = 12. 

Question. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer : Here, a = 5
l = 45 = T_n
S_n = 400
∵ T_n = a + (n − 1) d
∴ 45 = 5 + (n − 1) d
⇒ (n − 1) d = 45 – 5
⇒ (n − 1) d = 40 
Also S_n = n/2 (a + l)
⇒ 400 = n/2 (5 + 45)
⇒ 400 × 2 = n × 50
⇒ n = 400×2/50 = 16
From (1), we get
(16 − 1) d = 40
⇒ 15d = 40
⇒ d = 40/15 = 8/3

Question. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer : We have,
First term a = 17
Last term l = 350 = T_n
Common difference d = 9
Let the number of terms be ‘n’
∵ T_n = a + (n − 1) d
∴ 350 = 17 + (n − 1) × 9
⇒ (n − 1) × 9 = 350 − 17 = 333
⇒ n − 1 = 333/9 = 37
⇒ n = 37 + 1 = 38
Since, S_n = n/2 (a + l)
∴ S₃₈ = 38/2 (17 + 350)
= 19 (367) = 6973
Thus, n = 38 and S_n = 6973

Question. Find the sum of first 22 terms of an A.P. in which d = 7 and 22nd term is 149.
Answer : Here, n = 22, T₂₂ = 149 = l
d = 7
Let the first term of the A.P. be ‘a’.
∴ T_n = a + (n − 1) d
⇒ T_n = a + (22 − 1) × 7
⇒ a + 21 × 7 = 149
⇒ a + 147 = 149
⇒ a = 149 − 147 = 2
Now, S₂₂ = n/2 [a + l]
⇒ S₂₂ = 22/2 [2 + 149]
= 11 [151] = 1661
Thus S₂₂ = 1661 

Question. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Answer : Here, n = 51, T₂ = 14 and T₃ = 18
Let the first term of the A.P. be ‘a’ and the common difference is d.
∴ We have:

Question. Find the sum of the first 40 positive integers divisible by 6.
Answer : ∵ The first 40 positive integers divisible by 6 are:
6, 12, 18, ....., (6 × 40).
And, these numbers are in A.P. such that
a = 6
d = 12 − 6 = 6 and a_n = 6 × 40 = 240 = l
∴ S₄₀ = 40/2 [(2 × 6) + (40 − 1) × 6]
= 20 [12 + 39 × 6]
= 20 [12 + 234]
= 20 × 246 = 4920
OR
S_n = n/2 [a + l]
S₄₀ = 40/2 [6 + 240]
= 20 × 246 = 4920

Question. Find the sum of the first 15 multiples of 8.
Answer : The first 15 multiples of 8 are:
8, (8 × 2), (8 × 3), (8 × 4), ....., (8 × 15)
or 8, 16, 24, 32, ....., 120.
These numbers are in A.P., where
a = 8 and l = 120
∴ S₁₅ = 15/2 [a + l]
= 15/2 [8 + 120]
= 15/2 × 128
= 15 × 64 = 960
Thus, the sum of first positive 15 multiples of 8 is 960.

Question. Find the sum of the odd numbers between 0 and 50.
Answer : Odd numbers between 0 and 50 are:
1, 3, 5, 7, ....., 49
These numbers are in A.P. such that
a = 1 and l = 49
Here, d = 3 − 1 = 2
∴ T_n = a + (n − 1) d
⇒ 49 = 1 + (n − 1) 2
⇒ 49 − 1 = (n − 1) 2
⇒ (n − 1) = 48/2 = 24
∴ n = 24 + 1 = 25
Now, S₂₅ = 25/2 [1 + 49]
= 25/2 [50]
= 25 × 25 = 625
Thus, the sum of odd numbers between 0 and 50 is 625.

Question. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? 
Answer : Here, penalty for delay on
1st day = ₹ 200
2nd day = ₹ 250
3rd day = ₹ 300
...............
...............
Now, 200, 250, 300, ..... are in A.P. such that
a = 200, d = 250 − 200 = 50
∴ S₃₀ is given by
S₃₀ = 30/2 [2 (200) + (30 − 1) × 50]                          [using S_n = n/2[2a+(n−1)d]
= 15 [400 + 29 × 50]
= 15 [400 + 1450]
= 15 × 1850 = 27,750
Thus, penalty for the delay for 30 days is ₹ 27,750.

Question. A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performace. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Answer : Sum of all the prizes = ₹ 700
Let the first prize = a
∴ 2nd prize = (a − 20)
3rd prize = (a − 40)
4th prize = (a − 60)
........................................
Thus, we have, first term = a
Common difference = − 20
Number of prizes, n = 7
Sum of 7 terms S_n = 700
Since, S_n = n/2 [2a + (n − 1) d]
⇒ 700 = 7/2 [2 (a) + (7 − 1) × (− 20)]
⇒ 700 = 7/2 [2a + (6 × − 20)]
⇒ 700 × 2/7 = 2a − 120
⇒ 200 = 2a − 120
⇒ 2a = 200 + 120 = 320
⇒ a = 320/2 = 160
Thus, the values of the seven prizes are:
₹ 160, ₹ (160 − 20), ₹ (160 − 40), ₹ (160 − 60), ₹ (160 − 80), ₹ (160 − 100) and ₹ (160 − 120)
⇒ ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60 and ₹ 40.

Question. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students? 
Answer : Number of classes = 12
∵ Each class has 3 sections.
∴ Number of plants planted by class I = 1 × 3 = 3
Number of plants planted by class II = 2 × 3 = 6
Number of plants planted by class III = 3 × 3 = 9
Number of plants planted by class IV = 4 × 3 = 12
.......................................................................................................
Number of plants planted by class XII = 12 × 3 = 36
The numbers 3, 6, 9, 12, ..........., 36 are in A.P.
Here, a = 3 and d = 6 − 3 = 3
∵ Number of classes = 12
i.e., n = 12
∴ Sum of the n terms of the above A.P., is given by
S₁₂ = 12/2 [2 (3) + (12 − 1) 3]                                 [using S_n = n/2[2a+(n−1)d]
= 6 [6 + 11 × 3]
= 6 [6 + 33]
= 6 × 39 = 234
Thus, the total number of trees = 234.

Question. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Answer : Here, number of potatoes = 10
The up-down distance of the bucket:
From the 1st potato = [5 m] × 2 = 10 m
From the 2nd potato = [(5 + 3) m] × 2 = 16 m
From the 3rd potato = [(5 + 3 + 3) m] × 2 = 22 m
From the 4th potato = [(5 + 3 + 3 + 3) m] × 2 = 28 m
.................................. ...........................
∵ 10, 16, 22, 28, ..... are in A.P. such that
a = 10 and d = 16 − 10 = 6
∴ Using S_n = n/2 [2a + (n − 1) d], we have:
S₁₀ = 10/2 [2 (10) + (10 − 1) × 6]
= 5 [20 + 9 × 6]
= 5 [20 + 54]
= 5 [74]
= 5 × 74 = 370
Thus, the sum of above distances = 370 m.
⇒ The competitor has to run a total distance of 370 m.

Question. Which term of the A.P.: 121, 117, 113, ..., is its first negative term?
[Hint: Find n for an < 0] 
Answer : We have the A.P. having a = 121 and d = 117 − 121 = − 4
∴ a_n = a + (n − 1) d
= 121 + (n − 1) × (− 4)
= 121 − 4_n + 4
= 125 − 4_n
For the first negative term, we have
a_n < 0
⇒ (125 − 4_n) < 0
⇒ 125 < 4_n
⇒ 125/4 < n
⇒ 31 1/4 < n
or n > 31 1/4
Thus, the first negative term is 32nd term.