CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 04

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.

[LEVEL - 1]

Question 1. Find \(n^{\text{th}}\) term of -15, -18, -21, ..........
Answer: First term \(a = -15\)
Common difference \(d = -18 - (-15) = -18 + 15 = -3\)
\(a_n = a + (n - 1)d\)
\(a_n = -15 + (n - 1)(-3)\)
\(a_n = -15 - 3n + 3\)
\(a_n = -3n - 12 = -3(n + 4)\)
In simple words: Using the formula for the \(n^{\text{th}}\) term of an AP, we substitute the first term and common difference to get the expression.

Exam Tip: Always verify your common difference by checking \(a_3 - a_2\) as well as \(a_2 - a_1\).

Question 2. Find the common diff. of A.P 1, -2, -5, -8, ........
Answer: Common difference \(d = a_2 - a_1\)
\(d = -2 - 1 = -3\)
In simple words: Subtract the first term from the second term to find the common difference.

Exam Tip: Be careful with negative signs during subtraction.

Question 3. Find the A.P whose first term is 4 and common difference is -3
Answer: Given first term \(a = 4\) and common difference \(d = -3\)
\(a_1 = 4\)
\(a_2 = a + d = 4 + (-3) = 1\)
\(a_3 = a + 2d = 4 + 2(-3) = -2\)
\(a_4 = a + 3d = 4 + 3(-3) = -5\)
Thus, the A.P. is \(4, 1, -2, -5, \dots\)
In simple words: Add the common difference repeatedly to the first term to generate the terms of the AP.

Exam Tip: List at least 4 terms to clearly show the pattern of the AP.

Question 4. Find \(5^{\text{th}}\) term from end of the AP : 17, 14, 11 .......... -40.
Answer: Here, last term \(l = -40\)
First term \(a = 17\)
Common difference \(d = 14 - 17 = -3\)
Formula for \(n^{\text{th}}\) term from the end = \(l - (n - 1)d\)
\(5^{\text{th}}\) term from the end = \(-40 - (5 - 1)(-3)\)
\(= -40 - (4)(-3)\)
\(= -40 + 12 = -28\)
In simple words: You can either reverse the AP and find the 5th term from the beginning, or use the direct formula using the last term.

Exam Tip: Remember that \(d\) retains its original value when using the formula from the end.

Question 5. If 2p, p + 10, 3p + 2 are in AP then find p.
Answer: Since the terms are in AP, the difference between consecutive terms is equal:
\((p + 10) - 2p = (3p + 2) - (p + 10)\)
\(10 - p = 2p - 8\)
\(10 + 8 = 2p + p\)
\(3p = 18\)
\(p = 6\)
In simple words: Use the property that if \(a, b, c\) are in AP, then \(2b = a + c\).

Exam Tip: Using \(2b = a + c\) is often faster than equating differences.

Question 6. If arithmetic mean between 3a and 2a - 7 is a + 4, then find a.
Answer: The arithmetic mean of two numbers \(x\) and \(y\) is \(\frac{x + y}{2}\).
\(\frac{3a + (2a - 7)}{2} = a + 4\)
\(5a - 7 = 2(a + 4)\)
\(5a - 7 = 2a + 8\)
\(5a - 2a = 8 + 7\)
\(3a = 15\)
\(a = 5\)
In simple words: Set up the equation using the formula for the arithmetic mean and solve for \(a\).

Question 7. Find sum of all odd numbers between 0 & 50.
Answer: The odd numbers between 0 and 50 are \(1, 3, 5, 7, \dots, 49\).
This forms an AP with \(a = 1\), \(d = 2\), and \(l = 49\).
First find the number of terms \(n\):
\(a_n = a + (n - 1)d\)
\(49 = 1 + (n - 1)2\)
\(48 = 2(n - 1)\)
\(n - 1 = 24 \implies n = 25\)
Now find the sum \(S_n\):
\(S_n = \frac{n}{2}(a + l)\)
\(S_{25} = \frac{25}{2}(1 + 49)\)
\(S_{25} = \frac{25}{2}(50) = 25 \times 25 = 625\)
In simple words: The sum of the first \(n\) odd natural numbers is always equal to \(n^2\). Here \(n = 25\), so the sum is \(25^2 = 625\).

Exam Tip: Recognizing standard shortcuts like sum of first \(n\) odd numbers equals \(n^2\) saves time in competitive exams.

Question 8. If \(a = 5\), \(d = 3\) and \(a_n = 50\), then find n.
Answer: Using the \(n^{\text{th}}\) term formula:
\(a_n = a + (n - 1)d\)
\(50 = 5 + (n - 1)3\)
\(50 - 5 = 3(n - 1)\)
\(45 = 3(n - 1)\)
\(n - 1 = \frac{45}{3} = 15\)
\(n = 15 + 1 = 16\)
In simple words: Substitute the given values into the \(n^{\text{th}}\) term formula and solve linear equation for \(n\).

Exam Tip: Always isolate the term containing \((n - 1)\) first before dividing by \(d\).

Question 9. For what value of n are the \(n^{\text{th}}\) term of two AP, 63, 65, 67, ...... and 3, 10, 17, ....... equal?
Answer: For the first AP: \(63, 65, 67, \dots\)
First term \(a_1 = 63\), common difference \(d_1 = 2\)
\(n^{\text{th}}\) term = \(63 + (n - 1)2 = 63 + 2n - 2 = 2n + 61\)
For the second AP: \(3, 10, 17, \dots\)
First term \(a_2 = 3\), common difference \(d_2 = 7\)
\(n^{\text{th}}\) term = \(3 + (n - 1)7 = 3 + 7n - 7 = 7n - 4\)
Equating both \(n^{\text{th}}\) terms:
\(2n + 61 = 7n - 4\)
\(61 + 4 = 7n - 2n\)
\(55 = 5n\)
\(n = 13\)
In simple words: Write expressions for the \(n^{\text{th}}\) term of both APs separately, equate them, and solve for \(n\).

Exam Tip: Ensure you expand brackets carefully to avoid algebraic errors.

Question 10. If sum of n terms of an AP is \(2n^2 + 5n\), then find its \(n^{\text{th}}\) term.
Answer: Given \(S_n = 2n^2 + 5n\)
We know that \(a_n = S_n - S_{n-1}\)
\(S_{n-1} = 2(n - 1)^2 + 5(n - 1)\)
\(S_{n-1} = 2(n^2 - 2n + 1) + 5n - 5\)
\(S_{n-1} = 2n^2 - 4n + 2 + 5n - 5 = 2n^2 + n - 3\)
\(a_n = (2n^2 + 5n) - (2n^2 + n - 3)\)
\(a_n = 2n^2 + 5n - 2n^2 - n + 3\)
\(a_n = 4n + 3\)
In simple words: Subtract the sum of \((n-1)\) terms from the sum of \(n\) terms to get the \(n^{\text{th}}\) term.

Exam Tip: This is a standard property: \(a_n = S_n - S_{n-1}\) holds for any series.

[LEVEL - 2]

Question 1. Find \(n^{\text{th}}\) term of an AP is \(7 - 4n\). find its common difference.
Answer: Given \(a_n = 7 - 4n\)
To find the common difference, we calculate \(a_{n-1}\) and subtract it from \(a_n\), or find the first few terms.
\(a_1 = 7 - 4(1) = 3\)
\(a_2 = 7 - 4(2) = -1\)
\(a_3 = 7 - 4(3) = -5\)
Common difference \(d = a_2 - a_1 = -1 - 3 = -4\)
In simple words: Find the first two terms by substituting \(n = 1\) and \(n = 2\), then subtract them to get \(d\).

Exam Tip: In a linear expression in \(n\) like \(An + B\), the coefficient of \(n\) is always the common difference.

Question 2. Which term of an AP 5, 2, -1, ..... will be -22 ?
Answer: Here \(a = 5\), \(d = 2 - 5 = -3\), and \(a_n = -22\)
\(a_n = a + (n - 1)d\)
\(-22 = 5 + (n - 1)(-3)\)
\(-22 - 5 = -3(n - 1)\)
\(-27 = -3(n - 1)\)
\(n - 1 = \frac{-27}{-3} = 9\)
\(n = 9 + 1 = 10^{\text{th}}\text{ term}\)
In simple words: Set the general term formula equal to \(-22\) and solve for \(n\).

Question 3. Write the next term of an AP \(\sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\dots\dots\)
Answer: Simplify the given terms:
\(\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}\)
\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\)
\(\sqrt{32} = \sqrt{16 \n times 2} = 4\sqrt{2}\)
The AP is \(2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots\)
Common difference \(d = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}\)
Next term = \(4\sqrt{2} + \sqrt{2} = 5\sqrt{2}\)
In simple words: Simplify the surds into their simplest radical form to easily spot the common difference.

Exam Tip: Always simplify radical expressions before finding the common difference.

Question 4. Determine \(27^{\text{th}}\) term of an AP whose \(9^{\text{th}}\) term is -10 and common difference is \(1\frac{1}{4}\)
Answer: Given \(a_9 = -10\) and \(d = 1\frac{1}{4} = \frac{5}{4}\)
\(a_9 = a + 8d = -10\)
\(a + 8\left(\frac{5}{4}\right) = -10\)
\(a + 10 = -10\)
\(a = -20\)
Now, find \(27^{\text{th}}\) term (\(a_{27}\)):
\(a_{27} = a + 26d\)
\(a_{27} = -20 + 26\left(\frac{5}{4}\right)\)
\(a_{27} = -20 + \frac{130}{2} = -20 + 65 = 45\)
Wait, let's recheck with the answer key provided in the source: Ans. \(927 = \frac{25}{2}\) or similar notation, let's compute precisely:
If \(d = \frac{1}{4}\):
\(a + 8\left(\frac{1}{4}\right) = -10 \implies a + 2 = -10 \implies a = -12\)
\(a_{27} = -12 + 26\left(\frac{1}{4}\right) = -12 + \frac{13}{2} = \frac{-24 + 13}{2} = \frac{-11}{2}\).
Let us follow standard textbook values based on the OCR key: \(927\) or \(a_{27} = \frac{25}{2}\) assuming \(d = \frac{5}{4}\) or checking standard transcription. Let's write the exact verified solution steps for \(d = \frac{5}{4}\):
\(a_{27} = a + 26d = -20 + 26\left(\frac{5}{4}\right) = -20 + \frac{65}{2} = \frac{-40 + 65}{2} = \frac{25}{2}\).
In simple words: Find the first term \(a\) using the given \(9^{\text{th}}\) term, then use it to calculate the \(27^{\text{th}}\) term.

Exam Tip: Convert mixed fractions into improper fractions before performing calculations.

Question 5. Find the sum of series \(103 + 101 + 99 + \dots + 49\).
Answer: Here \(a = 103\), \(d = 101 - 103 = -2\), and \(l = 49\)
First find \(n\):
\(a_n = a + (n - 1)d\)
\(49 = 103 + (n - 1)(-2)\)
\(49 - 103 = -2(n - 1)\)
\(-54 = -2(n - 1)\)
\(n - 1 = 27 \implies n = 28\)
Now find sum \(S_n\):
\(S_{28} = \frac{n}{2}(a + l)\)
\(S_{28} = \frac{28}{2}(103 + 49)\)
\(S_{28} = 14 \times 152 = 2128\)
In simple words: Identify the first term, common difference, and last term, find the total number of terms, and use the sum formula.

Exam Tip: Use the formula \(S_n = \frac{n}{2}(a + l)\) when the last term is known to save time.

Question 6. Which term of the AP 3, 15, 27, 39, ..... will be 132 more than its \(54^{\text{th}}\) term ?
Answer: Given AP: \(3, 15, 27, 39, \dots\)
\(a = 3\), \(d = 15 - 3 = 12\)
First find \(a_{54}\):
\(a_{54} = a + 53d = 3 + 53(12) = 3 + 636 = 639\)
Let the required term be \(a_n\) which is 132 more than \(a_{54}\):
\(a_n = a_{54} + 132\)
\(a_n = 639 + 132 = 771\)
Now find \(n\):
\(a + (n - 1)d = 771\)
\(3 + (n - 1)12 = 771\)
\((n - 1)12 = 768\)
\(n - 1 = \frac{768}{12} = 64\)
\(n = 64 + 1 = 65^{\text{th}}\text{ term}\)
In simple words: Calculate the \(54^{\text{th}}\) term, add 132 to it, and then find the position \(n\) of that new value in the AP.

Question 7. How many three digit numbers are divisible by 7 ?
Answer: The first 3-digit number is 100. Dividing 100 by 7 gives a remainder of 2, so the first 3-digit number divisible by 7 is \(100 + (7 - 2) = 105\).
The last 3-digit number is 999. Dividing 999 by 7 gives a remainder of 4, so the last 3-digit number divisible by 7 is \(999 - 4 = 995\).
The sequence is \(105, 112, 119, \dots, 995\), which forms an AP with \(a = 105\), \(d = 7\), and \(a_n = 995\).
\(a_n = a + (n - 1)d\)
\(995 = 105 + (n - 1)7\)
\(995 - 105 = 7(n - 1)\)
\(890 = 7(n - 1)\) -> Wait, let's check \(994\) as last term since \(994 = 7 \times 142\).
Let's use \(a_n = 994\):
\(994 = 105 + (n - 1)7\)
\(889 = 7(n - 1)\)
\(n - 1 = \frac{889}{7} = 127\)
\(n = 127 + 1 = 128\)
In simple words: Find the first and last 3-digit numbers divisible by 7, then find the number of terms in that AP.

Exam Tip: Always double check whether 999 is divisible by 7 or if 994 is the actual multiple.

Question 8. Given \(a = 2\), \(d = 8\), \(S_n = 90\), find n and \(a_n\).
Answer: Using the sum formula \(S_n = \frac{n}{2}[2a + (n - 1)d]\):
\(90 = \frac{n}{2}[2(2) + (n - 1)8]\)
\(180 = n[4 + 8n - 8]\)
\(180 = n[8n - 4]\)
\(180 = 8n^2 - 4n\)
\(4n^2 - 2n - 90 = 0\)
\(2n^2 - n - 45 = 0\)
Factorizing the quadratic equation:
\(2n^2 - 10n + 9n - 45 = 0\)
\(2n(n - 5) + 9(n - 5) = 0\)
\((2n + 9)(n - 5) = 0\)
Since \(n\) must be a positive integer, \(n = 5\).
Now find \(a_n\):
\(a_5 = a + 4d = 2 + 4(8) = 2 + 32 = 34\)
In simple words: Form a quadratic equation using the sum formula, solve for positive \(n\), and then find the \(n^{\text{th}}\) term.

Exam Tip: Reject negative or fractional values of \(n\) as the number of terms must always be a positive integer.

[LEVEL - 3]

Question 1. Which term of the sequence -1, 3, 7, 11 .......... Is 95?
Answer: Here \(a = -1\), \(d = 3 - (-1) = 4\), and \(a_n = 95\)
\(a_n = a + (n - 1)d\)
\(95 = -1 + (n - 1)4\)
\(95 + 1 = 4(n - 1)\)
\(96 = 4(n - 1)\)
\(n - 1 = \frac{96}{4} = 24\)
\(n = 24 + 1 = 25^{\text{th}}\text{ term}\)
In simple words: Equate the general term formula to 95 and solve for \(n\).

Exam Tip: Watch out for negative signs when finding the common difference.

Question 2. How many terms are there in the sequence 3, 6, 9, 12, ...... 111?
Answer: Here \(a = 3\), \(d = 6 - 3 = 3\), and \(a_n = 111\)
\(111 = 3 + (n - 1)3\)
\(111 - 3 = 3(n - 1)\)
\(108 = 3(n - 1)\)
\(n - 1 = \frac{108}{3} = 36\)
\(n = 36 + 1 = 37\text{ terms}\)
In simple words: Substitute the last term into the \(n^{\text{th}}\) term formula to find the total number of terms.

Exam Tip: This is a straightforward application of the linear term formula.

Question 3. The first term of an AP is -7 and the common difference 5, find its \(18^{\text{th}}\) term and the general term.
Answer: Given \(a = -7\) and \(d = 5\)
General term (\(a_n\)):
\(a_n = a + (n - 1)d\)
\(a_n = -7 + (n - 1)5\)
\(a_n = -7 + 5n - 5 = 5n - 12\)
\(18^{\text{th}}\) term (\(a_{18}\)):
\(a_{18} = 5(18) - 12 = 90 - 12 = 78\)
In simple words: First find the general formula in terms of \(n\), then substitute \(n = 18\) to find the \(18^{\text{th}}\) term.

Exam Tip: Finding the general term first makes answering subsequent specific term questions very quick.

Question 4. How many numbers of two digits are divisible by 3?
Answer: Two-digit numbers start from 10 and end at 99.
The first two-digit number divisible by 3 is 12.
The last two-digit number divisible by 3 is 99.
This forms an AP: \(12, 15, 18, \dots, 99\)
Here \(a = 12\), \(d = 3\), and \(a_n = 99\)
\(99 = 12 + (n - 1)3\)
\(99 - 12 = 3(n - 1)\)
\(87 = 3(n - 1)\)
\(n - 1 = \frac{87}{3} = 29\)
\(n = 29 + 1 = 30\)
In simple words: Identify the smallest and largest 2-digit multiples of 3 and find the number of terms in that AP.

Exam Tip: 99 is divisible by 3 (\(99 = 3 \times 33\)), so it is included as the last term.

Question 5. If the \(n^{\text{th}}\) term of an AP is \((2n + 1)\), find the sum of first n terms of the AP
Answer: Given \(a_n = 2n + 1\)
First term \(a_1 = 2(1) + 1 = 3\)
Second term \(a_2 = 2(2) + 1 = 5\)
Last term \(l = a_n = 2n + 1\)
Sum of first \(n\) terms \(S_n = \frac{n}{2}(a + l)\)
\(S_n = \frac{n}{2}[3 + (2n + 1)]\)
\(S_n = \frac{n}{2}[2n + 4]\)
\(S_n = n(n + 2)\)
In simple words: Find the first term and use the sum formula with the given \(n^{\text{th}}\) term as the last term.

Exam Tip: Alternatively, you can use \(S_n = \sum a_n = \sum (2n + 1)\) using summation formulas.

Question 6. Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3.
Answer: The numbers between 250 and 1000 divisible by 3 start from 252 (\(252 = 3 \times 84\)) and end at 999 (\(999 = 3 \times 333\)).
AP is \(252, 255, 258, \dots, 999\)
Here \(a = 252\), \(d = 3\), and \(a_n = 999\)
\(999 = 252 + (n - 1)3\)
\(999 - 252 = 3(n - 1)\)
\(747 = 3(n - 1)\)
\(n - 1 = \frac{747}{3} = 249\)
\(n = 250\)
Now find sum \(S_{250}\):
\(S_{250} = \frac{250}{2}(252 + 999)\)
\(S_{250} = 125 \times 1251 = 156375\)
In simple words: Find the first and last multiples of 3 in the given range, count how many such numbers exist, and compute their sum.

Exam Tip: Check divisibility by summing digits (sum of digits of 252 is \(2+5+2=9\), which is divisible by 3).

[Problems for self evaluation]

Question 1. Show that the sequence defined by \(t_n = 4n + 7\) is an AP.
Answer: Given \(t_n = 4n + 7\)
Find \(t_{n-1}\):
\(t_{n-1} = 4(n - 1) + 7 = 4n - 4 + 7 = 4n + 3\)
Now find the difference \(t_n - t_{n-1}\):
\(t_n - t_{n-1} = (4n + 7) - (4n + 3) = 4n + 7 - 4n - 3 = 4\)
Since the difference between any term and its preceding term is a constant (4), the sequence is an AP.
In simple words: Show that \(t_n - t_{n-1}\) is independent of \(n\) and equals a constant value.

Exam Tip: Proving that the common difference is constant is the standard way to show a sequence is an AP.

Question 2. Find the number of terms for given AP : 7, 13, 19, 25, ......, 205.
Answer: Here \(a = 7\), \(d = 13 - 7 = 6\), and \(a_n = 205\)
\(a_n = a + (n - 1)d\)
\(205 = 7 + (n - 1)6\)
\(205 - 7 = 6(n - 1)\)
\(198 = 6(n - 1)\)
\(n - 1 = \frac{198}{6} = 33\)
\(n = 33 + 1 = 34\)
In simple words: Substitute the last term into the general term formula to find the total number of terms.

Question 3. The \(7^{\text{th}}\) term of an AP is 32 and it \(13^{\text{th}}\) term is 62. Find AP.
Answer: Given \(a_7 = 32 \implies a + 6d = 32\) -- (1)
Given \(a_{13} = 62 \implies a + 12d = 62\) -- (2)
Subtracting (1) from (2):
\((a + 12d) - (a + 6d) = 62 - 32\)
\(6d = 30 \implies d = 5\)
Substitute \(d = 5\) in (1):
\(a + 6(5) = 32\)
\(a + 30 = 32 \implies a = 2\)
Thus, the AP is \(2, 7, 12, 17, \dots\)
In simple words: Set up a pair of linear equations in \(a\) and \(d\), solve them simultaneously, and construct the AP.

Exam Tip: Elimination method is best for solving simultaneous linear equations involving AP terms.

Question 4. Find the sum of all two digit odd positive nos.
Answer: Two-digit odd numbers start from 11 and end at 99.
This forms an AP with \(a = 11\), \(d = 2\), and \(a_n = 99\)
First find \(n\):
\(99 = 11 + (n - 1)2\)
\(88 = 2(n - 1)\)
\(n - 1 = 44 \implies n = 45\)
Now find sum \(S_{45}\):
\(S_{45} = \frac{45}{2}(11 + 99)\)
\(S_{45} = \frac{45}{2}(110) = 45 \times 55 = 2475\)
In simple words: Find the first and last two-digit odd numbers, find how many such numbers there are, and apply the sum formula.

Exam Tip: Ensure you include both endpoints 11 and 99 since both are odd.

Question 5. Find the value of 'x' for AP. 1 + 6 + 11 + 16 + .... + X = 148.
Answer: Given AP series: \(1 + 6 + 11 + 16 + \dots + X = 148\)
Here \(a = 1\), \(d = 6 - 1 = 5\), and \(S_n = 148\)
Using \(S_n = \frac{n}{2}[2a + (n - 1)d]\):
\(148 = \frac{n}{2}[2(1) + (n - 1)5]\)
\(296 = n[2 + 5n - 5]\)
\(296 = n[5n - 3]\)
\(5n^2 - 3n - 296 = 0\)
Solving the quadratic equation using the quadratic formula:\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(n = \frac{3 \pm \sqrt{(-3)^2 - 4(5)(-296)}}{2(5)}\)
\(n = \frac{3 \pm \sqrt{9 + 5920}}{10} = \frac{3 \pm \sqrt{5929}}{10}\)
Since \(\sqrt{5929} = 77\):
\(n = \frac{3 + 77}{10} = \frac{80}{10} = 8\)
Now, \(X\) is the \(8^{\text{th}}\) term (\(a_8\)):
\(X = a_8 = a + 7d = 1 + 7(5) = 1 + 35 = 36\)
In simple words: Use the sum formula to find the number of terms \(n\), then use \(n\) to find the value of the last term \(X\).

Exam Tip: Memorizing squares of numbers up to 100 helps in simplifying large square roots quickly.

Question 6. Find the \(10^{\text{th}}\) term from the end of the AP 8, 10, 12, ... 126.
Answer: Here last term \(l = 126\), first term \(a = 8\), common difference \(d = 10 - 8 = 2\)
Formula for \(n^{\text{th}}\) term from the end = \(l - (n - 1)d\)
\(10^{\text{th}}\) term from the end = \(126 - (10 - 1)2\)
\(= 126 - (9)2\)
\(= 126 - 18 = 108\)
In simple words: Use the reverse AP formula directly with the given last term and common difference.

Question 7. The sum of three numbers of AP is 3 and their product is -35. Find the numbers.
Answer: Let the three numbers in AP be \((a - d), a, (a + d)\).
Given sum = 3:
\((a - d) + a + (a + d) = 3\)
\(3a = 3 \implies a = 1\)
Given product = -35:
\((a - d) \cdot a \cdot (a + d) = -35\)
\(a(a^2 - d^2) = -35\)
Substitute \(a = 1\):
\(1(1^2 - d^2) = -35\)
\(1 - d^2 = -35\)
\(d^2 = 36 \implies d = \pm 6\)
Case 1: When \(a = 1\) and \(d = 6\)
Numbers are: \(1 - 6 = -5\), \(1\), \(1 + 6 = 7\) (i.e., \(-5, 1, 7\))
Case 2: When \(a = 1\) and \(d = -6\)
Numbers are: \(1 - (-6) = 7\), \(1\), \(1 + (-6) = -5\) (i.e., \(7, 1, -5\))
Thus, the three numbers are \(-5, 1, 7\).
In simple words: Assume three terms in AP as \((a-d, a, a+d)\) so that their sum eliminates \(d\) instantly.

Exam Tip: Always choose symmetric terms like \((a-d, a, a+d)\) for three numbers and \((a-3d, a-d, a+d, a+3d)\) for four numbers.

Question 8. A man repays a loan of Rs 3250 by paying Rs 20 in the first month and then increase the payment by Rs 15 every month. How long will it take him to clear the loan ?
Answer: Total loan amount \(S_n = \text{Rs } 3250\)
First month payment \(a = \text{Rs } 20\)
Monthly increase \(d = \text{Rs } 15\)
Using the sum formula \(S_n = \frac{n}{2}[2a + (n - 1)d]\):
\(3250 = \frac{n}{2}[2(20) + (n - 1)15]\)
\(6500 = n[40 + 15n - 15]\)
\(6500 = n[15n + 25]\)
\(6500 = 15n^2 + 25n\)
Dividing by 5:
\(3n^2 + 5n - 1300 = 0\)
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(n = \frac{-5 \pm \sqrt{5^2 - 4(3)(-1300)}}{2(3)}\)
\(n = \frac{-5 \pm \sqrt{25 + 15600}}{6} = \frac{-5 \pm \sqrt{15625}}{6}\)
Since \(\sqrt{15625} = 125\):
\(n = \frac{-5 + 125}{6} = \frac{120}{6} = 20\)
(Ignoring the negative value \(n = \frac{-130}{6}\))
Thus, it will take 20 months to clear the loan.
In simple words: Model the monthly payments as an arithmetic progression and solve the quadratic equation for the total number of months \(n\).

Exam Tip: Simplify coefficients by dividing through by their common factor before applying the quadratic formula.

Question 9. The ratio of the sums of m and n terms of an AP is \(m^2 : n^2\). Show that the ratio of the \(m^{\text{th}}\) and \(n^{\text{th}}\) terms is \((2m - 1) : (2n - 1)\).
Answer: Let the first term be \(a\) and common difference be \(d\).
Given ratio of sums:\(\frac{S_m}{S_n} = \frac{m^2}{n^2}\)
\(\frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2}\)
Multiply both sides by \(\frac{n}{m}\):
\(\frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}\) -- (1)
We need the ratio of \(m^{\text{th}}\) and \(n^{\text{th}}\) terms:\(\frac{a_m}{a_n} = \frac{a + (m - 1)d}{a + (n - 1)d}\)
To match this with equation (1), rewrite the numerator and denominator by replacing \(m\) with \(2m - 1\):
Let \(m - 1 = \frac{2m - 1 - 1}{2}\) or substitute directly by taking \(2a + (2m - 1 - 1)d\).
In equation (1), replace \(m\) by \(2m - 1\) and \(n\) by \(2n - 1\):
\(\frac{2a + (2m - 1 - 1)d}{2a + (2n - 1 - 1)d} = \frac{2m - 1}{2n - 1}\)
\(\frac{2[a + (m - 1)d]}{2[a + (n - 1)d]} = \frac{2m - 1}{2n - 1}\)
\(\frac{a_m}{a_n} = \frac{2m - 1}{2n - 1}\)
Hence proved.
In simple words: Express the ratio of sums algebraically, simplify, and substitute suitable index values to establish the ratio of the terms.

Exam Tip: This is a standard textbook proof; memorize the substitution trick for replacing terms with \(2m-1\).

Question 10. In an AP, the sum of first n terms is \(\frac{3n^2}{2} + \frac{5n}{2}\). Find its \(25^{\text{th}}\) term.
Answer: Given \(S_n = \frac{3n^2}{2} + \frac{5n}{2}\)
We know that \(a_{25} = S_{25} - S_{24}\)
First, find \(S_{25}\):
\(S_{25} = \frac{3(25)^2}{2} + \frac{5(25)}{2} = \frac{3(625)}{2} + \frac{125}{2} = \frac{1875 + 125}{2} = \frac{2000}{2} = 1000\)
Next, find \(S_{24}\):
\(S_{24} = \frac{3(24)^2}{2} + \frac{5(24)}{2} = \frac{3(576)}{2} + \frac{120}{2} = \frac{1728 + 120}{2} = \frac{1848}{2} = 924\)
Now, calculate \(a_{25}\):
\(a_{25} = S_{25} - S_{24} = 1000 - 924 = 76\)
In simple words: Use the relation \(a_n = S_n - S_{n-1}\) by calculating the sum of 25 terms and subtracting the sum of 24 terms.

Exam Tip: This method avoids finding the general \(a_n\) formula first, saving computation time for specific terms.