Looking for accurate Class 10 Maths Chapter 5 AP Exercise 5.2 Solutions? This step-by-step guide covers all NCERT questions from Arithmetic Progressions to help you master the concepts and score high in exams.
Class 10 Maths Chapter 5 AP Exercise 5.2 Solutions-textbook
Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and the nth term of the AP:
Solution:
(i) Find an when a = 7, d = 3, n = 8
- First term (a) = 7
- Common difference (d) = 3
- Number of terms (n) = 8
- Use the formula for the nth term of an AP: an = a + (n – 1)d.
- an = 7 + (8 – 1) x 3
- an = 7 + (7) x 3
- an = 7 + 21
- an = 28
(ii) Find d when a = -18, n = 10, an = 0
- First term (a) = -18
- Number of terms (n) = 10
- nth term (an) = 0
- Use the formula for the nth term of an AP: an = a + (n – 1)d.
- 0 = -18 + (10 – 1)d
- Step 4: Simplify and solve for d.
- 0 = -18 + 9d
- 18 = 9d
- d = 18 / 9
- d = 2
(iii) Find a when d = -3, n = 18, an = -5
- Common difference (d) = -3
- Number of terms (n) = 18
- nth term (an) = -5
- Use the formula for the nth term of an AP: an = a + (n – 1)d.
- -5 = a + (18 – 1) x (-3)
- -5 = a + (17) x (-3)
- -5 = a – 51
- a = -5 + 51
- a = 46
(iv) Find n when a = -18.9, d = 2.5, an = 3.6
- First term (a) = -18.9
- Common difference (d) = 2.5
- nth term (an) = 3.6
- Use the formula for the nth term of an AP: an = a + (n – 1)d.
- 3.6 = -18.9 + (n – 1) x 2.5
- 3.6 + 18.9 = (n – 1) x 2.5
- 22.5 = (n – 1) x 2.5
- (n – 1) = 22.5 / 2.5
- (n – 1) = 9
- n = 9 + 1
- n = 10
(v) Find an when a = 3.5, d = 0, n = 105
- First term (a) = 3.5
- Common difference (d) = 0
- Number of terms (n) = 105
- Use the formula for the nth term of an AP: an = a + (n – 1)d.
- an = 3.5 + (105 – 1) x 0
- an = 3.5 + (104) x 0
- an = 3.5 + 0
- an = 3.5
- Answer: an = 3.5
Ex 5.2 Class 10 Maths Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …, is
(a) 97
(b) 77
(c) -77
(d) -87
Answer: (i)
First term (a) = 10
Common difference (d) = 7 – 10 = -3
Step 2: Use the formula for Tₙ
Tₙ = a + (n – 1) × d
T₃₀ = 10 + (30 – 1) × (-3)
T₃₀ = 10 + 29 × (-3)
T₃₀ = 10 – 87 = -77
(ii) 11th term of the AP: -3, −12 , 2, …, is
(a) 28
(b) 22
(c) -38
(d) -48
Solution:
First term (a) = -3
Second term = -1/2
Common difference (d) = (-1/2) – (-3) = -1/2 + 3 = 5/2
Tₙ = a + (n – 1) × d
We need to find the 11th term, so:
T₁₁ = -3 + (11 – 1) × (5/2)
T₁₁ = -3 + 10 × (5/2)
T₁₁ = -3 + 50/2
T₁₁ = -3 + 25 = 22
Ex 5.2 Class 10 Maths Question 3.
In the following APs, find the missing terms in the boxes:
Which term of the AP: 3, 8, 13, 18, …, is 78?
(i) 2, ___, 26
Solution:
- First term (T₁) = 2
- Third term (T₃) = 26
- Determine the common difference (d).
- The difference between T₃ and T₁ is 2 times the common difference.
- T₃ = T₁ + (3-1)d => 26 = 2 + 2d
- 26 – 2 = 2d
- 24 = 2d
- d = 24 ÷ 2 = 12
- Step 3: Calculate the missing term (Second term, T₂).
- T₂ = T₁ + d
- T₂ = 2 + 12
- T₂ = 14
- Answer: 2, 14, 26
(ii) ___, 13, ___, 3
- Second term (T₂) = 13
- Fourth term (T₄) = 3
- Determine the common difference (d).
- The difference between T₄ and T₂ is 2 times the common difference.
- T₄ = T₂ + (4-2)d => 3 = 13 + 2d
- 3 – 13 = 2d
- -10 = 2d
- d = -10 ÷ 2 = -5
- Step 3: Calculate the first term (T₁).
- T₁ = T₂ – d
- T₁ = 13 – (-5)
- T₁ = 13 + 5 = 18
- Step 4: Calculate the third term (T₃).
- T₃ = T₂ + d
- T₃ = 13 + (-5)
- T₃ = 13 – 5 = 8
- Answer: 18, 13, 8, 3
(iii) 5, ___, ___, 9 1/2
- Step 1: Identify given terms and their positions.
- First term (T₁) = 5
- Fourth term (T₄) = 9 1/2 = 19/2
- Step 2: Determine the common difference (d).
- The difference between T₄ and T₁ is 3 times the common difference.
- T₄ = T₁ + (4-1)d => 19/2 = 5 + 3d
- 19/2 – 5 = 3d
- 19/2 – 10/2 = 3d
- 9/2 = 3d
- d = (9/2) ÷ 3 = 9/(2*3) = 9/6 = 3/2
- Step 3: Calculate the second term (T₂).
- T₂ = T₁ + d
- T₂ = 5 + 3/2
- T₂ = 10/2 + 3/2 = 13/2
- Step 4: Calculate the third term (T₃).
- T₃ = T₂ + d
- T₃ = 13/2 + 3/2
- T₃ = 16/2 = 8
- Answer: 5, 13/2, 8, 9 1/2
(iv) -4, ___, ___, ___, ___, 6
- Step 1: Identify given terms and their positions.
- First term (T₁) = -4
- Sixth term (T₆) = 6
- Step 2: Determine the common difference (d).
- The difference between T₆ and T₁ is 5 times the common difference.
- T₆ = T₁ + (6-1)d => 6 = -4 + 5d
- 6 + 4 = 5d
- 10 = 5d
- d = 10 ÷ 5 = 2
- Step 3: Calculate the missing terms sequentially.
- Second term (T₂) = T₁ + d = -4 + 2 = -2
- Third term (T₃) = T₂ + d = -2 + 2 = 0
- Fourth term (T₄) = T₃ + d = 0 + 2 = 2
- Fifth term (T₅) = T₄ + d = 2 + 2 = 4
- Answer: -4, -2, 0, 2, 4, 6
(v) ___, 38, ___, ___, ___, -22
Answer: 53, 38, 23, 8, -7, -2
Step 1: Identify given terms and their positions.
Second term (T₂) = 38
Sixth term (T₆) = -22
Step 2: Determine the common difference (d).
The difference between T₆ and T₂ is 4 times the common difference.
T₆ = T₂ + (6-2)d => -22 = 38 + 4d
-22 – 38 = 4d
-60 = 4d
d = -60 ÷ 4 = -15
Step 3: Calculate the first term (T₁).
T₁ = T₂ – d
T₁ = 38 – (-15)
T₁ = 38 + 15 = 53
Step 4: Calculate the remaining missing terms sequentially.
Third term (T₃) = T₂ + d = 38 + (-15) = 23
Fourth term (T₄) = T₃ + d = 23 + (-15) = 8
Fifth term (T₅) = T₄ + d = 8 + (-15) = -7
Ex 5.2 Class 10 Maths Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19, …, 205
(ii) 18, 1512, 13, …, -47
Solution:
(i) AP: 7, 13, 19, …, 205
Step 1: Identify the first term, common difference, and last term
First term (a) = 7
Common difference (d) = 13 – 7 = 6
Last term (Tₙ) = 205
Use the formula:
Tₙ = a + (n – 1) × d
Substitute the values:
205 = 7 + (n – 1) × 6
205 – 7 = (n – 1) × 6
198 = (n – 1) × 6
n – 1 = 198 ÷ 6 = 33
n = 34
(ii) AP: 18, 15, 12, …, -47
Step 1: Identify a, d, and Tₙ
First term (a) = 18
Common difference (d) = 15 – 18 = -3
Last term (Tₙ) = -47
Use the formula:
Tₙ = a + (n – 1) × d
-47 = 18 + (n – 1) × (-3)
-47 = 18 – 3(n – 1)
-47 = 18 – 3n + 3
-47 = 21 – 3n
-47 – 21 = -3n
-68 = -3n
n = 68 ÷ 3 = 22.67
Since n is not a whole number, -47 is not a term of this AP.
Answer for (ii): -47 is not a term of the AP, so number of terms cannot be found.
Ex 5.2 Class 10 Maths Question 6.
Check, whether -150 is a term of the AP: 11, 8, 5, 2, ….
Solution:
Step 1: Identify the first term and common difference
First term (a) = 11
Common difference (d) = 8 – 11 = -3
Step 2: Use the formula for the nth term of an AP
Tₙ = a + (n – 1) × d
We want to check if there exists an n such that Tₙ = -150
So,
-150 = 11 + (n – 1) × (-3)
-150 = 11 – 3(n – 1)
-150 = 11 – 3n + 3
-150 = 14 – 3n
-150 – 14 = -3n
-164 = -3n
n = 164 ÷ 3
n = 54.67
Since n is not a whole number, -150 is not a term of the AP.
Ex 5.2 Class 10 Maths Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
We are given:
- 11th term (T₁₁) = 38
- 16th term (T₁₆) = 73
We are to find the 31st term (T₃₁).
Step 1: Use the formula for the nth term of an AP:
Tₙ = a + (n – 1) × d
From T₁₁ = 38:
a + 10d = 38 …(1)
From T₁₆ = 73:
a + 15d = 73 …(2)
Step 2: Subtract equation (1) from equation (2):
(a + 15d) – (a + 10d) = 73 – 38
a + 15d – a – 10d = 35
5d = 35
d = 7
Step 3: Substitute d = 7 into equation (1):
a + 10 × 7 = 38
a + 70 = 38
a = 38 – 70 = -32
Step 4: Find the 31st term:
T₃₁ = a + (31 – 1) × d
T₃₁ = -32 + 30 × 7
T₃₁ = -32 + 210 = 178
Ex 5.2 Class 10 Maths Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
We are given:
- Total number of terms = 50
- 3rd term (T₃) = 12
- 50th term (last term, T₅₀) = 106
We have to find the 29th term (T₂₉).
Step 1: Use the formula for the nth term of an AP:
Tₙ = a + (n – 1) × d
From T₃ = 12:
a + 2d = 12 …(1)
From T₅₀ = 106:
a + 49d = 106 …(2)
Step 2: Subtract equation (1) from equation (2):
(a + 49d) – (a + 2d) = 106 – 12
a + 49d – a – 2d = 94
47d = 94
d = 2
Step 3: Substitute d = 2 in equation (1):
a + 2 × 2 = 12
a + 4 = 12
a = 8
Step 4: Find the 29th term:
T₂₉ = a + (29 – 1) × d
T₂₉ = 8 + 28 × 2
T₂₉ = 8 + 56 = 64
Ex 5.2 Class 10 Maths Question 9.
If the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
We are given:
- 3rd term (T₃) = 4
- 9th term (T₉) = -8
We are to find which term of this AP is zero, i.e., find n such that Tₙ = 0.
Step 1: Use the general formula of AP
Tₙ = a + (n – 1) × d
From T₃ = 4:
a + 2d = 4 …(1)
From T₉ = -8:
a + 8d = -8 …(2)
Step 2: Subtract equation (1) from equation (2)
(a + 8d) – (a + 2d) = -8 – 4
a + 8d – a – 2d = -12
6d = -12
d = -2
Step 3: Substitute d = -2 into equation (1)
a + 2(-2) = 4
a – 4 = 4
a = 8
Step 4: Find n such that Tₙ = 0
Tₙ = a + (n – 1)d
0 = 8 + (n – 1)(-2)
0 = 8 – 2(n – 1)
-8 = -2(n – 1)
Divide both sides by -2:
4 = n – 1
n = 5
Ex 5.2 Class 10 Maths Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let the first term be a and the common difference be d.
The 17th term is:
T₁₇ = a + 16d
The 10th term is:
T₁₀ = a + 9d
Given:
T₁₇ – T₁₀ = 7
(a + 16d) – (a + 9d) = 7
a + 16d – a – 9d = 7
7d = 7
d = 1
Ex 5.2 Class 10 Maths Question 11.
Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:
First term (a) = 3
Common difference (d) = 15 – 3 = 12
Step 2: Find the 54th term
T₅₄ = a + (54 – 1) × d
T₅₄ = 3 + 53 × 12
T₅₄ = 3 + 636 = 639
Step 3: Let the required term be the nᵗʰ term
We are told this term is 132 more than the 54th term.
So,
Tₙ = T₅₄ + 132
Tₙ = 639 + 132 = 771
Now use the formula for Tₙ:
Tₙ = a + (n – 1) × d
771 = 3 + (n – 1) × 12
771 – 3 = (n – 1) × 12
768 = (n – 1) × 12
n – 1 = 64
n = 65
Ex 5.2 Class 10 Maths Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the first term of the first AP be a₁ and the first term of the second AP be a₂.
Let the common difference of both APs be d.
The 100th term of the first AP is:
T₁₀₀ = a₁ + 99d
The 100th term of the second AP is:
T₁₀₀ = a₂ + 99d
Given:
(a₁ + 99d) – (a₂ + 99d) = 100
Simplifying:
a₁ – a₂ = 100
Now find the difference between the 1000th terms:
The 1000th term of the first AP is:
T₁₀₀₀ = a₁ + 999d
The 1000th term of the second AP is:
T₁₀₀₀ = a₂ + 999d
Now,
(a₁ + 999d) – (a₂ + 999d) = a₁ – a₂ = 100
Ex 5.2 Class 10 Maths Question 13.
How many three-digit numbers are divisible by 7?
Solution:
Step 1: Find the smallest three-digit number divisible by 7
The smallest three-digit number is 100.
Divide 100 by 7:
100 ÷ 7 = 14.28 → Round up to next whole number → 15
15 × 7 = 105 (smallest 3-digit number divisible by 7)
Step 2: Find the largest three-digit number divisible by 7
The largest three-digit number is 999.
999 ÷ 7 = 142.71 → Round down to 142
142 × 7 = 994 (largest 3-digit number divisible by 7)
Step 3: Count the number of terms
We now have an arithmetic progression:
105, 112, 119, …, 994
This is an AP with:
First term (a) = 105
Common difference (d) = 7
Last term (l) = 994
Use the formula for nth term:
Tₙ = a + (n – 1) × d
994 = 105 + (n – 1) × 7
889 = (n – 1) × 7
n – 1 = 127
n = 128
Ex 5.2 Class 10 Maths Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
We are asked to find how many multiples of 4 lie between 10 and 250.
Step 1: Find the first multiple of 4 greater than 10
The first multiple of 4 after 10 is 12.
Step 2: Find the last multiple of 4 less than 250
The last multiple of 4 before 250 is 248.
So, the multiples of 4 between 10 and 250 are:
12, 16, 20, …, 248
This is an arithmetic progression (AP) with:
First term (a) = 12
Common difference (d) = 4
Last term (l) = 248
Use the formula for the nth term:
Tₙ = a + (n – 1) × d
248 = 12 + (n – 1) × 4
236 = (n – 1) × 4
n – 1 = 59
n = 60
Final Answer: There are 60 multiples of 4 between 10 and 250.
Ex 5.2 Class 10 Maths Question 15.
For what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?
Solution:
Step 1: Find the first term and common difference of both APs
For the first AP:
First term, a₁ = 63
Common difference, d₁ = 65 – 63 = 2
So, nth term = Tₙ = 63 + (n – 1) × 2
For the second AP:
First term, a₂ = 3
Common difference, d₂ = 10 – 3 = 7
So, nth term = Tₙ = 3 + (n – 1) × 7
63 + (n – 1) × 2 = 3 + (n – 1) × 7
Now simplify:
Left side: 63 + 2(n – 1) = 63 + 2n – 2 = 61 + 2n
Right side: 3 + 7(n – 1) = 3 + 7n – 7 = -4 + 7n
So,
61 + 2n = -4 + 7n
⇒ 61 + 2n = -4 + 7n
⇒ 61 + 4 = 7n – 2n
⇒ 65 = 5n
⇒ n = 13
Ex 5.2 Class 10 Maths Question 16.
Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
Solution:
Let the first term be a and the common difference be d
Step 1: Use the formula for the nth term of an AP:
Tₙ = a + (n – 1) × d
3rd term:
T₃ = a + 2d = 16 (Equation 1)
7th term:
T₇ = a + 6d
5th term:
T₅ = a + 4d
Given: T₇ = T₅ + 12
⇒ (a + 6d) = (a + 4d) + 12
⇒ a + 6d = a + 4d + 12
⇒ 2d = 12
⇒ d = 6
Step 2: Put d = 6 in Equation 1
a + 2 × 6 = 16
a + 12 = 16
a = 16 – 12 = 4
So, the first term a = 4 and common difference d = 6
The AP is:
a, a + d, a + 2d, … = 4, 10, 16, …
Final Answer: The required AP is 4, 10, 16, 22, 28, …
Ex 5.2 Class 10 Maths Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
First term (a) = 3
Common difference (d) = 8 – 3 = 5
Last term (l) = 253
Use the formula for the nth term:
Tₙ = a + (n – 1) × d
253 = 3 + (n – 1) × 5
250 = (n – 1) × 5
n – 1 = 50
n = 51
So, the AP has 51 terms.
Step 3: 20th term from the last
The 20th term from the last is the (51 – 20 + 1) = 32nd term.
T₃₂ = a + (32 – 1) × d
T₃₂ = 3 + 31 × 5
T₃₂ = 3 + 155 = 158
Ex 5.2 Class 10 Maths Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
We are given:
- The sum of the 4th and 8th terms of an AP is 24
- The sum of the 6th and 10th terms is 44
We need to find the first three terms of the AP.
Let the first term be a and the common difference be d.
The 4th term is a + 3d
The 8th term is a + 7d
So,
(a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 [Equation 1]
The 6th term is a + 5d
The 10th term is a + 9d
So,
(a + 5d) + (a + 9d) = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 [Equation 2]
Now subtract Equation 1 from Equation 2:
(a + 7d) – (a + 5d) = 22 – 12
2d = 10
⇒ d = 5
Substitute d = 5 into Equation 1:
a + 5 × 5 = 12
a + 25 = 12
a = 12 – 25 = -13
Now, find the first three terms:
1st term = a = -13
2nd term = a + d = -13 + 5 = -8
3rd term = a + 2d = -13 + 10 = -3
Answer: The first three terms of the AP are -13, -8, -3.
Ex 5.2 Class 10 Maths Question 19.
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 ?
Solution:
This situation forms an Arithmetic Progression (AP), where:
- First term (a) = 5000
- Common difference (d) = 200
- nth term (Tₙ) = 7000
Using the formula for the nth term of an AP:
Tₙ = a + (n – 1) × d
Substitute the values:
7000 = 5000 + (n – 1) × 200
⇒ 7000 – 5000 = (n – 1) × 200
⇒ 2000 = (n – 1) × 200
⇒ n – 1 = 10
⇒ n = 11
So, in the 11th year of service, his salary became ₹7000.
He started in 1995, so the 11th year is:
1995 + 10 = 2005
Ex 5.2 Class 10 Maths Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving become ₹ 20.75, find n.
Solution:
We are given:
- First week’s saving = ₹5 → a = 5
- Weekly increase in saving = ₹1.75 → d = 1.75
- nth week’s saving = ₹20.75 → Tₙ = 20.75
We have to find the value of n.
Use the formula for the nth term of an AP:
Tₙ = a + (n – 1) × d
Substitute the values:
20.75 = 5 + (n – 1) × 1.75
20.75 – 5 = (n – 1) × 1.75
15.75 = (n – 1) × 1.75
n – 1 = 15.75 ÷ 1.75 = 9
n = 9 + 1 = 10
Final Answer: n = 10
So, Ramkali’s weekly saving becomes ₹20.75 in the 10th week.
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