# Arithmetic Progressions -Class 10

**IMPORTANT QUESTIONS FOR CBSE CLASS 10 MATHEMATICSArithmetic Progressions – Chapter 5**

- If the common difference of an AP is 7, then find the difference between the fifteenth term and twelfth term?

Solution:

Given d = 7

n th term = a + (n-1)d

Fifteenth term = a + (15 – 1) d = a + 14d

Twelfth term = a + (12 -1) d = a + 11d

Difference = (a + 14d) – (a + 11d) = 3d = 3 x 7 = 21. - The sum of first 20 terms of the AP: 24, 20, 16, ——————-

Solution:

Here a = 24, d = 20 – 24 = -4, n = 20.

Sum of first 20 terms = n/2[2a + (n-1)d]

= 20/2[2 x 24 + (20 -1) (-4)]

= 10[48 + (-76)]

= 10 x (-28)

= -280. - Find the sum of first seven multiples of 7?

Solution:

Multiples of 7 are 7, 14, 21, 28, 35, 42, 49, ————

Here a = 7, d = 14 -7 = 7, n = 7.

Sum of seven terms = 7/2[2 x 7 + (7 -1)7]

= 7/2[14 + 42]

= 196. - Find the common difference of an AP in which the difference between the 20 th term and 5 th term is 60.

Solution:

We know that n th term = a + (n-1) d

20 th term = a + (20 -1) d = a + 19d

5 th term = a + (5 -1) d= a + 4d

Given difference = (a + 19d) – (a + 4d) = 60

15 d = 60

d = 60/15 = 4. - Find the 10 th term from end of the AP: 3, 8, 13…..…253?

Solution:

Here a = 253, d = 3 -8 = -5, n = 10.

10 th term = 253 + 9 x (-5) = 253 + (-45) = 208. - Find the common difference of the AP whose 11 th term is 38 and the 16 th term is 73?

Solution:

Given 11 th term = a + 10d = 38——————– (1)

16 th term = a + 15d = 73————————- (2)

Solving these two equations, we get

By subtraction, -5d = -35

d = (-35)/ (-5) = 7. - The n th term of an AP is (5 – 2n), then find the common difference?

Solution:

Given n th term = 5 – 2n

First term = 5 – 2 x 1 = 5 -2 = 3.

Second term = 5 – 2 x 2 = 5 – 4 = 1

Third term = 5 – 2 x 3 = 5 – 6 = -1

Common difference = 1 – 3 = -2. - Write the 19 th term of the AP: -60, -57, -54, ——————

Solution:

Here a = -60, d = (-57) – (-60) = 3

19 th term = a + (19 -1) d = (-60) + 18 x 3 = (-60) + 54 = -6. - How many terms of the AP: 24, 21, 18 …………… must be taken so that their sum is 78?

Solution:

Here a = 24, d = 21 – 24 = -3 and sum = 78.

Sum of n terms = n/2[2a + (n-1) d]

78 = n/2[2 x 24 + (n – 1) (-3)]

156 = n [48 -3n +3]

156 = n [51 – 3n]

156 = 3n [17 –n]

52 = n (17 – n), which is a quadratic equation.

By solving these, we get n = 4 or n = 13.

So the number of terms is either 4 or 13. - Which term of the AP: 21, 18, 15, ———– is -81?

Solution:

Here a = 21, d = 18 – 21 = -3 and n th term = -81 and we have to find n.

As n th term = a + (n -1) d, we have -81 = 21 + (n -1) (-3)

-81 = 21 – 3n + 3 = 24 – 3n

-105 = -3n

n = (-105) / (-3) = 35.

Therefore, the 35 th term of the given AP is -81.