This page 10th CBSE math solution for exercise 5.2 part 1 is going to provide you solution for every problems that you find in the exercise no 5.2 part 1

(1) Fill in the blanks in the following table, given that a is the first term,d the common difference and a n the nth term of the AP.

Solution:

a = 7 d = 3 n = 8

from the given details we have to find nth term

a n = a + (n - 1) d

a n = 7 + (8 - 1) (3)

a n = 7 + 7 (3)

= 7 + 21

= 28

The general term = 28

(ii) a = -18 n = 10 a n = 0

Solution:

we need the find the common difference

a n = 0

a + (n - 1) d = 0

-18 + (10 - 1) d = 0

-18 + 9 d = 0

9 d = 18

d = 18/9

d = 2

(iii) d = -3 n = 18 a n = -5

Solution:

a n = -5

a + (n - 1) d = -5

a + (18 - 1) (-3) = - 5

a + 17 (-3) = -5

a - 51 = -5

a = -5 + 51

a = 46

(iv) a = -18.9 d = 2.5 a n = 3.6

Solution:

**a n = 3.6**

**a + (n - 1) d = 3.6**

**-18.9 + (n - 1) (2.5) = 3.6**

**(n -1) 2.5 = 3.6 + 18.9**

**(n - 1) 2.5 = 22.5**

** n - 1 = 22.5/2.5**

** n - 1 = 9**

** n = 9 + 1 **

** n = 10**

(v) a = 3.5 d = 0 n = 105

Solution:

**a n = a + (n - 1) d **

**a n = 3.5 + (105 - 1) 0**

**a n = 3.5 + 104 (0)**

**a n = 3.5**

HTML Comment Box is loading comments...