GIVEN : the sum of 5th and 7th term of an A.P is 52.

a_{n} = a+(n-1)d

a_{5 }= a+(5-1)d

5th term = a+4d ..........1

similarly, a+6d.....................2

solving eqn. 1 and 2

(a+4d) + (a+6d) = 52

a+4d+a+6d =52

2a+10d = 52 ..........3

taking 2 as common in eqn. 3 we get,

2(a+5d) = 52

a+5d = 52/2

a+5d = 26.................4

now given , 10th term = 46

ie, 10th term = a+9d = 46...................5

solving eqn.4 and 5 we get,

a+9d = 46

a+5d = 26

on subtraction

a+9d = 46

-a-5d = -26

'a' gets cancel.

we get, 4d = 20

d = 20/4

d = 5.

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