Question

In an A. P. the 10th term is 46, sum of the 5th and 7th term is 52.Find the A. P.

Answer 1

Let a and d be first term and common difference of ap
10th term=46
a+9d=46------------------------eq(1)
5th +7th term=52
a+4d+a+6d=52
2a+10d=52
a+5d=26--------------------------eq(2)
Eq(1)-eq(2)
4d=20
d=5
Put d=5 in eq(1)
a+45=46
a=1
Hence reqd ap=1,6,11,16,21,...............

Answer 2

Solution:-

Given : Sum of the 5th and 7th term is 52.

∴ a + 4d + a + 6d = 52

2a + 10d = 52 ...............(1)

Also given : 10th term is 46

∴ a + 9d = 46

Multiplying it by 2, we get

2a + 18d = 92 ................(2)

 Now, subtracting equation (1) from equation (2), we get.

  2a + 18d = 92

  2a + 10d = 52

-      -          -

_______________

         8d = 40 

_______________

8d = 40

d = 40/8

d = 5

 Substituting the value of d in equation (1),we get.

2a + 10d = 52

2a + 10*5 + 52

2a = 52 - 50

2a = 2

a = 2/2

a = 1

Therefore, the required AP is 1, 6, 11, 16, 21, 26, 31....