Question

If 8th term of AP is 31 and 15th term is 16 more than 11 th term find AP

Answer 1

In an AP,

8th term = a + 7d = 31

15th term = a + 14d = 16 + 11th term , that is , 16 + a + 10d

AP = ?

Now,

a + 14d = 16 + a + 10d

=> a - a + 14d - 10d = 16

=> 4d = 16

=> d = 16/4 = 4

So,

a + 7d = 31

=> a + 7×4 = 31

=> a = 31 - 28

=> a = 3

So,

AP : a, (a + d), (a + 2d), ......

= 3, (3 + 4), (3 + 2×4),.....

= 3, 7, 11, .......

Answer 2

Given

T 8 = 31

a + 7d = 31 ---(1)

T15 = T11 + 16.

T15 = a + 10d + 16

a + 14d = a + 10d + 16

4d = 16

d = 4

put value of d in eq.(1) , we get

, a = 31 - 7 (4)

a = 3

so now , the A.P will be a = 3

a +d = 3+ 4 = 7

a + 2d = 3 + 2(4) = 11

a + 3d = 3 + 3(4) = 15

hence ,

sequence of A.P :3 , 7, 11, 15..................