Question

The 8th term of an AP is 31 and it 15th term exceeds it's 11th term by 16.Find that value​

Answer 1

Answer:

3,7,11,15,19,23

Step-by-step explanation:

, (a+d), (a+2d), (a+3d), (a+4d), ……………………………. , {a+(n-1)d}

and nth term is given by = {a+(n-1)d}

so 8th term will be a+7d ,

11th term will be a+10d and

15th term will be a+14d

so as per qn.

a+7d= 31 ……………… (1)

& a+14d = a+10d +16

therefore 4d = 16

& d = 4 ……………….. (2)

use by using (2) in (1)

a + 7 *4 = 31

we get a = 31–28 = 3

so the required A.P will be 3, 7, 11, 15, 19, 23, ……………………