Question

if the 8th term of the AP s 31, 15th term is 16 morethan the 11th term,find the ap.

Answer 1

Heya !!


Given , 8th term = 31
=> a+7d = 31 __(1)

and 15th term = 16 + (11th term)
=> a + 14d = 16 + a + 10d
=> 14d – 10d = 16
=> 4d = 16
=> d = 16/4
=> d = 4

Substituting d = 4 in (1)
a + 7(4) = 31
=> a = 31 – 28
=> a = 3

So, A.P. will be a , a+d , a+2d , a+3d , ........

i.e., 3 , 3+4 , 3+2(4) , 3+3(4) , .........

=> 3 , 7 , 11 , 15 , ........

Answer 2

Hi there!

We know,
$a_n$ = a + (n-1) × d

ATQ,

$a_8$ = a + 7d = 31 ----(i)

$a_{11}$ = a + 10d

[ The 16th term being 16 more than the 11th term. ]

$a_{16}$ = a + 15d = 16 + a + 10d

a + 15d = 16 + a + 10d

14d - 10d = 16

4d = 16

d = 16/4

d = 4

Substitute d = 4 in Eqn. (i)

a + 7 × 4 = 31

a = 31 - 28

a = 3

Therefore,
A.P. is - 3, (3+4), (3+4+4) .... Or 3, 7, 11, 15 ...

Hope it helps! :D

Cheers,
Ishaan Singh