Question

If the 8th term of an A.P. is 31 and the 15th term is 16 more than the 11 th term, find the A.P.

Answer 1

Given that,

➜ 8th term of an AP is 31

➜ 15th term is 16 more than the 11th term

To Find :-

➜ AP

Solution :-

Let the first term of the AP be a and the common Difference be d.

Given in the question,

⇒ 8th term = 31

⇒ a + (8 - 1)d = 31 [ ∵ aₙ = a + (n - 1)d ]

⇒ a + 7d = 31 ...(1)

Also,

⇒ 15th term = 11th term + 16

⇒ a + (15 - 1)d = a + (11 - 1)d + 16

⇒ a + 14d = a + 10d + 16

⇒ 14d - 10d = 16

⇒ 4d = 16

⇒ d = 4

Substituting d = 4 in (1), we get

⇒ a + 7×4 = 31

⇒ a = 31 - 28

⇒ a = 3

So, we have found the first term to be 3 and the common Difference to be 4.

The standard form of an AP is:

⇒ a , a + d , a + 2d, ... , a + (n - 1)d

⇒ 3 , 3 + 4 , 3 + 8 , ... , 3 + 4n - 4

⇒ 3 , 7 , 11 , ... , 4n - 1

Hence, The required AP is:

3 , 7 , 11 , ... , 4n - 1

Answer 2

Given ,

The 8th term of an A.P. is 31 and the 15th term is 16 more than the 11th term

We know that , the nth term of an AP is given by

$\boxed{ \sf{a_{n} = a + (n - 1)d }}$

Thus ,

a + (8 - 1)d = 31

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

And

a + (15 - 1)d = a + (11 - 1)d + 16

a + 14d = a + 10d + 16

4d = 16

d = 16/4

d = 4

Put the value of d = 4 in eq (i) , we get

a + 7(4) = 31

a + 28 = 31

a = 3

Therefore ,

The required AP is 3 , 7 , 11 , 15 .....