Question

The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

Answer 1

In the given problem, we are given 6^th and 17^th term of an A.P.

We need to find the 40^th term

Here

`a_6 = 19`

`a_17 = 41`

Now we will find `a_6` and `a_17` using the formula `a_n = a + (n - 1)d`

So

`a_6 = a + (6 - 1)d`

19 = a + 5d   .......(1)

Also

`a_17 = a + (17 - 1)d`

41 = a + 16d .....(2)

So to solve for a and d

On subtracting (1) from (2) we get

`a + 16d - a - 5d = 41 - 19`

11d = 22

`d = 22/11`

d = 2    .....(3)

Substituting (3) in (1), we get

`19 = a + 5(2)`

19 - 10 = a

a = 9

Thus 

a = 9

d = 2

n = 40

Substituting the above values in the formula `a_n = a + (n -1)d`

`a_40 = 9 + (40 - 1`)2`

`a_40 = 9 + 80 - 2`

`a_40 = 87`

Therefore `a_40 = 87`