Question

The first amd third term of an Arithmatic progression are x and y respectively. Find the sixteenth term of x and y


`tn = a + (n - 1)d`
​

Answer 1

The sixteenth term is
`x+(15(y-x))/(2)`.

Explanation:

Given,

`a_1=x`

`a_3=y`

And also given,
`T_n=a+(n-1)d`

We have to find out the 16th term of given A.P.

Firstly, we will find out the common difference(d).

Common difference(d) is calculated by given formula.

`T_3=a+(n-1)d`

On putting the values, we get;

`y=x+(3-1)d\\\\y=x+2d\\\\2d=y-x\\\\d=(y-x)/(2)`

Now the value of 'd' is calculated, so we can find out the 16th term by using the formula.

`T_(16)=x+(16-1)(y-x)/(2)\\\\T_(16)=x+15*(y-x)/(2)\\\\ T_(16)=x+(15(y-x))/(2)`

Hence The sixteenth term is
`x+(15(y-x))/(2)`.

Answer 2

The sixteenth term is (7.5y - 6.5x).

Explanation:

The first term of the A.P. is given to be x and the common difference is assumed to be d.

So, the third term = y = x + (3 - 1)d {Given that the third term is y}

Then, 2d = y - x

⇒
`d = (y - x)/(2)`

Therefore, the sixteenth term of the A.P. will be = x + (16 - 1)d

=
`x + 15 * (y - x)/(2)`

= x + 7.5y - 7.5x

= 7.5y - 6.5x (Answer)