Question

If an AP consists of n terms and sum of first three terms is x and sum of last three terms is y then show that sum of all terms is equal to (n/6)(x+y)?

Answer 1

See explanation.

Explanation:

Let `a_1, a_2, ..., a_n` be the terms of the arithmetic progression.
Let `x` be the sum of the first 3 terms: `x=a_1+a_2+a_3`.
Let `y` be the sum of the last 3 terms: `y=a_(n-2)+a_(n-1)+a_n`.

The sum of all `n` terms in the arithmetic progression is `n` times the average of the first and last terms:

`S_n=n/2(a_1+a_n)`

Since the terms in the sequence have a constant difference (which we'll call `d`), we can rewrite this sum to use the 2nd term and 2nd last term:

`S_n=n/2(a_1color(orange)(+ d)+a_n color(orange)(-d))`

`color(white)(S_n)=n/2(a_2+a_(n-1))`

And we can continue this to include the 3rd terms from the start and end, too:

`color(white)(S_n)=n/2(a_3+a_(n-2))`

Since all three of these are equal to `S_n`, we can sum them all together to get

`3S_n = n/2(a_1+a_n)+n/2(a_2+a_(n-1))+n/2(a_3+a_(n-2))`

`color(white)(3S_n) = n/2[(a_1+a_2+a_3)+(a_(n-2)+a_(n-1)+a_n)]`

`color(white)(3S_n) = n/2(x+y)`

Finally, we divide both sides by `3` to arrive at

`S_n=n/6(x+y)`.