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)?

1 Answer
Aug 24, 2017

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)#.