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
See explanation.
Explanation:
Let
Let
Let
The sum of all
#S_n=n/2(a_1+a_n)#
Since the terms in the sequence have a constant difference (which we'll call
#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
#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
#S_n=n/6(x+y)# .