Question

a be the first term and d, the
difference and Sn the sum of
,
the first n term of
an AP, prove that
Sn=n/2[2a+(n-1)d] =n/2(a+l), where I is the n term
​

Answer 1

Step-by-step explanation:

The nth term of an arithmetic progression can be defined as:

Tn=a+(n−1)d

where a is the first term of the A.P and d is the constant difference between one term and another.

Proof 1

If we sum up the first n terms starting from a , we get:

Sn=a+(a+d)+(a+2d)+…+(a+(n−3)d)+(a+(n−2)d+(a+(n−1)d)(1)

Since addition is commutative, this sum is the same as:

Sn=(a+(n−1)d)+(a+(n−2)d)+(a+(n−3)d)+…+(a+2d)+(a+d)+a(2)

As the sum is finite, we can add the two series by pairing the terms up like this - the first term from (1) to the first term from (2) , the second term from (1) to the second term from (2) and so on:

2Sn=[a+(a+(n−1)d)]+[(a+d)+(a+(n−2)d)]+[(a+2d)+(a+(n−3)d)]+…+[(a+(n−1)d)+a]

Notice that each term enclosed around the brackets above adds up to 2a+(n−1)d , and since there is a one-to-one correspondence while paring the terms up, we have n number of 2a+(n−1)d . Hence,

2SnSn=n(2a+(n−1)d)=n2(2a+(n−1)d)