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**Deriving the Formula for Sum of n Terms**

Consider a standard arithmetic sequence, where

a = first term

d = common difference, and

n = number of terms

S = sum of the terms

Consider the reverse of this AP. In the reversed AP, the last term, [a + (n-1)] d, appears in the first position, and the first term, a, in the last position. The sum of the terms of this AP also will be S.

Corresponding terms of of both the APs are added, that is, first term to first term, second term to second term and so on.

The sum of the corresponding terms is equal in every case.

The sum of the terms is [2a + (n-1)] d in every case.

= a + a +(n - 1)d

= 2a +(n - 1)d

= a + d + a +(n - 2)d

= 2a +(n - 1)d

= a + d + a +(n - 2)d

= 2a +(n - 1)d

= a + a +(n - 1)d

= 2a +(n - 1)d

â‡’ 2S = n x [2a + (n - 1)d]

â‡’ S = n/2 x [2a + (n - 1)d]

â‡’ S = n/2 x [a + a + (n - 1)d]

â‡’ S = n/2 [First term + Last term]

â‡’ S = n/2 [a + l]

**Deriving the Formula for Sum of n Terms**

Consider a standard arithmetic sequence, where

a = first term

d = common difference, and

n = number of terms

S = sum of the terms

Consider the reverse of this AP. In the reversed AP, the last term, [a + (n-1)] d, appears in the first position, and the first term, a, in the last position. The sum of the terms of this AP also will be S.

Corresponding terms of of both the APs are added, that is, first term to first term, second term to second term and so on.

The sum of the corresponding terms is equal in every case.

The sum of the terms is [2a + (n-1)] d in every case.

= a + a +(n - 1)d

= 2a +(n - 1)d

= a + d + a +(n - 2)d

= 2a +(n - 1)d

= a + d + a +(n - 2)d

= 2a +(n - 1)d

= a + a +(n - 1)d

= 2a +(n - 1)d

â‡’ 2S = n x [2a + (n - 1)d]

â‡’ S = n/2 x [2a + (n - 1)d]

â‡’ S = n/2 x [a + a + (n - 1)d]

â‡’ S = n/2 [First term + Last term]

â‡’ S = n/2 [a + l]