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If the mth term of an AP is a and its nth term is b, show that the sum of its
`( m +n) " term is " ((m +n))/2 . {a +b+ (( a-b))/((m-n))}`

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Let the first of and the common difference of the given AP be A and d respectively. Then ,
` T_(m) =a Rightarrow A + (m-1) d=a`
` and T_(n) =b Rightarrow A + (n-1) d=b`
On subtracting (ii) from (i), we get,
` (m-n) d= ( a-b) Rightarrow d = ((a -b))/((m-n))`
Adding (i) and (ii) , we get
` 2A + ( m+n-2) = a+b`
` Rightarrow 2A + (m +n -1) d-d = a +b`
` Rightarrow 2A + (m+n -1) d= (a +b+d)= (a +b + (a-b)/(m-n) )`
` T_(m+n) = ((m+n))/2. [2A + ( m+n-1)d]`
` Rightarrow T_(m+n) = ((m+n))/2 . [a +b+ (a -b)/m-n)]`

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