Show that sum `S_n` of n terms of an AP with first term a and common difference d is `S_n=n/2(2a+(n-1)d)`

Theorem: The sum of nth terms of an AP with first term a and common difference d is S_(n)=(n)/(2)(2a+(n-1)d)

A sequence is called an A.P if the difference of a term and the previous term is always same i.e if a_(n+1)- a_(n)= constant ( common difference ) for all n in N For an A.P whose first term is 'a ' and common difference is d is S_(n) = n/2 (2a +(n-1)d)=n/2 (a+a+(n-1)d)= n/2 (a+l) A sequence whose n^(th) term is given by t_(n) = An + N , where A,B are constants , is an A.P with common difference

Let S_n denote the sum of n terms of an AP whose first term is a. If common difference d is given by d=Sn-kS_(n-1)+S_(n-2) , then k is :

Let S_n denote the sum of n terms of an AP whose first term is a. If common difference d is given by d=Sn-kS_(n-1)+S_(n-2) , then k is :

Let S_n denote the sum of n terms of an AP whose first term is a. If common difference d is given by d=Sn-kS_(n-1)+S_(n-2) , then k is :

Let S_(n) denote the sum of n terms of an AP whose first term is a.If common difference d is given by d=Sn-kS_(n-1)+S_(n-2), then k is :

If S_(n) denote the sum to n terms of an A.P. whose first term is a and common differnece is d , then S_(n) - 2S_(n-1) + S_(n-2) is equal to

S_(n) denots the sum of first n terms of an A.P. Its first term is a and common difference is d. If d= S_(n)-k S_(n-1) + S_(n-2) then k= ……….

A sequence is called an A.P. if the difference of a term and the previous term is always same i.e. if a_(n+1)-a_(n) = constant (common difference) for all n in N . For an A.P. whose first term is 'a' and common difference is 'd' has its n^("th") term as t_(n)=a+(n-1)d Sum of n terms of an A.P. whose first is a, last term is I and common difference is d is S_(n)=(n)/(2)(2a+(n-1)d) =(n)/(2)(a+a+(n-1)d)=(n)/(2)(a+l) . S_(r) denotes the sum of first r terms of a G.P., then S_(n),S_(2n)-S_(n),S_(3n)-S_(2n) are in