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)

If S_n denote the sum of n terms of an A.P. with first term a and common difference d such that (S_x)/(S_(k x)) is independent of x , then d=a (b) d=2a (c) a=2d (d) d=-a

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 :

The sum of n terms of an A.P.is 3n^(2)-n. Find the first term and common difference of A.P.

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) . If a_(1),a_(2),............,a_(n) are in G.P. then log_(a_(1))a,log_(a_(2))a,log_(a_(3))a,...log_(a_(n)) a are in

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-3) - 3S_(n+1) - S_(n) is equal to