Explain the terms in the formula.
`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 has n^(th) term as t_n=a+(n-1)d Sum of n terms of an A.P. whose first term is a, last term is l 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 sum of n terms S_n for a sequence is given by S_n=An^2+Bn+C , then sequence is an A.P. whose common difference is

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 n^(th) term as t_n=a+(n-1)d Sum of n terms of an A.P. whose first term is a, last term is l 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_1,a_2,a_3,....a_8 is an A.P. with common difference d. Then a_(8-2k)-a_k(1 leklt4 k in N) is equal to

Second term of the AP if the S_(n) = n^(2) +2n is ………

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

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)

The absolute value of the sum of first 20 terms of series, if S_(n)=(n+1)/(2) and (T_(n-1))/(T_(n))=(1)/(n^(2))-1 , where n is odd, given S_(n) and T_(n) denotes sum of first n terms and n^(th) terms of the series