[a+(a+d)+(a+2d)+...,+[a+(n-1)d]],[,=(n)/(2){2a+(n-1)d}]

Statement-1: The sum of n terms of the series a+(a+d)+(a+2d)+...(a+(n-1)d)=(n)/(2)[2n+(n-1)d] Statement-2:- Mathematical induction is valid only for natural numbers.

For all ninNN , prove by principle of mathematical induction that, a+(a+d)+(a+2d)+ . . . to n terms =(n)/(2)[2a+(n-1)d] .

In the formulae S_(n)=(n)/(2){2a+(n-1)d} make d as the subject The following steps are involved in solving the above problem. Arrange them in sequential order. (A) (n-1)d=(2S_(n))/(n)-2a (B) Given, S_(n)=(n)/(2)[2a+(n-1)d]rArrn[2a+(n-1)d] = 2S_(n) (C) rArr d=(2)/(n-1)[(S_(n))/(n)-a] (D) 2a+(n-1)d=(2S_(n))/(n)

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)

The sum of first n terms of an A.P. S_n = …..A) n/2 [ t_1 + t_n] B) n/2 [a+(n-1)d] C) n/2 [2 + (n-1)d] D)none of these

(a) In the formula S_(n)=(n)/(2){2a+(n-1)d} , make d as the subject. (b) Find the value of d, when S_(n)=240,n=10, and a=6 .

Prove that a+(a+d)+(a+2d)+........n "terms"=n/2(2a+(n-1)d)

Show that Lt_(ntooo)[1/(a(a+d))+1/((a+d)(a+2d))+...+1/([a+(n-1)d][a+nd])]=1/(a.d) where a and d are constants. Hence find Lt_(ntooo)[1/1.4+1/4.7+...+1/((3n-1)(3n+2))]