Question

Prove that a+(a+d)+(a+2d)+........upto n times =n/2[2a+(n-1)d]

Answer 1

LHS= a + (a+d)+ (a+2d)...n
RHS=n/2[2a+(n-1)d]
Basic step :
For n=1 , LHS = a
RHS=1/2[2a]=a ....... (true)
For n=k , LHS = a+ (a+d) + (a+2d) + .... (a+kd)
RHS=k/2[2a+(k-1)d]=k/2[2a+kd-d]

To obtain the (k+1)th term , we add (k+1) to both sides of the equation.
LHS= a+(a+d)...(a+kd)+(a+kd+d)

RHS= k/2[2a+kd-d]+a+kd+d
=ak+k2d/2-dk/2+a+kd+d
=1/2[2ak+k2d-dk+2a+2dk+2d]
=1/2[2ak+dk2+dk+d+d+a]

Answer 2

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Let ,

first term = a

common Difference = d

Number of terms = n

Thus ,

By formula is Sn ,

$\tt{ Sn = \frac{n}{2} [ 2a + ( n - 1 ) d ] }$

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