Question

a+(a+d)+(a+2d)+....upto n term=n/2 [2a+(n-1)d]​

Answer 1

Answer:

Let n=1

Then

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] = a

On the other hand,

n/2[2a+(n-1)d] = 1/2 * 2a = a

for k=n we have that

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

k=n+1,

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] + [a+nd] = (n+1)/2 * [2a+nd]

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] + [a+nd]

= n/2[2a+(n-1)d] + [a+nd]

=na + n(n-1) d/2 + a + nd

=(n+1) a + (n^2/2 - n/2 + n) d

=(n+1) a + (n^2/2 + n/2) d

=(n+1) a + (n+1) n d /2

=(n+1)/2 [ 2 a + n d]

Hence proved.

Answer 2

Answer:

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