1.

The sum of first n terms of an AP is oppation (1). S=n/2{2a+(n+1)d}. (2). S=a+(n-1)d. (3). S=n{2a+(n-1)d} (4) . S=n/2{2a+(n-1)d}​

Answer»

Answer:

Option 4:

S = \frac{<klux>N</klux>}{2}[2a + (n - 1)d]

Step-by-step EXPLANATION:

Let's assume that, the ARITHMETIC Progression is:

a, (a + d), (a + 2d), (a + 3d),  . . . . . . . . .

The nth term = a + (n - 1)d . . . . . . . . . . . . . (i)

If the sum of first n terms = S, then,

S =       a   +     (a+d) +      (a+2d) + . . . . . . . . . + [a+(n-3)d]+[a+(n-2)d]+[a+(n-1)d]

S =[a+(n-1)d]+[a+(n-2)d]+[a+(n-3)d] + . . . . . . . . . + (a+2d)   +   (a+d)     +    a

Adding up the above two equations:

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

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

∴ S = \frac{1}{2}[2an + n(n - 1)d]  = \frac{n}{2}[2a + (n - 1)d]

Option 4 is the correct answer.


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