Question

Choose any values of a and d and write an A.P.
Find the sum of the first 100 terms (S100). (OEQ)​

Answer 1

Answer:

$S100 = \frac{n}{2} (2a+99d)$

Step-by-step explanation:

Let's take the following Arithmetic Progression:

a, a+d, a+2d, a+3d, . . . . . . . . .

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

100th term = a + (100 - 1)d = a + 99d [using equation (i)]

The sum of the first 100 terms:

S100 =      a      +    a+d    +   a+2d   + . . . . . . . . + (a+97d) + (a+98d) + (a+99d)

S100 = (a+99d) + (a+98d)+ (a+97d) + . . . . . . . . + (a + 2d) +  (a + d)   +    a

Adding the above two equations:

2S100 = (2a+99d)+(2a+99d)+(2a+99d)+ . . . . .+(2a+99d)+(2a+99d)+(2a+99d)

$S100 = \frac{n.(2a+99d)}{2} = \frac{n}{2} (2a+99d)$