Question

4. 1+2+3+4 + ......... +99 = ?

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Answer 1

Answer:

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

= 99/2 [ 2×1 + (99-1) 1]

= 99/2(2 + 98)

= 99/2 ×(100)

= 99× 50

= 4950

Answer 2

Answer:

4950

Step-by-step explanation:

Method 1.

Sum of n natural numbers = n(n+1)/2

So, Here in this case ,

Sum = 99(99+1)/2

=> Sum= 99 × 50=4950 ..

..

Method 2.

Sum of n terms of an Arithmetic progression = n/2{ 2a + (n-1)d}

Here, a = 1 , d=1 , n= 99

So, Sum = 99/2{2×1 +(99-1)×1}

=> Sum = 4950