Find the sum of f

The progression for odd numbers looks like 1, 3, 5, 7, 9, ……..

Now, for this A.P

First term, a = 1

Common difference, d = 3 – 1 = 2

Number of terms = 100

We know that the sum of n terms is given by,

Therefore,

S₁₀₀ = 50[2 + 198]

S₁₀₀ = 50 × 200

S₁₀₀ = 10000

OR

Let a be the first term, of an A.P with a_n as the nth term and d as the common difference.

We know that,

a_n = a + (n – 1)d

Similarly,

a_n-1 = a + (n – 1 – 1)d = a + (n – 2)d

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

a_n – a_n-1 = a + nd – d – a – nd + 2d

a_n – a_n-1 = 2d – d

a_n – a_n-1 = d

Hence, Proved.