If the mth term of an A.P. is and nth term is , then show that its (mn)th term is 1.

OR

Find the sum of all odd numbers between 0 and 50.

Given:

Now, a_m = a + (m – 1)d

⇒ an + n(m – 1)d = 1

⇒ an + mnd – nd = 1 …(i)

⇒ am + mnd – md = 1 …(ii)

From eq. (i) and (ii), we get

an + mnd – nd = am + mnd – md

⇒ an – am – nd + md = 0

⇒ a(n – m) – d (n – m) = 0

⇒ a = d

Now, putting the value of a in eq. (i), we get

dn + mnd – nd = 1

⇒ mnd = 1

Hence,

Now, (mn)^th terms of AP is

a_mn = a + (mn – 1)d

⇒ a_mn = 1

Hence Proved

OR

To find: Sum of Odd numbers

Proof:

We know that Odd numbers are 1, 3, 5, …49

Now, we have to find the sum of these odd numbers .i.e.

1 + 3 + 5 + … + 49

Here, we can see that these are in AP

First term, a = 1

Last term, l = 49

& Common difference, d = a₂ – a₁ = 3 – 1 = 2

We know that,

Firstly, we find n

a_n = a + (n – 1)d

Here, a_n = l = 49

⇒ 49 = 1 + (n – 1)×2

⇒ 49 – 1 = 2(n – 1)

⇒ 48 = 2(n – 1)

⇒ n – 1 = 24

⇒ n = 24 + 1

⇒ n = 25

∴ n = 25

Now, we find the sum

⇒ S_n = 25 × 25

⇒ S_n = 625

Hence, the sum of all odd numbers from 0 to 50 is 625