Question

the sum of all the two digit odd numbers is​

Answer 1

Solution :

The series of two digit odd numbers is

    11 + 13 + 15 + ... + 97 + 99

First term (a₁) = 11

Common difference (d) = 2

Let, the nth term (aₙ) = 99

  ⇒ a₁ + (n - 1) d = 99

  ⇒ 11 + (n - 1) * 2 = 99

  ⇒ 11 + 2n - 2 = 99

  ⇒ 2n + 9 = 99

  ⇒ 2n = 99 - 9 = 90

  ⇒ n = 45

Thus, there are 45 odd numbers in two digits

Hence, the required sum

  = $\frac{n}{2}$ (a₁ + aₙ)

  = $\frac{45}{2}$ (11 + 99)

  = $\frac{45 * 110}{2}$

  = 45 * 55

  = 2475

Answer 2

Answer: 2475

Step-by-step explanation:

Odd numbers having two digits are,

11 , 13 , 15 , 17  ......................97 , 99

We have  to find,

S = 11 + 13 + 15 + 17 ........... + 97 + 99

This  series is an AP, Having,

First term (a) = 11

Last term (l ) = 99

Common Difference(d) = 13 - 11 = 15 - 13 . .... ... = 2

If there are n  terms in the series, Then,

l = a + (n - 1)d

99 = 11 + (n - 1)2

99 - 11 = 2(n - 1)

88 = 2(n - 1)

88/2 = n - 1

44 = n - 1

n = 44 + 1

n = 45

Now,

We know that,

S = n/2{a + l}

S = 45/2( 11 + 99)

S = 45/2(110)

S = 45 × 55

S = 2475