Question

Sum of all two digit odd numbers is

Answer 1

step-1: first of all find the no. of two digit odd positive numbers (n) to be added by using the formula: tn= a + (n -1)d

where tn=lartgest two digit odd number.

a= smallest two digit odd number

d=common difference between two consecutive odd numbers

n=total no. of two digit odd numbers

step-2: now after finding the no. of two digit odd positive numbers present in the number system go for finding their sum (S) by using following formula:

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

That’s it. ‘S’ is your answer.

Now finding the sum of two digit odd numbers by following above steps:

Two digit odd positive numbers are 11,13,15,17...........99 are in A.P.

Here a = 11 and d = 2, tn= 99, n = ?

Sum of the n terms = (n/2)[2a+(n -1)d]

But tn= a + (n -1)d

⇒ 99 = 11+ (n-1)2

⇒ 99 -11 = (n-1)2

⇒ 88/2 = (n-1)

∴ n = 45.

substitute n = 45 in sum of the n terms we obtain

⇒ S = (45/2)(2×11 + (45 -1)2)

⇒ S =(45/2)(110)

⇒S= 45×55.

⇒ S= 2475.

∴ sum of all two digit odd positive numbers = 2475.

Answer 2

Step-by-step explanation:

sum of all two digit odd no. is even

ex:

17+19=36