Question

find the sum of all two digit even number​

Answer 1

a = 10 ; d = 2 ; l = 98

l = a + (n-1) d

98 = 10 + ( n - 1) 2

n - 1 = 88/2

n = 45

sum = n/2 (2a + ( n - 1 ) d)

       = 45/2 ( 2*10 + 88/2 * 2)

       = 45/2 ( 20 + 88)

       = 45/2 * 108

       = 2430

Answer 2

Answer:

Sum of  all two digit even number is 2430

Step-by-step explanation:

Explanation:

There 45 two digit even numbers .

These are , 10, 12 ,14 ,16 .......... 98 .

And we know  that   , Sum of n term of A.P =  $\frac{n}{2} [2a + (n-1)d]$

Where a stand for first term , d is the common difference and

n is the number of term .

Step 1:

We  have , a = 10 , d = 12 - 10 = 2

and n = 45

$S_n = \frac{n}{2} [2a + (n-1)d]$

Put the value of a , d and n in the above formula we get ,

$S_{45}$ = $\frac{45}{2} [2(10) + (45-1)2]$

⇒$S_{45} =$ $\frac{45}{2}[20 + 88]$ = $\frac{45}{2} [108]$ = 45 ×54 = 2430

Final answer :

Hence , sum of all two digit even number is 2430 .

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