Question

SUM OF INTEGERS BETWEEN 1 TO 500 DIVISIBLE BY 2 AS WELL AS 5

Answer 1

Hi ,

1 ) 2 , 4 , 6 ,.... , 498 are the numbers

divisible by 2 between 1 to 500

They are in A.P

first term = a = 2

Common difference = d = 4 - 2 = 2

last term = l = 498

Let number of terms = n

a + ( n - 1 )d = l

2 + ( n - 1 ) 2 = 498

( n - 1 ) 2 = 496

n - 1 = 248

n = 249

Therefore ,

Sum of the terms = n/2 [ a + l ]

= 249 /2 [ 2 + 298 ]

= 249 /2 × 300

= 74700/2

= 37350

2 ) 5 , 10 , 15 , .... , 495 are the

numbers divisible by 5 between 1 and

500

They are in A.P

a = 5 , d = 5

l = 495

a + ( n - 1 )d = 495

5 + ( n - 1 ) 5 = 495

( n - 1 )5 = 490

n - 1 = 490/5

n - 1 = 98

n = 99

Therefore b,

Sum = n/2 [ a +l ]

= 99/2 [ 5 + 495 ]

= 99/2 × 500

= 50500/2

= 25250

3 ) 10 , 20 , 30 , ....490 are

numbers which are divisible by 2 and 5 .

There are in A.P

a = 10

d = 20 - 10 = 10

l = 490

let number of terms = n

a + ( n - 1 )d = l

10 + ( n - 1 ) 10 = 490

( n - 1 ) 10 = 480

n - 1 = 480/10

n - 1 = 48

n = 49

sum = n/2 [ a + l ]

= 49 /2 [ 10 + 490 ]

= 49/2 × 500

= 24500/2

= 12250

I hope this helps you.

:)

Answer 2

Step-by-step explanation:

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