Math, asked by thasnithasni383, 21 days ago

find how many integers between 100 and 400 are divisible by 8​

Answers

Answered by tennetiraj86
1

Step-by-step explanation:

Given :-

The numbers are 100 and 400

To find :-

Find how many integers between 100 and 400 are divisible by 8 ?

Solution :-

Method -1:-

Given numbers are 100 and 400

The list of integers between 100 and 400

= 101,102,...,399.

The list of integers between 100 and 400 which are divisible by 8

= 104, 112, 120, ..., 392.

First term (a) = 104

Common difference = d

=112-104 = 8

= 120-112 = 8

Since the common difference is same throughout the series

They are in the Arithmetic Progression.

The last term = 392

Let an = 392

We know that

The nth term of an AP =an = a+(n-1)d

We have,

a = 104

d = 8

an = 392

On Substituting these values in the above formula then

=> 104+(n-1)(8) = 392

=> 104+8n-8 = 392

=> 8n+96 = 392

=> 8n = 392-96

=> 8n = 296

=> n = 296/8

=> n = 37

Number of terms = 37

Method -2:-

Given numbers are 100 and 400

Let a = 100

Let b = 400

The integer which is divisible by 8 then the common difference between every two consecutive integers = 8

d = 8

Let the number of AM's between two numbers be n

We know that

d = (b-a)/(n+1)

=> 8 =(400-100)/(n+1)

=> 8 = 300/(n+1)

=> 8(n+1) = 300

=> 8n+8 = 300

=> 8n = 300-8

=> 8n = 292

=> n = 292/8

=> n = 36.5 ~ 37

=> n = 37

Since n must be a natural number.

So the required integers = 37

Answer :-

The number of integers between 100 and 400 which are divisible by 8 is 37

Check:-

The integers between 100 and 400 which are multiples of 8(divisible by 8)

104,112,120,128,136,144,152,160,168,176,184,

192,200,208,216,224,232,240,248,256, 264, 272, 280, 288, 296, 304, 312, 320, 328,336,344,352,360,368,376,384,392.

Total number of integers = 37

Verified the given relations in the given problem.

Used formulae:-

  • The nth term of an AP =an = a+(n-1)d
  • a = First term
  • d = Common difference
  • n = Number of terms
  • If n AM's between two numbers a and b in an AP then d = (b-a)/(n+1)
  • a = first number
  • b = last number
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