Question

3. How many natural numbers upto 150 are divisible by 7?​

Answer 1

There are 21 natural numbers upto 150 are divisible by 7

Given :

The natural numbers upto 150 are divisible by 7

To find :

The number of natural numbers upto 150 are divisible by 7

Concept :

If in an arithmetic progression

First term = a

Common difference = d

Then nth term of the AP

= a + (n - 1)d

Solution :

Step 1 of 3 :

Write down the given numbers

First we find natural numbers upto 150 are divisible by 7

The natural numbers upto 150 are divisible by 7 are 7 , 14 , 21 , ... , 147

This is an arithmetic progression

Step 2 of 3 :

Write down first term and common difference

First term = a = 7

Common Difference = d = 14 - 7 = 7

Step 3 of 3 :

Find the number of terms

Let 147 is the nth term of the AP

⇒ a + (n - 1)d = 147

⇒ 7 + (n - 1) × 7 = 147

⇒ 7 + 7n - 7 = 147

⇒ 7n = 147

⇒ n = 147/7

⇒ n = 21

Hence there are 21 natural numbers upto 150 are divisible by 7

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$1.$ If the middle term of a finite AP with 7 terms is 21 find the sum of all terms of the AP

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Answer 2

Given :

natural numbers up to 150 are divisible by 7

Concept :

in an arithmetic progression

first term $=$ $a$

common difference $d$

Solution :

between 3 and 200, a first number divisible by 7 is 7 and last number divisible by 7 is 196 herefore, a number divisible by 7 are 7, 14, 21,..., 196 which is an arithmetic progression,

where

First-term a = 7

Common difference d= 7

Last term a,,=196

a = a+ (n-1) d

->1967+ (n-1) x 7 ->196-7-7 (n-1)

(n-1)= 189 189

7

=>n-1=27 => n = 28

Hence, 28 natural numbers between 3 and 200 are divisible by 7.

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