Question

how many integers are there between 25 and 129 which are divisible by 7

Answer 1

In Arithmetic Progressions,  

a = first term

d = common difference

a₂ = second term = a + d  

aₓ = x th term

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Smallest number between 25 and 129 which is divisible by 7 is 28

Largest number between 25 and 129 which is divisible by 7 is 126

It means that 28 is the first term and 126 is the last of the Arithmetic progress. Common difference between the terms is 7 as the numbers should be divisible by 7.

From the identities of arithmetic progression, $T_{n} = a + ( n - 1 )$

Where Tn is the last term and n is the number of terms.

∴ 126 = 28 + ( n - 1 )7

⇒ 126 - 28 = ( n - 1 )7

⇒ 98 = ( n - 1 )7

⇒ $\dfrac{98}{7}= n - 1$  

⇒ 14 = n - 1

⇒ 14 + 1 = n

⇒ 15 = n

Therefore, there are 15 number which are divisible by 7 between 25 and 129.