Question

3. The sum of all 2-digit numbers divisible by 5 is : ​

Answer 1

Given  :

• all the 2 digit numbers divisible by 5 :
• the numbers are: 10,15,20...95

To find :

we need to find the sum of the sequence we got :

10+15+20+...+95

Concept :

As we can see that the sequence is forming an Arithmetic Progression or an AP.

so, we will use the formula of the sum of AP to find the sum of the sequence. The formulae are as follows :

Formula To be Applied :

the general formula of AP :

$\displaystyle \rm \bull a_n = a+(n-1)d$

now, the formula of the sum of AP :

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

here, the abbreviations used :

a = first term

$\displaystyle \rm a_n = final \: term$

n = no. of terms

d = common difference

S = sum of the sequence

Solution :

now, in this AP :

first term = 10

last term = 95

common difference = 5

so, using the formula, we can find out the number of terms in this AP :

$\displaystyle \rm \implies a_n = a+(n-1)d$

$\displaystyle \rm \implies 95 = 10+(n-1)5$

$\implies \displaystyle\rm 10+5n-5 = 95$

$\implies \displaystyle \rm 5n+5 = 95$

$\implies \displaystyle \rm 5n = 95-5$

$\implies \displaystyle \rm 5n = 90$

$\implies \displaystyle \rm n = \frac{90}{5}$

$\implies \displaystyle \rm n = 18$

so, the number of terms of AP = 18

Now, we also know the number of terms of the AP.

So, using the sum formula, we can easily find the sum :

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

$\implies \displaystyle \rm S_n = \frac{18}{2}[2(10)+(18-1)5]$

$\implies \displaystyle \rm S_n = 9[20+(17)5]$

$\implies \displaystyle \rm S_n = 9[20+85]$

$\implies \displaystyle \rm S_n = 9[105]$

$\red{\underline{\boxed{\implies \displaystyle \rm S_n = 945}}}$

so, the sum of the sequence = 945

Final Answer :

the answer we got is :

• the sum of all 2 digit number divisible by 5 = 945