**Question:**

Find the sum of

(i) the first 15 multiples of 8

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(iii) all 3 − digit natural numbers which are divisible by 13.

(iv) all 3 - digit natural numbers, which are multiples of 11.

**Solution:**

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of *n* terms of an A.P.,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

(i) First 15 multiples of 8.

So, we know that the first multiple of 8 is 8 and the last multiple of 8 is 120.

Also, all these terms will form an A.P. with the common difference of 8.

So here,

First term (*a*) = 8

Number of terms (*n*) = 15

Common difference (*d*) = 8

Now, using the formula for the sum of *n* terms, we get

$S_{\pi}=\frac{15}{2}[2(8)+(15-1) 8]$

$=\frac{15}{2}[16+(14) 8]$

$=\frac{15}{2}(16+112)$

$=\frac{15}{2}(128)$

$=960$

Therefore, the sum of the first 15 multiples of 8 is 960

(ii) (a) First 40 positive integers divisible by 3

So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (*a*) = 3

Number of terms (*n*) = 40

Common difference (*d*) = 3

Now, using the formula for the sum of *n* terms, we get

$S_{n}=\frac{40}{2}[2(3)+(40-1) 3]$

$=20[6+(39) 3]$

$=20(6+117)$

$=20(123)$

$=2460$

Therefore, the sum of first 40 multiples of 3 is 2460

(b) First 40 positive integers divisible by 5

So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.

Also, all these terms will form an A.P. with the common difference of 5.

So here,

First term (*a*) = 5

Number of terms (*n*) = 40

Common difference (*d*) = 5

Now, using the formula for the sum of *n* terms, we get

$S_{n}=\frac{40}{2}[2(5)+(40-1) 5]$

$=20[10+(39) 5]$

$=20(10+195)$

$=20(205)$

$=4100$

Therefore, the sum of first 40 multiples of 3 is 4100

(c) First 40 positive integers divisible by 6

So, we know that the first multiple of 6 is 6 and the last multiple of 6 is 240.

Also, all these terms will form an A.P. with the common difference of 6.

So here,

First term (*a*) = 6

Number of terms (*n*) = 40

Common difference (*d*) = 6

Now, using the formula for the sum of *n* terms, we get

$S_{\pi}=\frac{40}{2}[2(6)+(40-1) 6]$

$=20[12+(39) 6]$

$=20(12+234)$

$=20(246)$

$=4920$

Therefore, the sum of first 40 multiples of 3 is

(iii) All 3 digit natural number which are divisible by 13

So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.

Also, all these terms will form an A.P. with the common difference of 13.

So here,

First term (*a*) = 104

Last term (*l*) = 988

Common difference (*d*) = 13

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

$a_{e}=a+(n-1) d$

So, for the last term,

$988=104+(n-1) 13$

$988=104+13 n-13$

$988=91+13 n$

Further simplifying,

$n=\frac{988-91}{13}$

$n=\frac{897}{13}$

$n=69$

Now, using the formula for the sum of *n* terms, we get

$S_{n}=\frac{69}{2}[2(104)+(69-1) 13]$

$=\frac{69}{2}[208+(68) 13]$

$=\frac{69}{2}(208+884)$

On further simplification, we get,

$S_{n}=\frac{69}{2}(1092)$

$=69(546)$

$=37674$

Therefore, the sum of all the 3 digit multiples of 13 is $S_{n}=37674$.

(iv) all 3-digit natural numbers, which are multiples of 11.

We know that the first 3 digit number multiple of 11 will be 110.

Last 3 digit number multiple of 11 will be 990.

So here,

First term (*a*) = 110

Last term (*l*) = 990

Common difference (*d*) = 11

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$990=110+(n-1) 11$

$990=110+11_{n}-11$

$990=99+11 n$

$891=11 n$

$81=n$

Now, using the formula for the sum of *n* terms, we get

$S_{n}=\frac{81}{2}[2(110)+(81-1) 11]$

$S_{n}=\frac{81}{2}[220+80 \times 11]$

$S_{n}=\frac{81}{2} \times 1100$

$S_{m}=81 \times 550$

$S_{n}=44550$

Therefore, the sum of all the 3 digit multiples of 11 is 44550.