Find the sum of

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.

990=110+(n-1)11990=110+11n-11990=99+11n891=11n81=n">