Question

find the sum of first 40 positive integers
divisible by
6​

Answer 1

Answer:

4920

Step-by-step explanation:

6+ 12+....

6(1+2+3+....40)

6 n(n+1)/2

6×40×41/2

Answer 2

Answer :-

Required to find :-

Sum of first 40 positive integers divisible by 6

Formulae used :-

$\huge{\dagger{\boxed{\rm{ {a}_{nth} = a + ( n - 1 ) d }}}}$

$\huge{\dagger{\boxed{\rm{ {S}_{nth} = \dfrac{n}{2} [ first \; term + last \; term ] }}}}$

Solution :-

We can solve this problem by using the arithmetic progession

So,

Let's form an arithmetic progession from the given data

Given :- First 40 positive integers divisible by 6

So,

The arithmetic progession is

6 , 12 , 18 - - - - - - -

we need to find the sum of first 40 positive integers

So,

First term of the AP = 6

Common difference = ( 2nd term - 1st term ) = ( 3rd term - 2nd term )

=> ( 12 - 6 ) = ( 18 - 12 )

=> 6 = 6

Hence,

Common difference = 6

We need to find the 40th term in order to find the sum of the first 40 positive integers .

Using the formula ,

$\huge{\dagger{\boxed{\rm{ {a}_{nth} = a + ( n - 1 ) d }}}}$

here,

$\boxed{\boxed{\begin{minipage}{10cm} \\ \\ \sf{ a = first term } \\ \\ \sf{ d = common difference } \\ \\ \sf{n = the term number which you want to find } \end{minipage}}}$

Hence,

$\longrightarrow{\sf{ {a}_{nth} = {a}_{40}}}$

$\longrightarrow{\sf{ {a}_{40} = 6 + ( 40 - 1 ) 6 }}$

$\longrightarrow{\sf{ {a}_{40} = 6 + ( 39 ) 6 }}$

$\longrightarrow{\sf{ {a}_{40} = 6 + 234 }}$

$\longrightarrow{\sf{ {a}_{40} = 240 }}$

So,

$\huge{\leadsto{\boxed{\tt{ 40th \; term = 240 }}}}{\bigstar}$

Similarly,

Using the formula ,

$\huge{\dagger{\boxed{\rm{ {S}_{nth} = \dfrac{n}{2} [ first \; term + last \; term ] }}}}$

This formula enables us to find the sum of the nth terms .

Here,

$\boxed{\boxed{\begin{minipage}{10cm} \\ \\ \sf{a = first term } \\ \\ \sf{d = common difference } \\ \\ \sf{ n = the term number till which you want to find the sum } \end{minipage}}}$

So,

$\longrightarrow{\rm{ {S}_{nth} = {S}_{40} }}$

$\longrightarrow{\rm{ {S}_{40} = \dfrac{40}{2} [ 6 + 240 ] }}$

$\longrightarrow{\rm{ {S}_{40} = \dfrac{40}{2} [ 246 ] }}$

$\longrightarrow{\rm{ {S}_{40} = 20 [ 246 ] }}$

$\implies{\rm{ {S}_{40} = 4,920 }}$

$\small{\boxed{\tt{\therefore{ Sum \; of \; first \; 40 \; positive \; integers \; divisible \; by \; 6 = 4,920 }}}}{\bigstar}$