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Hint: Here first we will write the sequence of first 40 integers that are divisible by 6. This sequence will form an A.P. After that we can find the sum using the formula for sum of n terms of an A.P.

Complete step-by-step answer:

The first 40 positive integers that are divisible by 6 are given as:

6, 12, 18, 24, 30, 36 ,……………. ,240

To find the find the last term i.e. the 40th integer we can use:

$6\times 40=240$

So, the last term will be 240.

So, this sequence is forming an AP and the first term of this AP is 6.

Now we will find the common difference of this AP. So, to find the common difference we may subtract the 1st term from the 2nd term:

So, Common difference (d) =12-6=6

So, for this AP we have the d = 6 and also a=6.

Now to find the sum we will use the formula for sum of n terms of an AP which is given as:

${{S}_{n}}=\dfrac{n}{2}${ 2a+( n-1 )d}..........(1)

So, on substituting the values of n, a and d in equation (1) we get:

${{S}_{40}}=\dfrac{40}{2}\left\{ 2\times 6 + \left( 40-1 \right)\times 6 \right\}$

$\begin{align}

& {{S}_{40}}=20\left( 12+39\times 6 \right) \\

& \,\,\,\,\,\,\,\,=20\left( 12+234 \right) \\

& \,\,\,\,\,\,\,\,=20\left( 246 \right) \\

& \,\,\,\,\,\,\,\,=4920 \\

\end{align}$

So, the sum of the 40 terms of this AP is 4920.

Hence, the sum of the first 40 positive integers divisible by 6 is = 4920.

Note: Students should note here that the sum of n terms of an AP can also be found by applying the formula ${{S}_{n}}=\dfrac{n}{2}$( first term + last term). And in this question both the first term and the last term is given so using this formula will be time saving.

Complete step-by-step answer:

The first 40 positive integers that are divisible by 6 are given as:

6, 12, 18, 24, 30, 36 ,……………. ,240

To find the find the last term i.e. the 40th integer we can use:

$6\times 40=240$

So, the last term will be 240.

So, this sequence is forming an AP and the first term of this AP is 6.

Now we will find the common difference of this AP. So, to find the common difference we may subtract the 1st term from the 2nd term:

So, Common difference (d) =12-6=6

So, for this AP we have the d = 6 and also a=6.

Now to find the sum we will use the formula for sum of n terms of an AP which is given as:

${{S}_{n}}=\dfrac{n}{2}${ 2a+( n-1 )d}..........(1)

So, on substituting the values of n, a and d in equation (1) we get:

${{S}_{40}}=\dfrac{40}{2}\left\{ 2\times 6 + \left( 40-1 \right)\times 6 \right\}$

$\begin{align}

& {{S}_{40}}=20\left( 12+39\times 6 \right) \\

& \,\,\,\,\,\,\,\,=20\left( 12+234 \right) \\

& \,\,\,\,\,\,\,\,=20\left( 246 \right) \\

& \,\,\,\,\,\,\,\,=4920 \\

\end{align}$

So, the sum of the 40 terms of this AP is 4920.

Hence, the sum of the first 40 positive integers divisible by 6 is = 4920.

Note: Students should note here that the sum of n terms of an AP can also be found by applying the formula ${{S}_{n}}=\dfrac{n}{2}$( first term + last term). And in this question both the first term and the last term is given so using this formula will be time saving.

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