Question

★Question:-

Find the sum of the first 40 positive integers which give a remainder 1 when divided by 6.


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Answer 1

Answer-

Hint- here first we will write the sequence of first 40 integers that are divisible by 6 this sequence from an A.P. after that we can find the sum using the formula for sum of n terms of an A.P.

step by step explanation-

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

6, 12, 18, 24, 30, 36,..,240

to 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:

$\large Sn = \frac{n}{2} {2a + (n - 1)d}...(1)$

so on subs.tituting the values of n,a and D in equation (1) we get:

$\large s40 = \frac{40}{2} {2 \times 6 + (40 - 1) \times 6}$

$\large s40 = 20 \: (12 + 39 \times 6)$

$\large = 20 \: (12 + 234)$

$\large = 20 \: (246)$

$4920$

so the some of the 40 terms of this AP is 4920.

hence, the sum of the first 40 positive integers divisible by 6= 4920

Note- everyone should not here that the sum of n terms of an AP can also be found by applying

$\sf{the \: formula} \: Sn \: = \frac{n}{2} (first \: term + last \: term)$

and in this question both the first term and the last term is given to using this formula will be time saving

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Answer 2

Answer:

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