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

$\huge\pink{Answer}$

First number = 7

Last such number = 7+(39x6)

= 241

Sum of such numbers = (N/2)x{A+L}

where N is total number ,

A is first number

and L is last number

Here ,

N = 40 ,

A = 7 and L = 241

Therefore , sum = (40/20){7+241}

= 20 x 248

= 4960

$\huge\blue{hope \: it \: will \: help \: you}$

Answer 2

Answer:

4920

Step-by-step explanation:

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:

6×40=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:

Sn=n2{ 2a+( n-1 )d}..........(1)

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

S40=402{2×6+(40−1)×6}

S40=20(12+39×6)=20(12+234)=20(246)=4920

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.