Question

find the sum of the first 40 positive integen
divisible by 6.​

Answer 1

ANSWER:

• Sum of first 40 Positive integer divisible by 6 = 4920

GIVEN:

• 40 Positive integer divisible by 6.

TO FIND:

• Sum of first 40 Positive integer divisible by 6.

SOLUTION:

A.P :

6 , 12 , 18 , 24 ......................

Here:

=> a = 6 (first term)

=> d = (12- 6) = 6 ( common difference)

=> l = 240 ( last term)

=> n = 40 ( number of terms)

=>Last term = a +(n-1)d

=> l = 6 + (40-1) 6

=> l = 6 + 39*6

=> l = 6+ 234

=> l = 240

Formula

Sum of nth term : ( when last term is not given)

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

Sum of nth term : ( when last term is given)

$= \dfrac{n}{2} (a + l)$

Putting the values we get;

$= \dfrac{40}{2} (6 + 240) \\ \\ = 20(246) \\ \\ = 4920$

• Sum of first 40 Positive integer divisible by 6 = 4920

NOTE:

Some important formulas:

• => nth term = a+(n-1)d
• => nth term from last = l -(n-1)d

where

• a =first term
• d = common difference
• l = last term
• n = number of terms

Answer 2

GIVEN:

40 Positive integers divisible by 6

TO FIND:

Sum of the first 40 integers.

SOLUTION:

The numbers form an AP

6, 12, 18, 24,..........................

Where,

First term = a = 6

Common Difference = 6

Number of terms = n = 40

We know that,

Sum of terms in an AP = n/2 [ 2a + (n - 1)d ]

= 40/2 [ 2(6)+ (40 - 1)6]

= 20 [ 12 + (39)6 ]

= 20 [ 12 + 234 ]

= 20 ( 246)

= 4920

Therefore, the sum of first 40 positive integers which divisible 6 is 2920.