Question

find the sum of first 40 positive integers between 10 and 450 divisible by 6​

Answer 1

$\text{The first number which is divisible by 6 is 12}$

$\text{The second number which is divisible by 6 is 18}$

$\text{Now, we get an A.P, 12, 18, 24......}$

$\text{The 40 th term in this A.P}$

$t_{40}=a+39d$

$t_{40}=12+39(6)$

$t_{40}=246$

$\text{The required sum is}$

$\text{12+18+..........+246}$

$=\frac{n}{2}[a+l]$

$=\frac{40}{2}[12+246]$

$=20[258]$

$=5160$

$\therefore\textbf{The sum of first 40 intgers divisible by 6 is 5160}$

Answer 2

Sum of first 40 positive integers of series is 5160

•Positive integer between 10 to 450 which are divisible by 6 are 12 ,18 ,24.......... 444

• series is clearly an AP with a= 12 & d = 6

•Now sum of n terms of AP is Sn = n(2a+(n-1)d)/2

•sum of n terms of AP is

S = (40)(2(12)+(40-1)6)/2S = (40)(24+(39)6)/2

S = (40)(24+(39)6)/2

S = (40)(24+234)/2

S = (20)(258)

S = 5160