find the sum of all natural number between 300 and 500 which are divisible by 11
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find sum of all natural numbers between 300 and 500 which are divisible by 11 ?
find first number,
300/11 which gives 3 as remainder
so 3+8 is 11
so 300+8= 308 which is divisible by 11 gives no remainder
so 308 is first number
find last number,
500/11 gives 5 as remainder
so, 5+6= 11
then 500-5 = 495 which is divisible by 11 gives no remainder
so last number is 495
sum from 308 to 495 is n(n+1)/2 =
find first number,
300/11 which gives 3 as remainder
so 3+8 is 11
so 300+8= 308 which is divisible by 11 gives no remainder
so 308 is first number
find last number,
500/11 gives 5 as remainder
so, 5+6= 11
then 500-5 = 495 which is divisible by 11 gives no remainder
so last number is 495
sum from 308 to 495 is n(n+1)/2 =
sibhiamar:
last step is (no of terms((first number of series * 2)+((no of terms -1) * 11)/2
so no of terms = (495-308)/11 +1 = 17+1 = 18
first number is 308
last step = (18((308*2)+((18-1)*11))/2 = (18(616+(17*11))/2 = (18/2)(616+187) = 9(803) = 7227
so sum of terms from 300 to 500 which are divisible by 11 is 7227
sorry for inconvenient because last night i slept in the middle of answer to your question
Answered by
4
As per the question,
a=308. (First number divisible by 11 in the range)
d= 11.
l=495. (Last number divisible by 11 in the range)
A(n)= a+(n-1)d.
=>495=308+(n-1)11.
=>495-308=(n-1)11.
=>187=(n-1)11.
=>n-1= 187/11= 17.
=> n= 17+1 =18.
S(n)= n/2 (a+l).
S(18)= 18/2(308+495).
= 9 × 803.
= 7227.
Sum of all natural numbers between 300 and 500 which are divisible by 11 is 7227.
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