Math, asked by lkshmsarath, 5 months ago

find the sum of all numbers between 1000 and 2500 which are divisible by 6​

Answers

Answered by Anonymous
0

Answer:

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know that

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)n=250

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)n=250We know that

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)n=250We know thatS=n/2(a+l)

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)n=250We know thatS=n/2(a+l)=250/2(252+999)

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)n=250We know thatS=n/2(a+l)=250/2(252+999)=125*1251

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)n=250We know thatS=n/2(a+l)=250/2(252+999)=125*1251=156375

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999We Know thatl=a+(n-1)d999=252+(n-1)3999 - 252 = 3(n - 1)n=250We know thatS=n/2(a+l)=250/2(252+999)=125*1251=156375Hence, the required sum is 156375.

Step-by-step explanation:

HOPE IT HELPS..........✨

HOPE IT HELPS..........✨❥❥❥PLZ MARK AS BRAINLIEST IF IT'S HELPS YOU

Similar questions