Fund the sum of all the natural numbers less than 100 which are divisible by 6
Answers
Answer:
816
Step-by-step explanation:
Numbers which are divisible by 6 are the 6th table itself.
There is no specific formula. When you add them all you get,
If you add all the multiples from 6 to 96 you will get 816
Answer:
816 will be ur answer
Explanation:
6,12,18,…96 are all natural less than 100 and divisible by 6
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=96
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6n-1=15
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6n-1=15n=15+1
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6n-1=15n=15+1n=16
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6n-1=15n=15+1n=16sum = n/2[a+ nth term]
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6n-1=15n=15+1n=16sum = n/2[a+ nth term]=16/2[6+96]
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6n-1=15n=15+1n=16sum = n/2[a+ nth term]=16/2[6+96]=8×102
6,12,18,…96 are all natural less than 100 and divisible by 6first term =6common difference= d = a2 -a1=12-6 =6nth term = 96a+(n-1)d=966+(n-1)6=96(n-1)6=96-6(n-1)6=90n-1=90/6n-1=15n=15+1n=16sum = n/2[a+ nth term]=16/2[6+96]=8×102=816
hope it helps you
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