Find the sum of all the odd integers between 1 and 1000 wch are divisible by 3
Answers
Answered by
4
Let the numbers are 3,9,12,15.................... 999
If the initial term of arithmetic progression
Is a=3 here and the common difference of the successive numbers is =6here....
Then the nth term of the sequence is given by
An =a +(n-1)d
999=3+(n-1)6
999-3 =(n-1)6
996/6=n-1
166=n-1
166+1=n
167=nterms
Now sum of the progression with the first term
a=3 and n=167 and last term is An = 999
Sn =n/2[a +an]
Sn =167/2[3+999]
Sn =167/2[1002]
Sn=167×501
Sn=83,667answer
I hope it helps you!!!!!!!!!!!!!!!!!!!!!
If the initial term of arithmetic progression
Is a=3 here and the common difference of the successive numbers is =6here....
Then the nth term of the sequence is given by
An =a +(n-1)d
999=3+(n-1)6
999-3 =(n-1)6
996/6=n-1
166=n-1
166+1=n
167=nterms
Now sum of the progression with the first term
a=3 and n=167 and last term is An = 999
Sn =n/2[a +an]
Sn =167/2[3+999]
Sn =167/2[1002]
Sn=167×501
Sn=83,667answer
I hope it helps you!!!!!!!!!!!!!!!!!!!!!
Similar questions