Question

Show that the sum of all odd integers between 1 to 1000 wch are divisible by 3 is 83667

Answer 1

Hey!
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Let the numbers be 3,9,15,21...........999

Let be (a1)=3

and d=6

Put 999 in place of a(n)

Use this formula :-

=>a(n)= (a1)+(n-1)×d

=>(999) =3+(n-1)×6 

=>996/6 =n-1 

=>166=n-1

=>n=166+1

=>n=167 terms 

Now we are find the value n Now start with the 1st term a(1)=3 and (n) = 167 and last term is a(n) =999.

S(n) = n/2×{ a(1) + a(n) }

S(n) = 167/2×[3+999] 

S(n) = (167/2)×(1002)

S(n)= 167×501 

S(n) = 83,667

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Regards :)

Yash Raj ✌

Answer 2

Answer:

It is proved that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

Step-by-step explanation:

Given :  

Odd integers between 1 and 1000 which are divisible by 3 are  3, 9 ,15 ..........999.

Here, a = 3 , d = 9 - 3 = 6 , an ,(l) = 999

By using the formula ,an = a + (n - 1)d

999 = 3 + (n - 1)6

999 = 3 + 6n - 6

999 = 6n - 3

999 + 3 = 6n

1002 = 6n

n = 1002/6

n = 167

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

S167 = 167/2 [3 + 999]

S167 = 167/2 × 1002

S167 = 167 (501)  

S167 = 83667

Hence Proved that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

HOPE THIS ANSWER WILL HELP YOU….