Question

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 166833

Answer 1

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

If the initial term of an arithmetic progression is a1 (= 3 here) and the common difference of successive members is d (= 6 here), then the nth term of the sequence is given by: 
a(n) = a(1) + (n - 1)*d 
999 = 3 + (n - 1)*6 
996/6 = n - 1 
166 = n - 1 or n = 167 terms 
Now sum of the progression with first term is a(1) = 3 and number of term (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) or 167*501 
S(n) = 83,667 ................. Answer 
but how u said 166833.....
i had solved the same question having value 83667...??m?
dont know if u got that then plz tell me correct
& if my ans is correct then mark me as brainliest dear....
☺☺☺

Answer 2

Step-by-step explanation:

Question should be like :Show that the sum of all integers between 1 and 1000 which are divisible by 3 is 166833.

The numbers are 3,6,9... 999

here a= 3

d=3

To find total terms n?

let us find which term is 999

$\boxed{a_n = a + (n - 1)d} \\ \\ 999 = 3 + (n - 1)3 \\ \\ 999 - 3 = 3(n - 1) \\ \\ 996 = 3(n - 1) \\ \\ \frac{996}{3} = n - 1 \\ \\ 332 = n - 1 \\ \\ n = 333$

To find the sum:

$\boxed{S_n = \frac{n}{2} (a + l)} \\ \\ S_n = \frac{333}{2} (3 + 999) \\ \\ S_n = \frac{333}{2} \times 1002 \\ \\ S_n = 333 \times 501 \\ \\ S_n = 166833$

Hence proved