Find the sum of all the number between 250 and 1000 which are equally divisible by 3
Answers
Natural numbers between 250 and 1000 which are divisible by 3:
252...... 999
First term (a) = 252
Last term (tn) = 999
Common difference (d) = 3
tn = a + (n - 1)d
⇒ 252 + 3(n - 1) = 999
⇒ 3(n - 1) = 999 - 252
⇒ 3(n - 1) = 747
⇒ 3n - 3 = 747
⇒ 3n = 750
⇒ n = 250
Therefore, there are 250 numbers which are divisible by 3 between 250 and 1000.
Sn = n/2(a + l)
= 250/2(252 + 999)
= 125(1251)
= 1376
Sum of all the natural numbers between 250 and 1000 which are divisible by 3 = 1376
i think this is the answer
Answer:
Step-by-step exp
To find a:
On dividing 250 by 3, the remainder is 1.
Subtract the remainder from the dividend (3-1) and add it to the divisor.
⇒ a = 250 + (3-1) = 252
To find aₙ:
On dividing 1000 by 3, the remainder is 1.
Subtract the remainder from the divisor (1000-1)
⇒ aₙ = 1000 - 1 = 999
To find n:
aₙ = a + (n-1)d
aₙ = 252 + (n-1)3 = 999
(n-1)3 = 999 - 252 = 747
(n-1) = 747÷3 = 249
n = 249 + 1 = 250
Sₙ = n÷2 [ 2a + (n-1)d ]
S₂₅₀ = 250÷2 [ 2x252 + (252-1)x3 ]
= 125 [ 504 + 753 ]
S₂₅₀ = 157125