Find the sum of all the multiples of 3 between 250 and 1000
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Solution. Clearly, the numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, …., 999. This is an A.P. with first term a = 252, common difference = 3 and last term = 999.
Step-by-step explanation:
Let there be n terms in this A.P. Then,
⇒ an = 999
⇒ a + (n – 1)d = 999
⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250
∴ Required sum = Sn= n/2(9+1
= 250/2 (252+999) = 156375
hope it's helpful to you
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