Question

find sum of all three digits numbers which are divisible by 7

Answer 1

Hello!


105, 112, 119, 126... 994

Above sequence forma an A.P.

Then,

✓ First term, a = 105
✓ Common difference, d = 7

Let aₙ = 994

a + (n - 1).d = 994

⇒ 105 + 7.(n - 1) = 994

⇒ 7.(n - 1) = 994 - 105

⇒ n - 1 = 889 / 7

⇒ n = 127 + 1 ⇒ n = 128

Now :


S₁₂₈ = (128 / 2) [2 × 105 + (128 - 1). 7]

 = 64 [210 + 889]

= 64 × 1099 = 70, 336


Cheers!

Answer 2

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