find sum of all three digits numbers which are divisible by 7
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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!
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!
vaibhav585:
thank you 4 ur help
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