The three digit natural numbers which are multiples of 7 are
105, 112, 119,…, 994
a2 – a1 = 112 – 105 = 7
a3 – a2 = 112 – 105 = 7
∵ a3 – a2 = a2 – a1 = 7
Therefore, the series is in AP
Here, a = 105, d = 7 and an = 994
We know that,
an = a + (n – 1)d
⇒ 994 = 105 + (n – 1)7
⇒ 994 – 105 = (n – 1)7
⇒ 889 = (n – 1)7
⇒ 127 = (n – 1)
⇒ n = 128
Now, we have to find the sum of this AP
⇒ S128 = 64[210 + 127 × 7]
⇒ S128 = 64[1099]
⇒ S128 = 70336
Hence, the sum of all three digit numbers which are multiples of 7 are 70336.
