The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97

Here a = 10, d = 13 - 10 = 3

t_{n} = 97

nth term of an AP is t_{n} = a + (n – 1)d

97 = 10 + (n – 1)3

⇒ 97 = 10 + 3n – 3

⇒ 97 = 7 + 3n

⇒ 3n = 97 – 7 = 90

∴ n = 90/3 = 30

Recall sum of n terms of AP,

= 15[20 + 87] = 15 × 107 = 1605

Hence sum of 2-digit numbers which when divided by 3 yield 1 as remainder is 1605.