find the sum of all 2-digit natural numbers which when divided by 3 yield 1 as remainder.
Answer.
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.
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.
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