Question

Find sum of all 3 digit number which leave remainder 3 on division by 5 in the above sequence. How many terms are tgere which has 8 in the units place

Answer 1

Answer:

Three digit numbers which leave the remainder 3 when divided by 5 are 103, 108, 113,......., 998.

103, 108, 113,...... 998 is an A.P

First term of the A.P, a = 103

Common different of the A.P, d = 5

Let 998 be the nth term of the A.P.

an = a + (n – 1) d

∴ 103 + (n – 1) × 5 = 998

⇒ 5 (n – 1) = 998 – 103 = 895

⇒ (n – 1) = 179

⇒ n = 180

Sum of all three digit numbers which leaves remainder 3 when divided by 5

= 180 / 2 ( 2 ×103 + ( 180 − 1 ) × 5 )

= 90 ( 206 + 895 )

= 90 × 1101

= 99090 .

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