Question

find the sum of all three digit numbers each of which leave the remainder 3 when divided
by 5.​

Answer 1

HERR IS YOUR ANS

first 3 Digit no. which leave remainder=3

is 103 ..

so, a = 103 , d = 5

An = 998

so, now we have to find Sn:-

An = a + (n-1)d

998 = 103+(n-1)5

895/5 = n-1

179 +1 = n

n = 180

we know that :-

Sn = n/2(a + An)

= 90(103+998)

= 99090 ans.

hope it helps you

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Answer 2

Answer:

The least three digit number is 100 &  is divisible by 5.

First term = 103                      

Second term = 108

Third term = 113 ......

The greatest three digit number = 999

If we divided 999 by 5,  

then we get 4 as the remainder.

Last term = 998  [We want the remainder 3]

Let the nth term of the progression be 998.

$t_{n} =998\\103+(n-1)5=998\\103+5n-5=998\\5n=998-98\\5n=900\\n=180$

Now lets find $S_{180}$.

$S_{180}=\frac{180}{2}[2*103+(180-1)5]\\  S_{180}=90[206+179 * 5]\\S_{180}=90[206+895]\\S_{180}=90* 1101]\\S_{180}=99090$