Question

Find the sum of all four digit numbers which leave remainder 2 when divided by 5​

Answer 1

Answer:

Three digit number which leaves remainder 2 when divided by 5 is

102,107,112,117......997

This form an AP whose first term is =102, d=5

Let 997 is the n th term of AP i.e. a

n

=997

a

n

=a+(n−1)d

997=102+(n−1)5

5n=900

n=180

Sum of all three digits which leaves remainder 2 when divided by 5 is

S

n

=

2

n

[2a+(n−1)d]

=

2

180

[2×102(180−1)5]

=90[204+179×5]

=90[204+895]

=90×1099

=98910

$\huge \: hope \: it \: help$

Please note these are the three digits .....

Please mark me as brainilest ✔️

Answer 2

Step-by-step explanation:

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hope it helps you