Question

write the sequence of all three digit number which get remainder 3 on divided by 7 and what is the last term of the sequence​

Answer 1

Answer:

997

Step-by-step explanation:

Sequence will be as:

• aₙ=7n+3

Let's find limitations for n so that only 3 digit numbers included.

• 100 is the lowest and 999 is the highest 3-digit numbers.

• 100/7= 14 rem 2, so n>14

• 999/7=142 rem 6, so n<143

• a₁₄₂=7*142+3=997 is the last term

So with the limitations the formula will look as:

aₙ=7n+3,  14<n<143,  n∈N