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Question

Write down the sequence of natural numbers which leave a remainder 3 on division by 6. What is the 10th term of this sequence? How many terms of this sequence are between 100 and 400?

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Solution

(1)

1st natural number which leaves the remainder 3 when divided by 6 = 9

2nd natural number which leaves the remainder 3 when division by 6 = 15

3rd natural number which leaves the remainder 3 when divided by 6 = = 21

So, the arithmetic sequence is 9, 15, 21, …


(2)

Common difference = 15 − 9 = 6

To get the 10th natural number from the 1st natural number, we must add the common difference nine times to the 1st natural number.

10th term = 9 + 9 × 6 = 9 + 54 = 63


(3)

1st natural number between 100 and 400 which leaves the remainder 3 when divided by 6 = 105

2nd natural number between 100 and 400 which leaves the remainder 3 when divided by 6 = 111

3rd natural number between 100 and 400 which leaves the remainder 3 when divided by 6 = 117

nth natural number between 100 and 400 which leaves a remainder 3 on division by 6 = 399

So, the arithmetic sequence is 105, 111, 117, …, 399

Common difference = 111 − 105 = 6

To get the nth natural number from the 1st natural number, we must add the common difference (n 1) times to the 1st natural number.

399 = 105 + (n − 1) × 6

399 − 105 = 6n − 6

6n − 6 = 294

6n = 300

n = 50

Thus, there are 50 terms between 100 and 400 which leave the remainder 3 when divided by 6.


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