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?
(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.