Question

write the sequences of natural numbers leave remainder 3 on divided by 6 what is the 10th term of this arithmetic sequence. How many terms are the 100 and 400​

Answer 1

$\huge\mathfrak\color {red}ANSWER$

(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.