Question

9,15,21,... a)12th term .b)50th term.nth term?​

Answer 1

Answer:

given sequence is 9, 15, 21,....

d=a2-a1 =15-9 =6

first term a=9

$a12 = a1 + (n - 1)d$

$a12 = 9 + (12 - 1)6 = 9 + 66 \\ = 75$

$a50 = 9 + (50 - 1)6 \\ = 9 + (49 \times 6) \\ = 9 + 294 \\ = 303$

$an = 9 + (n - 1)6 \\ = 9 + 6n - 6 \\ = 3 + 6n$

sum of first 50 terms

Step-by-step explanation:

$s = \frac{n}{2} (2a + (n - 1)d) \\ = \frac{50}{2} (2 \times 9 + (50 - 1)6) \\ = 25(18 + 49 \ \times 6) \\ = 25(18 + 294 ) \\ = 25 \times 312 \\ = 7800$