Question

The 6th and 9th term of an Ap is 12 and 18 respectively.find 20th term

Answer 1

20th term of an A.P. is 40

Step-by-step explanation:

GIVEN

$t_{6} = 12$

$t_{9} = 18$

___________________________

We know that,

$t_{n} = a + (n - 1)d$

$\therefore t_{6} = a + (6 - 1)d$

$\therefore 12 = a + 5d...(1)$

$similarly$

$t_{9} = a + (9 - 1)d$

$\therefore 18 = a + 8d...(2)$

$12 = a + 5d...from(1)$

$a = 12 - 5d$

$substituting \: this \: value \: in \: equation \: 2$

$\therefore equation(2) \: a + 8d = 18$

$\therefore 12 - 5d + 8d = 18$

$3d = 18 - 12$

$3d = 6$

$d = \frac{6}{3}$

$d = 2$

$substituting \: d = 2 \: in \: equation \: 1$

$a + 5d = 12$

$\therefore a + 5 \times 2 = 12$

$\therefore a + 10 = 12$

$\therefore a = 12 - 10$

$\therefore a = 2$

$Now$

$Let's \: find \: t_{20}$

$t_{n} = a + (n - 1)d$

$t_{20} = 2 + (20 - 1)2$

$t_{20} = 2 + 19 \times 2$

$t_{20} = 2 +38$

$t_{20} =40$

20th term of an A.P. is 40

Answer 2

Answer:

Step-by-step explanation: