Question

The first term of a g. P is 15 and the common ratio is 2 find the 12th term and the sum of the first 9 term

Answer 1

EXPLANATION.

First term of a G. p = 15 = a

common ratio of Gp = r = 2

Find 12th term and the sum of first 9th

term.

According to the question,

Formula of Nth term of an Gp

$\boxed{\bold{ \implies{t_{n} = ar {}^{(n \: - \: 1)} }}}$

=> 12 th term of an Gp

$12th \: \: \: term \: \: = ar {}^{11} = 15(2) {}^{11}$

=> 12 th term = 30,720

Formula of sum of Nth term of an Gp

$\boxed{\bold{ \implies{s_{n} \: = \frac{a(r {}^{n \: } - 1) }{(r \: - 1)}} }}$

$\boxed{\implies{\bold{s_{9} \: = \frac{15(2 {}^{9} - 1) }{(2 \: - 1)}} }}$

$\implies{ s_{9} \: = 15(512 \: - 1) }$

=> S9 = 15 X 511

=> S9 = 7665

Therefore,

12th term = 30720

sum of 9th term = 7665

Answer 2

Step-by-step explanation:

EXPLANATION.

First term of a G. p = 15 = a

common ratio of Gp = r = 2

Find 12th term and the sum of first 9th

term.

According to the question,

Formula of Nth term of an Gp

\boxed{\bold{ \implies{t_{n} = ar {}^{(n \: - \: 1)} }}}

⟹t

n

=ar

(n−1)

=> 12 th term of an Gp

12th \: \: \: term \: \: = ar {}^{11} = 15(2) {}^{11}12thterm=ar

11

=15(2)

11

=> 12 th term = 30,720

Formula of sum of Nth term of an Gp

\boxed{\bold{ \implies{s_{n} \: = \frac{a(r {}^{n \: } - 1) }{(r \: - 1)}} }}

⟹s

n

=

(r−1)

a(r

n

−1)

\boxed{\implies{\bold{s_{9} \: = \frac{15(2 {}^{9} - 1) }{(2 \: - 1)}} }}

⟹s

9

=

(2−1)

15(2

9

−1)

\implies{ s_{9} \: = 15(512 \: - 1) }⟹s

9

=15(512−1)

=> S9 = 15 X 511

=> S9 = 7665

Therefore,

12th term = 30720

sum of 9th term = 7665