Question

The 3rd term of a G.P is 27 and the fifth term is 3. Which term is
$\frac{1}{9}$
?
​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

3rd term of the G.P = 27

5th term of the G.P = 3

TO FIND :

The term whose value is 1/9.

PROCESS :

$\bf{ \implies \: a_{3} \: = 27}$

$\implies \: a {r}^{2} \: = \: 27$

$\implies \: a \: = \: \frac{27}{ {r}^{2} } \: \: \: \: \: \: \: \: \: \rightarrow(1)$

$\bf{ \implies \: a_{5} \: = \: 3}$

$\implies \: a {r}^{4} \: = \: 3$

Putting the value of 'a', we get

$\implies \: \frac{27}{ {r}^{2}} \times {r}^{4} \: = \:3$

$\implies \: 27 {r}^{2} \: = \: 3$

$\implies \: {r}^{2} = \: \frac{1}{9}$

$\implies \: r = \: \frac{1}{3}$

Now, Putting the value of 'r' in equation (1) we get,

$\implies \: a \: = \: \frac{27}{ { (\frac{1}{3}) }^{2} } \: \:$

$\implies \: a \: = \: {3}^{3} \times {3}^{2}$

$\implies \: a \: = \: {3}^{5}$

Now , A/Q

$\implies \: a_{n} \: = \: \frac{1}{9}$

$\implies \: a {r}^{n - 1} \: = \: \frac{1}{9}$

$\implies \: {3}^{5} {( \frac{1}{3} )}^{n - 1} \: = \: \frac{1}{9}$

$\implies \: {3}^{5} \times {3}^{2} = {3}^{n - 1}$

$\implies \: {3}^{7} = {3}^{ n- 1}$

$\implies \: 7 = \: n - 1$

$\implies \: n \: = \: 8$

$\boxed{ \sf{ \therefore \: Our \: required \: term \: is \: 8}}$

Answer 2

answers

3rd term of the G.P = 27

5th term of the G.P = 3

TO FIND :

The term whose value is 1/9.

PROCESS :

\bf{ \implies \: a_{3} \: = 27}⟹a

3

=27

\implies \: a {r}^{2} \: = \: 27⟹ar

2

=27

\implies \: a \: = \: \frac{27}{ {r}^{2} } \: \: \: \: \: \: \: \: \: arrow(1)⟹a=

r

2

27

arrow(1)

\bf{ \implies \: a_{5} \: = \: 3}⟹a

5

=3

\implies \: a {r}^{4} \: = \: 3⟹ar

4

=3

Putting the value of 'a', we get

\implies \: \frac{27}{ {r}^{2}} \times {r}^{4} \: = \:3⟹

r

2

27

×r

4

=3

\implies \: 27 {r}^{2} \: = \: 3⟹27r

2

=3

\implies \: {r}^{2} = \: \frac{1}{9}⟹r

2

=

9

1

\implies \: r = \: \frac{1}{3}⟹r=

3

1

Now, Putting the value of 'r' in equation (1) we get,

\implies \: a \: = \: \frac{27}{ { (\frac{1}{3}) }^{2} } \: \:⟹a=

(

3

1

)

2

27

\implies \: a \: = \: {3}^{3} \times {3}^{2}⟹a=3

3

×3

2

\implies \: a \: = \: {3}^{5}⟹a=3

5

Now , A/Q

\implies \: a_{n} \: = \: \frac{1}{9}⟹a

n

=

9

1

\implies \: a {r}^{n - 1} \: = \: \frac{1}{9}⟹ar

n−1

=

9

1

\implies \: {3}^{5} {( \frac{1}{3} )}^{n - 1} \: = \: \frac{1}{9}⟹3

5

(

3

1

)

n−1

=

9

1

\implies \: {3}^{5} \times {3}^{2} = {3}^{n - 1}⟹3

5

×3

2

=3

n−1

\implies \: {3}^{7} = {3}^{ n- 1}⟹3

7

=3

n−1

\implies \: 7 = \: n - 1⟹7=n−1

\implies \: n \: = \: 8⟹n=8

\boxed{ \sf{ \therefore \: Our \: required \: term \: is \: 8}}

∴Ourrequiredtermis

explanation

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