Question

15. Find the 10th term of the GP 16, 8, 4, 2....
O A. 32
B. 1/64
O c. 1/32
OD. 64​

Answer 1

Answer:

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She wore a beautiful dress.

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Answer 2

Given : GP :- 16, 8, 4, 2 ...

To find : 10th term of GP

Solution :

Whenever we are asked to find nth term of a GP, we have to find the common ratio of GP. In the given GP, first term $\rm \: a_1$ = 16, second term $\rm \: a_2$ =8. So it's common ratio will be ::

${ \longrightarrow \sf \: Common \: ratio(r) \: = \dfrac{a_2}{a_1}}$

${ \longrightarrow \sf \: Common \: ratio(r) \: = \dfrac{8}{16}}$

>> Divide by 8 from both nr. and dn.

${\longrightarrow \sf \: Common \: ratio(r) \: = \dfrac{8 \div 8}{16 \div 8}}$

${ \boxed{ \longrightarrow \sf \: \blue{Common \: ratio(r) \: = \dfrac{1}{2}}}}$

~Now we have formula for nth term of GP ::

$\boxed{ \longrightarrow\sf{a_n = a_1 r^{n-1}}}$

>> Substitute given values

$\longrightarrow\sf{a_{10} = (16) \times \Big( \dfrac {1}{2} \Big)}^{10-1}$

$\longrightarrow\sf{a_{10} = (16) \times \Big( \dfrac {1}{2} \Big)}^{9}$

$\longrightarrow\sf{a_{10} = (16) \times \Big( \dfrac {1}{2^9} \Big)}$

$\longrightarrow\sf{a_{10} = {2}^{4} \times \dfrac {1}{2^9} }$

$\longrightarrow\sf{a_{10} = \dfrac{ {2}^{4}} {2^9} }$

$\longrightarrow\sf{a_{10} = {2}^{(4 - 9)}}$

$\longrightarrow\sf{a_{10} = {2}^{( - 5)}}$

$\longrightarrow\sf{a_{10} = \dfrac{1}{ {2}^{ 5}}}$

$\boxed{ \longrightarrow\sf{ \purple{a_{10} = \dfrac{1}{ 32}}}}$

>> Hence 10th term of GP is 1/32