Question

Find $ 10th $ and $ 16th $ terms of the G.P. $ 256,128,64,...... $

Answer 1

Hint: The geometric progression (G.P.) is defined as the sequence in which succeeding element is obtained by multiplying the preceding number by the constant and the same continues for the series. The ratio between the two terms remains the same. Here we use the standard formula for geometric progression is $ {T_n} = a.{r^{n - 1}} $

Complete step-by-step answer:
Given Expression: $ 256,128,64,...... $
First term here is $ a = 256 $
Ratio, $ r = \dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{128}}{{256}} = \dfrac{1}{2} $
Ratio between any term remains the same, therefore find between the second and the third data.
 $ r = \dfrac{{{T_3}}}{{{T_2}}} = \dfrac{{64}}{{128}} = \dfrac{1}{2} $
Now, for the tenth term in Geometric progression use the standard formula –
 $ {T_{10}} = a.{r^{10 - 1}} $
Place the identified values in the above expression –
 $ {T_{10}} = 256.{\left( {\dfrac{1}{2}} \right)^9} $
Simplify the above expression –
 $ {T_{10}} = {2^8}.\left( {\dfrac{1}{{{2^9}}}} \right) $
When bases are the same then powers are subtracted when in division by using the negative quotient rule for power and exponent.
 $ {T_{10}} = \left( {\dfrac{1}{{{2^{9 - 8}}}}} \right) $
Simplify the above expression finding the difference in the powers of the exponent.
 $ {T_{10}} = \left( {\dfrac{1}{{{2^1}}}} \right) $
 $ {T_{10}} = \left( {\dfrac{1}{2}} \right) $ ….. (A)
Now, for the sixteenth term in Geometric progression use the standard formula –
 $ {T_{16}} = a.{r^{16 - 1}} $
Place the identified values in the above expression –
 $ {T_{16}} = 256.{\left( {\dfrac{1}{2}} \right)^{15}} $
Simplify the above expression –
 $ {T_{16}} = {2^8}.\left( {\dfrac{1}{{{2^{15}}}}} \right) $
When bases are the same then powers are subtracted when in division by using the negative quotient rule for power and exponent.
 $ {T_{16}} = \left( {\dfrac{1}{{{2^{15 - 8}}}}} \right) $
Simplify the above expression finding the difference in the powers of the exponent.
 $ {T_{16}} = \left( {\dfrac{1}{{{2^7}}}} \right) $
 $ {T_{16}} = \left( {\dfrac{1}{{128}}} \right) $ ….. (B)
Hence, the required values are $ {T_{10}} = \left( {\dfrac{1}{2}} \right) $ and $ {T_{16}} = \left( {\dfrac{1}{{128}}} \right) $

Note: Know the difference between the arithmetic and geometric progression and apply its concepts accordingly. In arithmetic progression, the difference between the numbers remains the constant in the series whereas, in the geometric progression the ratio between the two remains the same.