Question

Find 7th term and nth term of the series 0.2+0.02+0.002+...

Answer 1

Answer:

a=0.2

d= t2-t1

d= 0.02 - 0.2

d= -0.18

n= 7

tn = a + (n-1) d

t7 = 0.2 + (7-1) - 0.18

t7 = 0.2 - 6×0.18

= 0.2-1.08

= -0.88

Answer 2

Topic :-

Sequence and Series

Given :-

Series :

0.2, 0.02, 0.002, . . . . . . .

To Find :-

$\sf {n^{th}\:term\:and\:7^{th}\:term\:of\:series.}$

Solution :-

Observe the relation between consecutive terms of series,

Relation between first term and second term :

$\sf {\dfrac{0.02}{0.2}=\dfrac{1}{10}}$

Relation between second term and third term :

$\sf {\dfrac{0.002}{0.02}=\dfrac{1}{10}}$

We observe that to find next term we need to mulyiply previous term with 0.1.

So,

$\sf{\bold{ n^{th}\:term\:is\:given\:by\: :}}$

$\sf{a_n=0.2\left( \dfrac{1}{10}\right)^{n-1}}$

$\sf{a_n=2\left( \dfrac{1}{10}\right)^{n}}$

where

$\sf{a_n=n^{th}\:term}$

Calculating 7th term :

Put n = 7,

$\sf{a_7=2\left( \dfrac{1}{10}\right)^{7}}$

$\sf{a_7=0.0000002}$

Answer :-

$\sf{a_n=2\left( \dfrac{1}{10}\right)^{n}}$

$\sf{a_7=0.0000002}$

Additional Information :-

Geometric Progression

Series in which ratio of successive terms is constant.

For example,

3, 9, 27, 81, 243, . . . . . .

Its 'n'th term is given by :

$\sf{a_n=a\left( r \right)^{n-1}}$

where

a = First term of series and

r = Common ratio between successive terms