Question

find the 12th term of arithmetic series 1 by 9 4 by 9 7 by 9​

Answer 1

Answer:

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Answer 2

Given,

Arithmetic series, $\[\frac{1}{9},\frac{4}{9},\frac{7}{9}, - - - \]$

Solution,

Consider the first term of A.P. is a, common difference is d, so nth term will be $\[{a_n} = a + \left( {n - 1} \right)d\]$

Calculate the 12th term of A.P.

$\[\frac{1}{9},\frac{4}{9},\frac{7}{9}, - - - \]$

First term, $\[a = \frac{1}{9}\]$

Common difference,

$\[\begin{array}{l}d = \frac{4}{9} - \frac{1}{9}\\ \Rightarrow d = \frac{3}{9}\\ \Rightarrow d = \frac{1}{3}\end{array}\]$

12th term,

$\[{a_n} = a + \left( {n - 1} \right)d\]$

$\[ \Rightarrow {a_{12}} = \frac{1}{9} + \left( {12 - 1} \right) \times \frac{1}{3}\]$

$\[ \Rightarrow {a_{12}} = \frac{1}{9} + \frac{{11}}{3}\]$

$\[ \Rightarrow {a_{12}} = \frac{{34}}{9}\]$

Hence the 12th term of A.P. is $\[\frac{{34}}{9}\]$.