Question

in an AP : given a=2 , d=8 , Sn=90 ,find n and An .
this is a problem from the chapter arithematic progressions from the 10th NCERT textbook 

Answer 1

An = a + d (n -1 )
Sn = n * [ a  +  a + d (n-1) ]  / 2  =  n * mean of 1st term and nth term
90 =  n * [ 4 + 8 n - 8 ] /2 =>  180 = 8 n² - 4 n    =>  2 n² - n - 45 = 0
n = 5    => An = 2 + 4 * 8 = 34

Answer 2

Answer:

The "number of terms", n be 5 and its value $a_5=34$  

Solution:

Given an AP with  

$a = 2 , d = 8 , S _ { n } = 90$

For an AP ,

$S _ { n } = \frac { n } { 2 } ( 2 a + ( n - 1 ) d )$

$\begin{array} { c } { 90 = \frac { n } { 2 } ( 4 + 8 ( n - 1 ) ) } \\\\ { 90 = \frac { n } { 2 } ( 4 + 8 n - 8 ) } \\\\ { 90 = \frac { n } { 2 } ( 8 n - 4 ) } \\\\ { 180 = n ( 8 n - 4 ) } \\\\ { 8 n ^ { 2 } - 4 n = 180 } \end{array}$

$\begin{aligned} 8 n ^ { 2 } - 4 n - 180 & = 0 \\\\ 2 \left( 4 n ^ { 2 } - 2 n - 90 \right) & = 0 \\\\ 4 n ^ { 2 } - 2 n - 90 & = 0 \\\\ 2 \left( 2 n ^ { 2 } - n - 45 \right) & = 0 \\\\ 2 n ^ { 2 } - n - 45 & = 0 \end{aligned}$

By factorising, we get,

$\begin{array} { c } { 2 n ^ { 2 } - 10 n + 9 n - 45 = 0 } \\\\ { 2 n ( n - 5 ) + 9 ( n - 5 ) = 0 } \\\\ { ( n - 5 ) ( 2 n + 9 ) = 0 } \end{array}$

Either,

$\begin{array} { c } { ( n - 5 ) = 0 } \\\\ { n = 5 } \end{array}$

Or,

$\begin{array} { c } { ( 2 n + 9 ) = 0 } \\\\ { 2 n = - 9 } \\\\ { n = - 4.5 } \end{array}$

Here the "number of terms", n cannot be negative.

Thus the "number of terms" be n=5  

The value of the nth term be,

$\begin{array} { c } { a _ { n } = a + ( n - 1 ) d } \\\\ { a _ { 5 } = 2 + ( 5 - 1 ) 8 } \\\\ { a _ { 5 } = 2 + 4 \times 8 } \\\\ { a _ { 5 } = 2 + 32 = 34 } \end{array}$

Thus, the "number of terms", n be 5 and its value $a_5=34$