Question

which term of the Ap 3,15,27,39.......will be 132 more than its 5th term

Answer 1

First term( a ) = 3

second term( $a_{2}$ = 15

third term ( $a_{3}$ ) = 27

fourth term ( $a_{4}[/text] = 39 </p><p>fifth term = [tex] \bold{a_{4} = a + ( 5 - 1 )d}$

                = 3 + 4d

common difference ( d ) = 3rd term - 2nd term = 2nd term - 1st term

 

                                          = 27 - 15 = 15 - 3

                                          = 12

Therefore, fifth term = 3 + 4( 12 )

                                    = 3 + 48

                                    = 51

According to the question we have to get the term which will 132 more than 5th term ( 51 ), as 5th term is 51 ,  132 more than 5th term = 51 + 132 = 183

Let 183 be the nth term,

$a_{n} = a+ ( n - 1 )d$

$183 = 3 + ( n - 1 ) 12$

$180 = ( n - 1 ) 12$

$\dfrac{180}{12} = n - 1$

$15 = n - 1$

16 = n

Hence, the term which will be 132 more than the 5th term of the AP is 16th term of the AP.