Question

If the nth term of the A.P.9,7,5,..is same as the nth term of the A.P.15,12,9,...find n.

Answer 1

Answer:

if terms of the A. P. =9, 7,5 then, a=9, d=7-9=-2

nth term=a+(n-1) d

=9+(n-1) -2

=9+(-2n+2)

=9-2n+2

=11-2n....... -(eqn.1)

Again,

if terms of the A. P. are 15,12,9

then, a=15, d=12-15=-3

nth term = a+(n-1) d

=15+(n-1) -3

=15+(-3n+3)

=15-3n+3

=18-3n........ -(eqn.2)

On solving eqn. (1) and (2) we get,

11-2n=18-3n

or, 11-18=-3n+2n

or, -7=-n

hence, n= 7 (minus minus will be cancelled on both sides)

Answer 2

The nth term of Arithmetic progression is 7

Step-by-step explanation:

Given as :

For Arithmetic Progression

The first set of A.P = 9 , 7 , 5 , ................, n terms

Here , first term = a = 9

          common difference = d = second term -  first term

i.e        d = 7 - 9 = - 2

Now, nth term for any A.P , $t_n$  = a + (n - 1) d

Or, $t_n$  = 9 + (n - 1) ( - 2)

Or, $t_n$  = 9 - 2 n + 2

∴    $t_n$   = 11 - 2 n                                ..........1

Again

The first set of A.P = 15 , 12 , 9 , ................, n terms

Here , first term = a = 15

          common difference = d = second term - first term

i.e        d = 12 - 15 = - 3

Now, nth term for any A.P , $T_n$  = a + (n - 1) d

Or, $T_n$  = 15 + (n - 1) ( - 3)

Or, $T_n$  = 15 - 3 n + 3

∴   $T_n$   = 18 - 3 n                                ..........2

According to question

nth term of the A.P 9, 7, 5,.......is same as the nth term of the A.P 15 ,12 , 9 , ..........

i.e  $t_n$  =  $T_n$

Or,  11 - 2 n  =  18 - 3 n                  ( from eq 1 and eq 2 )

Or, 3 n - 2 n = 18 - 11

Or,  n = 7

∴   nth term of the A.P = n = 7

Hence, The nth term of Arithmetic progression is 7 Answer