Answer:

12

Step-by-step explanation:

There are n terms in an  a.P. Where n is even. The sum of odd terms and even terms are 63 and 72 respectively. The last term is greater than FIRST term by 16.5 , calculate n

Let SAY  n = 2k

AP is

a  , a + d , a + 2d , +.............................................a + (2k-2)d  ,  a + (2k-1)d

The last term is greater than first term by 16.5

=> a + (2k-1)d - a = 16.5

=> (2k-1)d = 16.5  

=> 2kd  - d = 16.5 eq 1

Odd terms

a  , a + 2d  , ..............................................a + 2(k-1)d

in this Ap First Term = a  CD = 2d  & number of Terms = k

Sum = (k/2)(a + a + 2(k-1)d)  = 63

=> k (2a + 2(k-1)d) = 126  - eq 2

Even terms

a+d  , a + 3D  , .............................................,.a + (2k-1)d

in this Ap First Term = a+d  Cd = 2d  & number of Terms = k

Sum = (k/2)(a + d + a + (2k-1)d)  = 72

  => (k)( 2a + 2kd) = 144    eq 3

eq 3 - eq 2

=> k ( 2d) = 18

=> kd = 9

putting in eq 1

2*9 - d = 16.5

=> d = 1.5

kd = 9

=> k(1.5) = 9

=> k = 6

n = 2k = 2 * 6 = 12

Hence number of terms = 12

AP is

3 , 4.5 , 6 , 7.5 , 9 , 10.5 , 12 , 13.5 , 15 , 16.5 , 18 , 19.5