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.
Answers
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
Answer:
12
Step-by-step explanation:
(Number of terms / 2) * Common difference = (Sum of even terms - Sum of odd terms )
(n/2) * d = 72 - 63
nd= 18
d= common difference
Last term - first term = 16.5
(n-1)d = 16.5
nd - d = 16.5
d = 1.5
n = 12