 # k and d( d being variable) are the nth term and common difference of an A.P. if the multiplication of (n - 1)th , (n+3)th term of an A.P. is maximum, then k/d is equal to (a) 3 (b) 4 (c) 5 (d) 6 11 years ago

let k=a+(n-1)*d, where a is the first term

so a=[k-(n-1)*d]

now let p=a+(n-2)*d and q=a+(n+2)*d ,where p is the (n-1)th term and q is the (n+3)th term

now pq=max

so [a+(n-2)d][a+(n+2)d]=max=f(d)     (assumed as this varies as d varies)

now f(d)=[k-(n-1)d+(n-2)d][k-(n-1)d+(n+2)d]

now f(d)=[k-d][k+3d]=max

now f'(d)=0       (to find max value of f(d))

so (-1)(k+3d)+(k-d)(3)=0

so -k-3d+3k-3d=0

so 2k=6d

so k=3d

hence k/d=3

hence (a) is correct

11 years ago

If a is the first term of the AP

Then K=a+(n-1)d i.e the nth term

also (n-1)th term will, be K-d and (n+3)rd term will be K+3d.

ATQ

(K-d)(K+3d)=maxima of the function f(d)=(K-d)(K+3d)

so d(f(d))/dd=0

so Using Product Rule

(K+3d)d(K-d)/dd+(K-d)d(K+3d)/dd=0

or (K+3d)(-1)+(K-d)(3)=0 (As K is a constant term)

or -K-3d+3K-3d=0

or 2K=6d

or K=3d

or K/d=3

All the Best!!!