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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
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
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!!!
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