refers to the nth term of an arithmetic progression with common difference , satisfying .
Over all such sequences, what is the value of that would minimize
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Let the first term of the A.P. = a
and the common difference = d
so T(7)= a + 6d = 9
=> a = 9 − 6 d
T ( 1 ) × T ( 2 ) × T ( 7 ) = a ( a + d ) . 9 ...........................(1)
Putting a = 9 − 6 d in the equation ( 1 )
T ( 1 ) × T ( 2 ) × T ( 7 ) = ( 9 − 6 d ) ( 9 − 6 d + d ) . 9
Now T ( 1 ) × T ( 2 ) × T ( 7 ) is a function of d ..
.so let T ( 1 ) × T ( 2 ) × T ( 7 ) = f ( d ) [ f(d) means a function of d ]
so f ( d ) = 9 . ( 9 − 6 d ) ( 9 − 5 d )
So for getting the least value of f(d), lets differentiate it..and then equate to zero , from there we will get the value of d at which f(d) will be minimum.
f ′ ( d ) = 9 ( 9 − 6 d ) ( − 5 ) + 9 ( 9 − 5 d ) ( − 6 )
putting f ′ ( d ) = 0 , we get d = 2 0 3 3
and also f ′ ′ ( d ) > 0 , so at d = 2 0 3 3 . f(d) will attain minimum value.
Hence d = 2 0 3 3