Find the least value of number of terms of series 20,18,16, for which the series has maximum sum
Answers
Given :
A.P. series => 20 , 18 , 16 , ....
Common difference = - 2
To find :
Least value of the number of terms so that the sum of the series is maximum .
Solution :
Sum of series , S = ( n / 2 ) [ 2a + ( n - 1 ) d ]
where , n is number of terms
d is common difference
and , a is starting term
now ,
S = ( n / 2 ) [ 2*20 - 2 * ( n - 1 ) ]
S = 21 n - n^2
now , for S to be maximise , ( dS / dn ) should equal to zero and n will me minimised .
dS/dn = 21 - 2n = 0
=> n ≤ 10.5
for n = 10 , S will me maximised ( n is a whole number ) .
The least value of the number of terms is 10 so that the sum of the series is maximum .
Given : series 20,18,16
To find : least value of number of terms of series for which the series has maximum sum
Solution:
AP = 20 , 18 , 16 .................
a = 20
d = - 2
Nth term added must be grater than 0 to get maximum sum
a + (n-1)d > 0
=> 20 + (n-1)(-2) > 0
=> (n-1) < 10
=> n < 11
Hence n = 10
Another way
Sum of n terms = (n/2)(2a + (n-1)d)
= (n/2)(2*20 + (n-1)(-2)
= (n/2)(42 - 2n)
= n(21 - n)
= 21n - n²
dS/dn = 21 - 2n
d²S/dn² = - 2
sum is maximum when
21 - 2n = 0
=> n = 10.5
but n has to be integer lets check for
n = 10 & n = 11
S = n(21-n)
for n = 10 10(21-10) = 110
for n = 11 11(21-11) = 110
Sum is same for both
hence least value of n is 10
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