Math, asked by Akhand1945, 8 months ago

Find the least value of number of terms of series 20,18,16, for which the series has maximum sum​

Answers

Answered by Anonymous
0

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 .

Answered by amitnrw
0

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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