which term of the AP 40,36,32,.... is 0 ? the aum of how many terms of the AP is 0?
Answers
Answer:
AP = 40, 36, 32, ... 0
To find which term is 0.
Last term = 0
First term = 40
Common Difference = -4
Number of terms = ?
⇒ l = a + ( n - 1 ) d
⇒ 0 = 40 + ( n - 1 ) -4
⇒ 0 - 40 = ( n - 1 ) - 4
⇒ -40 = -4n + 4
⇒ -40 - 4 = -4n
⇒ -44 = -4n
⇒ 4n = 44
⇒ n = 44 / 4 = 11
Hence the 11th term of the AP is zero.
Also sum of n terms is given to be 0. We must find the value of n.
⇒ Sum of n terms = ( n / 2 ) ( 2a + ( n - 1 ) d )
⇒ 0 = ( n / 2 ) ( 2 ( 40 ) + ( n - 1 ) -4 )
⇒ 0 = ( n / 2 ) ( 80 - 4n + 4 )
⇒ 0 × 2 = n ( 84 - 4n )
⇒ 0 = n ( 84 - 4n )
⇒ 0 / n = ( 84 - 4n )
⇒ 0 = 84 - 4n
⇒ 4n = 84
⇒ n = 84 / 4 = 21
Hence the sum of first 21 terms is zero. Hence 21 terms must be added to get 0 as the sum.
This is the required answer.
Let the first term be a.
Let the common difference be d.
Given a=40
d=36-40=-4
Now,
Let the nth term be 0.
nth term=a+(n-1)d
Therefore
a+(n-1)d=0
=40+(-4)(n-1)=0
=40-4n+4=0
=-4n+44=0
=n=-44/-4
=11
Let n terms added be =0.
n/2*[2a+(n-1)d]=0
n*[2a+(n-1)d]=0
=n[2*40+(n-1)(-4)]=0
=n[80-4n+4]=0
=n[84-4n]=0
Either n=0 or 84-4n=0
But n cannot be zero.
So
84-4n=0
4n=84
n=84/4
=21
Answer:
11th term is 0
21 terms have to be added to get 0.
Hope it helps you.
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