Question

how many terms of the AP 22 20 18 should be taken so that their sum is zero

Answer 1

Let number of terms be n
So 0= n/2{22 ×2 + (n-1)(-2)}
or 0= n{46-2n}
or n = 23

Answer 2

Answer:

23 terms of the AP should be taken so that their sum is zero .

Step-by-step explanation:

AP : 22 20 18 ...

First term = a = 22

Common difference = d = 22-20 = 20-18 = 2

Sum of first n terms= $S_n=\frac{n}{2}(2a+(n-1)d)$

We are given that the sum is 0

So, $0=\frac{n}{2}(2(22)+(n-1)(-2))$

$0=n(44-2n+2)$

$0=n(46-2n)$

$n=0,23$

Hence 23 terms of the AP should be taken so that their sum is zero .