Question

How many terms of the A.P. 22,20,18. should be taken so that their sum is zero

Answer 1

Answer:

23 terms

Step-by-step explanation:

In the given A.P - 22,20,18....

a = 22

d = 20 - 22 = -2

we know that S = n/2 ( 2a + (n-1) d )

0 = n/2 ( 2(22) + (n-1) -2)

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

0 = n/2 ( 46 - 2n)

0 = n (23 - n)

$0 = { -n}^{2} + 23n$

$-n {}^{2} = - 23n$

-n = -23

or n = 23

Answer 2

Answer:

$Consider \: the \: \: given \: A.P. \: series.</p><p> \\ </p><p>27,24,21,......</p><p></p><p> \\ </p><p>Here, a=27,d=−3</p><p></p><p> \\ </p><p>Since, Sum=0</p><p> \\ </p><p></p><p>Therefore,</p><p></p><p>$

$sum = \frac{n}{2} [2a + (n - 1)d]$

$0= \frac{n}{2} [2 \times 27 + (n - 1) \times - 3]$

$54 - 3n + 3 = 0$

$57 - 3n = 0$

$57 = 3n$

$n = \frac{ \cancel{ 57}}{ \cancel 3} = 19$

$so, \: \boxed{n = 19}$