Question

How many terms of the A.P.27,24,21,..... should be taken so that their sum is zero ?

Answer 1

A=27

D=24-27=-3

N=?

Sn=0

Sn=n/2[2a+n-1]d

0= n/2[2(27)+(n-1)-3

0=54-3n+3

-54=-3(n-1)

8=n-1

n=9

So,9.terms should be taken.

Answer 2

Answer:

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

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

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

$54 - 3n + 3 = 0$

$57 - 3n = 0$

$57 = 3n$

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

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