Question

How many terms of an A.P. :20,19,18,....... should be taken so that their sum is 200?​

Answer 1

Step-by-step explanation:

a=20 a2=19

d=(-1)

S=200

S=n/2[2a+(n-1)d]

200=n/2[40+1-n]

400=41n-n^2

n^2-41n+400=0

Apply splitting the middle term

(n-16)(n-25)=0

So....n=16,25

Answer 2

Answer:

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

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

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

$54 - 3n + 3 = 0$

$57 - 3n = 0$

$57 = 3n$

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

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