how many terms of the sequence 18,16,14......should be taken so that their sum is  zero.

we have,

AP sequence 18,16,14....

First term,a=18

common difference,=-2

let  sum of nth term of given AP be 0.Then,

S_n=n/2.{2a+(n-1)d}

0=n/2{2*18+(n-1)(-2)}

0=n/2{36-2n+2}

0=38n-2n²

2n²-38n=0

2n(n-19)=0

so,n-19=0

n=19

hence sum of 19 terms is 0.

a = 54

d = -3

Sn = 513

Sn = n/2(2a + (n-1)d)

513 = n/2( 108 + 3 -3n)

1026 = 111n - 3n^2

n^2 -37n +342 = 0

on solving, we get

n= 18 , n=19

We are getting here 2 values , both positive so , Both are applicable here.
regards

let's see what all is given first:

To find:  n

d=a2-a1=16-18=-2

So, d=-2

use the Sum of n terms formula and then substitute,

Sn=n/2{2a+(n-1)d}
0=n/2{2*18 +(n-1)(-2)}
Tranpose 2 to the other side we'll have,
0*2=n{36-2n+2}
0=n(38-2n)
Now take 2 common out of the bracket,
0=2n(38-2n)
again transpose 2n to the other side,
0/2n=38-2n
0=38-2n
so, 2n=38
n=38/2
therefore n=19

So, Sum of 19 terms will give Sn as 0.

Hope it helps! Thumbs up please!