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

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!

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