Question

how many terms for ap 118,116,114 to get sum zero​

Answer 1

Step-by-step explanation:

a=118

d=116-118=-2

Sn=0

n=?

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

0=n/2{2(118)+(n-1)-2}

0=n/2(236-2n+2)

0×2/n=238-2n

238=2n

n=238/2

n=118

Answer 2

The value of n is equal to 119.

Step-by-step explanation:

The given A.P.

118, 116, 114 ,......

Here, first term (a) = 118, common difference(d) = 116 - 118 = - 2  and

$S_{n} =0$

To find, the value of n = ?

Let n be the number of terms.

We know that,

The sum of nth term of an AP

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

∴ $\dfrac{n}{2} [2(118)+(n-1)(-2)] = 0$

⇒  $\dfrac{n}{2} [236-2n+2] = 0$

⇒  n (238 - 2n) = 0

⇒  238 - 2n = 0

⇒ 2n = 238

⇒ n = $\dfrac{238}{2} =119$

Thus, the value of n is equal to 119.