Question

How many terms of ap-6,-11/2,-5,-9/2....Are neededcto give their sum zero?

Answer 1

Step-by-step explanation:

AP :

-6,-11/2,-5,-9/2.....

a=-6

d=-11/2-(-6)

-11/2+6/1=-11+12/2

1/2

Sn=N/2[2a+(n-1)×d

0= N/2[2×-6+(n-1)×1/2]

0= N/2[-12+1n/2-1/2]

0=N/2-[24-1/2 +1n/2]

0= N/2[25n/2 +1n/2]

0=25n/4 +n²

[cross multiply]

0=-25n +n²

n(n-25)=0

n=0/n-25=0

n-25=0

n=25

0 terms cannot be taken so answer is 25 terms.

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Answer 2

Given terms are

$- 6 \: \\ - 11 \div 2 \: \\ - 5 \: \\ - 9 \div 2$

These terms are in AP which means these terms have common difference between them.

Let

$a0 = - 6 \\ a1 = - 11 \div 2 \\ a2 = - 5 \\ a3 = - 9 \div 2$

Common difference between these terms is

$d = a1 - a0 \\ = - 11 \div 2 + 6 \\ = 1 \div 2$

Therefore, d=

$1 \div 2$

Now, we have the find the number of terms needed to make the sum zero.

We know the formula of sum of n terms of AP series.

Sn =

$n \div 2(2a + (n - 1)d)$

Here, we have to find n.

a= First term of the sequence=

$- 6$

d= common difference=

$1 \div 2$

Sn is the sum of numbers= Zero( we want the number of terms needed when the sum is zero).

Substituting all these values in the formula we get

$0 = n \div 2(2 \times - 6 + (n - 1)1 \div 2) \\ n \div 2( - 12 + (n - 1)1 \div 2 = 0 \\ n \div 2(( - 24 + n - 1) \div 4) = 0 \\ n( - 25 + n) = 0 \\ - 25n + {n}^{2} = 0 \\ {n}^{2} = 25n \\ n = 25$

Therefore,

$n = 25$

we need 25 terms to make the sum of given AP series sero.

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