Question

How many terms of ap -10,-7,-4,-1... Must be added to get the sum of -104

Answer 1

Answer:

a = -10

d = 3

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

104 = n/2 (-20+3n-3)

208 = -23n + 3n^2

3n^2 -39n+16n-208 = 0

(3n+16)(n-13) = 0

Since n is always positive

Hence , n=13

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Answer 2

Answer:

We cannot determine the value of n.

Step-by-step explanation:

Given : A.P.  -10,-7,-4,-1...

To find : How many terms of A.P must be added to get the sum of -104?

Solution :

In the given A.P  -10,-7,-4,-1...

The first term is a=-10

The common difference is d=-7-(-10)=-7+10=3

The sum of n terms must be -104

The sum formula of A.P is given by,

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

Substitute the value in the formula,

$-104=\frac{n}{2}[2(-10)+(n-1)3]$

$-104\times 2=n[-20+3n-3]$

$-208=n[-23+3n]$

$-208=-23n+3n^2$

$3n^2-23n+208=0$

The value of n comes into imaginary which means there is no real number.

We cannot determine the value of n.