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amid [387]
2 years ago
11

Find the number of terms, n, of the arithmetic series given a1=11, an=95, and Sn=689.

Mathematics
1 answer:
borishaifa [10]2 years ago
4 0

Answer:

13

Step-by-step explanation:

Its calculate the common difference first.

(95-11)/(n-1).

We also have the sum of these n terms is 689.

So we have the following:

11

+(11+(95-11)/(n-1))

+(11+2(95-11)/(n-1))

+...

+(11+(n-1)(95-11)/(n-1))

This can be re-expressed alittle:

There are (n) amount of 11's in the addition... also 1+2+3+...+(n-1)=(n-1)(n)/2.

So we have the sum is

11(n)+n(n-1)/2×(95-11)/(n-1)

But this equal to 689.

We need to solve the following equation: ​

11(n)+n(n-1)/2×(95-11)/(n-1)=689

The (n-1)'s in second term can cancel.

11(n)+n/2×84=689

11n+42n=689

53n=689

53n=689

n=689/53

n=13

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Answer:

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Step-by-step explanation:

Given the 2 equations

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4x - y = - 16 → (2)

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2y = 8 ( divide both sides by 2 )

y = 4

solution is (- 3, 4 )


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