Find the number of terms, n, of the arithmetic series given a1=11, an=95, and Sn=689. A. 16
B. 15
C. 13
D. 14

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

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)

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