The sum of 'n' terms of an arithmetic sequence is 4n^2+3n. What is the first term, the common difference, and the sequence?

The sequence has first term 7 and common difference is 8.

So the sequence is f(n)=7 + 8(n-1)

Step-by-step explanation:

Let a be the first term.

Let a+d be the second term where d is the common difference.

Then a+2d is the third

And a+(n-1)d is the nth term.

Adding these terms we get:

an+(n-1)(n)/2×d

For the first term of this sum I seen we had n amount of a's and for the second term I used the well known identity sum of the first n positive integers is n(n+1)/2.

Let's simplify:

an+(n-1)(n)/2×d

Distribute:

an+(n^2d/2)-(nd/2)

Find common denominator:

(2an/2)+(n^2d/2)-(nd/2)

Combine terms into one:

(2an+n^2d-nd)/2

Reorder terms:

(n^2d+2an-nd)/2

Regroup terms:

(n^2d+(2a-d)n)/2

We want the following sum though:

4n^2+3n

This means d/2=4 (so d=8) and (2a-d)/2=3.

So plug d=8 into second equation to solve for a.

(2a-8)/2=3

2a-8=6

2a=14

a=7

The sequence has first term 7 and common difference is 8.

So the sequence is f(n)=7 + 8(n-1).

the answer is c.

for continuous functions, they continue in the same pattern and you have to find that pattern kinda like slope

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

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