# Answer to Question #42508 in Calculus for shasha

Question #42508

Find an equation for the nth term of the arithmetic sequence.

a14 = -33, a15 = 9

an = -579 + 42(n + 1)

an = -579 + 42(n - 1)

an = -579 - 42(n + 1)

an = -579 - 42(n - 1)

Help me show work and explanation please

a14 = -33, a15 = 9

an = -579 + 42(n + 1)

an = -579 + 42(n - 1)

an = -579 - 42(n + 1)

an = -579 - 42(n - 1)

Help me show work and explanation please

Expert's answer

If the initial term of an arithmetic progression is a_1 and the common difference of successive members is d, then the nth term of the sequence (a_n) is given by:

a_n=a_1+(n-1)d and in general

a_n=a_m+(n-m)d.

So,using the formula above. Let's n=15 and m=14. We obtain a_15=a_14+(15-14)d or 9=-33+d. From this d=-42.

a_1=a_n-(n-1)d or a_1=a_15-(15-1)(-42). a_1=-579.

Thus, the right answer is a_n=-579-42(n-1).

a_n=a_1+(n-1)d and in general

a_n=a_m+(n-m)d.

So,using the formula above. Let's n=15 and m=14. We obtain a_15=a_14+(15-14)d or 9=-33+d. From this d=-42.

a_1=a_n-(n-1)d or a_1=a_15-(15-1)(-42). a_1=-579.

Thus, the right answer is a_n=-579-42(n-1).

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