Question

I need help plz....zzzzz

Answer 1

Problem 1

Answer: Choice C) -511

---------------------------------------

Work Shown:

• a1 = 9 = first term
• d = -7 = common difference

We decrease by 7 each time we need a new term

an = nth term

an = a1+d(n-1)

an = 9+(-7)(n-1)

an = 9-7n+7

an = -7n+16

a14 = -7*14+16

a14 = -82

Sn = sum of the first n terms of an arithmetic sequence

Sn = (n/2)*(a1+an)

S14 = (14/2)*(9+(-82))

S14 = -511

================================================

Problem 2

Answer: Choice C)
`a_n = -6(a_(n-1)), \ a_1 = -2.5`

--------------------------------------

Step-by-step explanation:

The nth term of a geometric sequence is a*r^(n-1)

• a = first term
• r = common ratio

We're given that

• a2 = 15 = second term
• a5 = -3240 = fifth term

Those two facts lead us to these two equations

• a*r = 15
• a*r^4 = -3240

Divide the second equation over the first equation

The left hand sides would divide to r^3

The right hand sides divide to -216

Solving r^3 = -216 leads to r = -6

This leads to...

a*r = 15

a*(-6) = 15

a = 15/(-6)

a = -2.5

This points us at choice C as the answer. Choice C says that the nth term is found by multiplying the previous term by -6, which is tied directly to the common ratio.