The 4th term in a geometric sequence is -15, the 5th term is 45 and the 6th term is -135. Find the sum of the first 7 terms.

2 answers:

7 0

Is the 4th term is -15 and the 5th term is 45, then you would set it up like a5=a4(r)^5-4
45=-15(r)^1
Then you would get -3=r so the common ratio is -3, then you have to solve for a1 which would copy the previous format like -15=a1(-3)^3 which would be
(5/9). The formula for a geometric sequence is an=a1(r)^n-1
The formula for a limited geometric sequence is Sn=a1((1-r^n)/(1-r))
(5/8)((1-(-3)^7)/(1-(-3))
(5/9)(16385/5)
2735/9=303 8/9
So D!

Andreas93 [3]2 years ago

3 0

Answer:

d) 303 8/9

Step-by-step explanation:

4th term = ar^3 and 5th = ar^4 where a = first term and r = common ratio.

So ar^4 / ar^3

= r = 45/-15 = -3.

Working back,:

The first term a = ar^3/ r^3

= -15 / (-3)^3

= -15/-27

= 5/9

Sum of n terms = a * (r^n - 1)/(r-1)

= 5/9 * ((-3)^7 - 1 ) / (-3 -1)

= 303 8/9

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