Answer:
what are the answer choices?
Explanation:
A sequence is a list of numbers.
A <em>geometric</em> sequence is a list of numbers such that the ratio of each number to the one before it is the same. The common ratio can be any non-zero value.
<u>Examples</u>
- 1, 2, 4, 8, ... common ratio is 2
- 27, 9, 3, 1, ... common ratio is 1/3
- 6, -24, 96, -384, ... common ratio is -4
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<u>General Term</u>
Terms of a sequence are numbered starting with 1. We sometimes use the symbol a(n) or an to refer to the n-th term. The general term of a geometric sequence, a(n), can be described by the formula ...
a(n) = a(1)×r^(n-1) . . . . . n-th term of a geometric sequence
where a(1) is the first term, and r is the common ratio. The above example sequences have the formulas ...
- a(n) = 2^(n -1)
- a(n) = 27×(1/3)^(n -1)
- a(n) = 6×(-4)^(n -1)
You can see that these formulas are exponential in nature.
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<u>Sum of Terms</u>
Another useful formula for geometric sequences is the formula for the sum of n terms.
S(n) = a(1)×(r^n -1)/(r -1) . . . . . sum of n terms of a geometric sequence
When |r| < 1, the sum converges as n approaches infinity. The infinite sum is ...
S = a(1)/(1-r)
In an arithmetic sequence:
Tn=t₁+(n-1)d
t₄=t₁+(4-1)d=t₁+3d
t₅=t₁+(5-1)d=t₁+4d
t₆=t₁+(6-1)d=t₁+5d
t₄+t₅+t₆=(t₁+3d) +(t₁+4d)+(t₁+5d)=3t₁+12d
Therefore:
3t₁+12d=300 (1)
t₁₅=t₁+(15-1)d=t₁+14d
t₁₆=t₁+(16-1)d=t₁+15d
t₁₇=t₁+(17-1)d=t₁+16d
t₁₅+t₁₆+t₁₇=(t₁+14d)+(t₁+15d)+(t₁+16d)=3t₁+45d
Therefore:
3t₁+45d=201 (2)
With the equations (1) and (2) we make an system of equations:
3t₁+12d=300
3t₁+45d=201
we can solve this system of equations by reduction method.
3t₁+12d=300
-(3t₁+45d=201)
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-33d=99 ⇒d=99/-33=-3
3t₁+12d=300
3t₁+12(-3)=300
3t₁-36=300
3t₁=300+36
3t₁=336
t₁=336/3
t₁=112
Threfore:
Tn=112+(n-1)(-3)
Tn=112-3n+3
Tn=115-3n
Now, we calculate T₁₈:
T₁₈=115-3(18)=115-54=61
Answer: T₁₈=61
0,3 3,3 2,3 and 1,2 (from A to D)
