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arsen [322]
3 years ago
7

The 3rd term of the geometrical sequence is larger than the 2nd

Mathematics
1 answer:
notsponge [240]3 years ago
3 0

Answer:the first term is 81

Step-by-step explanation:

The formula for determining the nth term of a geometric sequence is expressed as

Tn = ar^(n - 1)

Where

Tn represents the nth term

n represents the number of terms

r represents the common ratio

The 3rd term of the geometrical sequence is larger than the 2nd

term by 36. This means that

T3 - T2 = 36 - - - - - - - - - - 1

The product of these two terms is -243. This means that

T2 × T3 = - 243

T2 = - 243/T3

Substituting T2 = - 243/T3 into equation 1, it becomes

T3 - (- 243/T3) = 36

T3 + 243/T3 = 36

T3^2 + 243 = 36T3

T3^2 - 36T3 + 243 = 0

T3^2 - 27T - 9T3 + 243 = 0

T3(T3 - 27) - 9(T3 - 27) = 0

T3 - 9 = 0 or T3 - 27 = 0

T3 = 9 or T3 = 27

Substituting T3 = 9 or T3 = 27 into

T2 = - 243/T3 = 27, it becomes

T2 = - 243/9 or - 243/27

T2 = - 27 or T2 = - 9

Therefore,

The 3rd term is 9

The 2nd term is - 27

The common ratio, r would be

T3/T2 = 9/-27 = - 1/3

The expression for the third term would be

9 = a × - 1/3^(3 - 1)

9 = a × (- 1/3)^2

9 = a × 1/9

a = 9 × 9 = 81

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Answer:

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It is given that, two parallel lines l and m are intersected by a transversal t.

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8 0
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