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prisoha [69]
3 years ago
6

Can u please help me??

Mathematics
1 answer:
frez [133]3 years ago
3 0

Answer:

<u>Given GP:</u>

  • tₙ = (1/3)ⁿ⁻¹

<u>We see that:</u>

  • t₁ = 1, r = 1/3

<u>Sum of  n terms:</u>

  • Sₙ = t₁(1 - rⁿ)/(1 - r)
  • Sₙ = [1 - (1/3)ⁿ] / (1 - 1/3) = [1 - (1/3)ⁿ] / (2/3) = 3/2 [1 - (1/3)ⁿ]
  • Sₙ = 3/2 [1 - (1/3)ⁿ]
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Answer:

a_{n} = 12 -3(n-1)

Step-by-step explanation:

a_{1}  = 12\\a_{n}  = a_{n-1} - 3

we have to find formula for nth term of series for this problem

a_{1}  = 12\\a_{2}  = a_{2-1} - 3 = a_{1}  -3 =  12 -3 = 9\\a_{3}  = a_{3-1} - 3 = a_{2}  -3 =  9 -3 = 6

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and we know that when there is in increase or decrease in series by a a constant number then series is arithmetic progression series.

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In AP

nth term is given

nth term = a+(n-1)d

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common difference is given by = nth term - (n-)th term

lets take 2nd and 1st term to get common difference.

d = 9-12 = -3

(note: this was obvious by looking at series itself that d is -3 as series was decreasing by 3 unit)

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