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3 years ago

Find P 20 of the following arithmetic sequence: 206, 212, 218, 224....

Mathematics

1 answer:

Yanka [14]3 years ago

7 0

Answer:

$p_{20} = 1.25528$

P₀ = 206 and difference d = 6

Step-by-step explanation:

Step(i):-

Given that the arithmetic sequence

206,212,218,224,.....

P₀ = 206 and difference d = 212-206 = 6

The $n^{th}$ term of the sequence

$t_{n} = ar^{n-1}$

Step(ii):-

The $n^{th}$ term of the sequence

$P_{n} = ar^{n-1}$

Put n=20

$P_{20} = 206 ( 6) ^{20-1}$

$p_{20} = 1.25528$

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<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

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Let

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and subtracting this from $S_n$ gives

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Now, we're given

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Solving for r by substitution, we have

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$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

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which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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