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iren2701 [21]
3 years ago
13

These are the first six terms of a sequence with a = 2:

Mathematics
2 answers:
soldi70 [24.7K]3 years ago
6 0

Answer: The formular for this sequence is AR^n-1 (that is, A multiplied by R{raised to the power of n minus 1} )

Step-by-step explanation:This is a geometric progression in which every term is calculated by multiplying each previous term by a common ratio.

The common ratio here is 7, which is derived as

14/2, or 98/14, or 686/98, or 4802/686...

In simply put, R is derived as Tn/Tn-1, where Tn is the nth term and Tn-1 is the previous term.

Therefore the formular for this progression is given as

AR^n-1

Where A = 2, R = 7 and n = the nth term.

SpyIntel [72]3 years ago
3 0

Answer:

<h2>an+1 = 2×7ⁿ</h2>

Step-by-step explanation:

98÷14

=7

686÷98

=7

4 802÷686

=7

33 614÷4 802

=7

Then the common ratio q for this sequence is 7

recursive formula : an+1 = q×an = ?

an= a1 × qⁿ⁻¹

   =2×7ⁿ⁻¹

 

an+1 = q×an

        = 7×(2×7ⁿ⁻¹)

       = 2×7ⁿ

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Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega
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Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

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Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

To find the simplified form of the given expression :

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Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

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