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3 years ago

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:

Step-by-step explanation:

98÷14

686÷98

4 802÷686

33 614÷4 802

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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2 years ago

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

AlekseyPX

Answer:

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

That is $\frac{1}{3^8}$ is correct answer

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 :

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

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

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

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

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3 years ago

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