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yulyashka [42]
3 years ago
9

Write the first four terms of a geometric sequence. Using the values you have written, calculate the common ratio and write the

rule for the nth term. Show all calculations.
Mathematics
1 answer:
sukhopar [10]3 years ago
6 0

Answer:

FIRST FOUR (4) TERM = 6, 18, 54,162.

COMMON RATIO = 3

Step-by-step explanation:

In geometric expression.

Tn = the nth term

a = first term

r = common ratio.

Let's take the First term of the G.P as 6 and the third term as 54. Let's find the common ratio.

3rd term will be written as:

T3 = ar² = 54

Given the first term as 6.

T3 = 6r² = 54

6r² = 54

r² = 54 / 6

r² = 9

r= √9

r = 3

Therefore common ratio = 3.

Let's use this information to get the fourth term.

T4 = ar³

when a = 6

and r = 3.

T4 = 6 * 3³

T4 = 6 * 27

T4 = 162

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Step-by-step explanation:

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Let the original piece of wood be x

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the shortest length would be 5/24 * x

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the longest will be 7/24 * x = 7x/24

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In the expansion (ax+by)^7, the coefficients of the first two terms are 128 and -224, respectively. Find the values of a and b
madam [21]

Answer:

a = 2, b = 3.5

Step-by-step explanation:

Expanding (ax+by)^7 using Binomial expansion, we have that:

(ax+by)^7 =

(ax)^7(by)^0 + (ax)^6(by)^1 + (ax)^5(by)^2 + (ax)^4(by)^3 + (ax)^3(by)^4 + (ax)^2(by)^5 + (ax)^1(by)^6 + (ax)^0(by)^7

= (a)^7(x)^7+ (a)^6(x)^6(b)(y) + (a)^5(x)^5(b)^2(y)^2 + (a)^4(x)^4(b)^3(y)^3 + (a)^3(x)^3(b)^4(y)^4 + (a)^2(x)^2(b)^5(y)^5 + (a)(x)(b)^6(y)^6 + (b)^7(y)^7\\\\\\= (a)^7(x)^7+ (a)^6(b)(x)^6(y) + (a)^5(b)^2(x)^5(y)^2 + (a)^4(b)^3(x)^4(y)^3 + (a)^3(b)^4(x)^3(y)^4 + (a)^2(b)^5(x)^2(y)^5 + (a)(b)^6(x)(y)^6 + (b)^7(y)^7

We have that the coefficients of the first two terms are 128 and -224.

For the first term:

=> a^7 = 128

=> a = \sqrt[7]{128}\\ \\\\a = 2

For the second term:

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b = \frac{-224}{a^6}

b = \frac{-224}{2^6} \\\\\\b = \frac{-224}{64} \\\\\\b = 3.5

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