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2 years ago

15

Multiply Conjugates Using the Product of Conjugates Pattern

1 answer:

Harrizon [31]2 years ago

7 0

Answer:

The product is the difference of squares is $\left(4-6y\right)\left(4+6y\right)=16-36{{y}^2}$

Step-by-step explanation:

Explanation

The given expression is (4-6 y)(4+6y).

We have to multiply the given expression.

Square the first term 4. Square the last term 6y.

$\begin{aligned}&(4-6 y)(4+6 y)=(4)^{2}-(6 y)^{2} \\&(4-6 y)(4+6 y)=16-36 y^{2}\end{aligned}$

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The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

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Step-by-step explanation:

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