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3 years ago

The 3rd term of an arithinetic progression is-9

Mathematics

1 answer:

padilas [110]3 years ago

6 0

Answer:

-64

Step-by-step explanation:

The computation of the 10th term of the progression is shown below:

The 3rd term is -9

A_n = a + (n - 1)d

-9 = a + (3 - 1)d

-9 = a + 2d

-9 - 2d = a

Now the 7th term is -29

A_n = a + (n - 1)d

-29 = a + (7 - 1)d

-29 = a + 6d

Put the a value to the above equation

-29 = -9 - 2d + 6d

-29 + 9 = 4d

-20 = 4d

d = -5

Now

-9 - 2d = a

-9 - 2(-5) = a

-9 - 10 = a

a = -19

Now finally the 10th term of the progression is

= a + (n - 1)d

= -19 + (10- 1)-5

= -19 -45

= -64

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HELP ME ITS TIMED AND I WILL GIVE BRAINLIEST!!!!!

Olegator [25]

To find the gradient of a line we first hv to find the formula

so Gradient=√X1-x2and √y1-y2

((-2,1) and (0,-5)) line 1

line 2(0,-1) and (1,0)

first we will work for line 1

X1 is -2

X2 is -0

Y1 is 1

Y2 is -1

so y=x-1

so we substitute the values

1--1=-2--0

so,

1+1=-2+0

2=-2

-0

So the equation 1 line 1 is not true

we will now go to equation 2 line 2

the equation is 3x+y=-5

we substitute the values

3(-2--0)+1--1=-5

3(-2)+-2=-5

-6+-2=-5

-8= -5

group like terms

Ur answer should look like this -3

So the answer is 3x+y=-5

7 0

3 years ago

2^5×8^4/16=2^5×(2^a)4/2^4=2^5×2^b/2^4=2^c A= B= C= Please I'm gonna fail math

aleksley [76]

9514 1404 393

Answer:

a = 3, b = 12, c = 13

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

(a^b)^c = a^(bc)

___

You seem to have ...

$\dfrac{2^5\times8^4}{16}=\dfrac{2^5\times(2^3)^4}{2^4}\qquad (a=3)\\\\=\dfrac{2^5\times2^{3\cdot4}}{2^4}=\dfrac{2^5\times2^{12}}{2^4}\qquad (b=12)\\\\=2^{5+12-4}=2^{13}\qquad(c=13)$

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Additional comment

I find it easy to remember the rules of exponents by remembering that an exponent signifies repeated multiplication. It tells you how many times the base is a factor in the product.

$2\cdot2\cdot2 = 2^3\qquad\text{2 is a factor 3 times}$