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3 years ago

14

the first term of a geometric progression is 12 and the second term is -6. Find the (i) the tenth term of the progression (ii) t

he sum to infinity

1 answer:

lesantik [10]3 years ago

6 0

Step-by-step explanation:

a1=ar⁰=12 A2=ar¹=-6

a=12 ar=-6

12r=-6

r=-6/12

r=-1/2

a10=ar⁹

=12*(-1/2)⁹

=12*(-512)

=-3/128

S∞ = a1 / (1-r )

=12/(1-1/2)

=12*1/2

=6

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The third term of an arithmetic sequence is 15 and when 20 is added to the third term the

djyliett [7]

Answer:

Common difference = 4

Position of 103 = 25th term

Step-by-step explanation:

a)

According to the given information:

$a_3 + 20 = a_8$

$a+ (3-1)d + 20 = a+ (8-1)d$

$\cancel a+ 2d + 20 = \cancel a + 7d$

$2d + 20 = 7d$

$20 = 7d-2d$

$20 = 5d$

$d = \frac{20}{5}$

$\purple {\bold {d = 4}}$

Common difference = 4

b)

$\because a_3 = 15....(given)$

$\therefore a + (3-1)d = 15$

$\therefore a + 2d = 15$

$\therefore a + 2\times 4= 15$

$\therefore a + 8= 15$

$\therefore a = 15-8$

$\red{\bold {\therefore a = 7}}$

$\because a_n = a + (n - 1) d$

$\therefore 103 = 7 + (n - 1)4$

$\therefore 103 - 7 = (n - 1)4$

$\therefore 96 = (n - 1)4$

$\therefore \frac{96}{4}= n - 1$

$\therefore 24= n - 1$

$\therefore 24+1= n$

$\blue{\bold {\therefore n = 25}}$

So, the position of 103 in this sequence is 25th term.

8 0

3 years ago

What is 7.06 divide 0.353

solong [7]

7.06 ➗ 0.353= 20

Hope this helps! :3

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3 years ago

1 example of factoring sum of two cubes

hodyreva [135]

7^3+8^3=(7+8)(7^2−(7*8)+8^2)

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3 years ago

Suppose we have a binomial experiment with n= 40 trials and a probability of success p= 0.50.Is it appropriate to use a normal a

exis [7]

Answer:

We can approximate to normal.

The parameters of the approximate normal distribution are:

$\mu=np=40*0.5=20\\\\ \sigma=\sqrt{npq} =\sqrt{40*0.5*0.5} =\sqrt{10} =3.16$

$P_B(r>23)\approx 0.314$

It is not unusual to have 23 or more successes. The probabiltity of this event happening is 13.4%.

Step-by-step explanation:

To approximate the binomial distribution to a normal distribution, we can calculate n*p and n*q and if both are bigger than 5 we can do the approximation.

$n*p=n*q=0.5*40=20$

We can approximate to normal.

The parameters of the approximate normal distribution are:

$\mu=np=40*0.5=20\\\\ \sigma=\sqrt{npq} =\sqrt{40*0.5*0.5} =\sqrt{10} =3.16$

The continuity correction factor is used because the binomial is a discrete function and the normal a continous function.

The probability of 23 successes must be expressed in the normal distribution as:

$P_B(r>23)=P_N(r>23+0.5)=P_N(r>23.5)\\\\z=(23.5-20)/3.16=1.1076\\\\P(r>23.5)=P(z>1.1076)=0.134\\\\P_B(r>23)\approx 0.314$

It is not unusual to have 23 or more successes. The probabiltity of this event happening is 13.4%.

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horrorfan [7]

A is the prime factored form of the lowest common denominator

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