Given two terms in a geometric sequence find the common...

Given two terms in a geometric sequence find the common ratio. 14) $a_3 = 144$ and $a_6 = 31104$ 15) $a_6 = 2048$ and $a_1 = 2$ 16) $a_3 = -50$ and $a_2 = 10$ Evaluate each geometric series described. 17) $\sum_{n=1}^{8} 2^{n-1}$ 18) $4 - 12 + 36 - 108..., n=9$ 19) $a_1 = 2, a_n = 1024, r = 2$ 20) $a_1 = -2, r = -3, n = 7$ Determine the number of terms $n$ in each geometric series. 21) $3 + 9 + 27 + 81..., S_n = 3279$ 22) $\sum_{i=1}^{n} -2 \cdot (-5)^{i-1} = 208$ Determine the common ratio of the infinite geometric series. 23) $a_1 = -2, S = -\frac{5}{2}$ Evaluate each infinite geometric series described. 24) $a_1 = 1, r = -\frac{1}{5}$ 25) $-6 - \frac{6}{5} - \frac{6}{25} - \frac{6}{125}...$ 26) $\sum_{m=1}^{\infty} -160 \cdot \left(-\frac{1}{2}\right)^{m-1}$ Express as a fraction in lowest terms. SHOW ALL WORK! 27) $0.2626262626...$ (repeating) 28) $-2.075075075$ (repeating)

Transcript

00:01 Hello everyone, the given problem is in the concept of geometric sequence.

00:04 In the problem we have to find the fourth term that is a 4, fifth term a 5th term a5 and 6 term a 6 for three given series.

00:13 In the first part we have been given the series 5, 10, 20.

00:19 To find the next 3 terms we have to find the common ratio we can see that we are multiplying 2 with each term to get the next term we can see that we are multiplying 2 with 5 to get the next term then we are multiplying 2 with 10 to get the term 20 so the next three terms can be obtained by multiplying 2 again and again therefore we can find the fourth term by multiplying 2 with the third term that is 20 so we get 20 multiplied with 2 that will give us 40 similarly a 5 can be obtained by multiplying 2 with 40 that will give us 80 and the 6th term a 6 can be obtained by multiplying 2 with 80 that will give us 1 so the next three terms for the series 510 and 20 it will be 40 80 160 for the second series we have been given 256 192 and 144 if we are taking the terms as a n therefore we can say a 1 is equal to 256 192 is a 2 and a 3 the third term it is 144 we have to find the terms a4, a5 and a6 which are the fourth, fifth and six terms respectively.

01:51 Now to find the next three terms, we have to find the common ratio.

01:54 If we are taking the common ratio as r, then we can write r as a2 divided by a1, which is also equal to a3 divided by a2.

02:06 For example, we are putting a3 divided by a2.

02:09 Therefore, we get 144 divided by 190.

02:13 Which gives us 3 divided by 4 therefore the common issue value is 3 divided by 4 we multiply with a 3 we get a 4 we multiply 3 divided by 4 with a 4 we get a 5 and so on therefore the value of a 4 becomes 144 multiplied with 3 divided by 4 we get 108 we get the 5th by multiplying 3 divided by 4 with 108 therefore we get 100 multiplied with 3 divided by 4 which will give us 81.

02:51 Similarly, the 6 term a6 can be written as 81 multiplied with 3 divided by 4 which will give us 243 divided by 4.

03:04 Therefore the next 3 digits of the given series will be 108, 81 and 243 divided by 4.

03:19 In the third series, we have been given 0 .5 minus 3, 18.

03:27 We apply the same formula.

03:29 These three terms in the series can be considered as a1, a2 and 3.

03:34 Therefore, 0 .5, it is a1, minus 3, it is a2, and 18, it is a3.

03:40 We have to find the fourth, fifth and six terms which are a4, a5 and a6 respectively.

03:46 Therefore, we have to find the common ratio which you can write.

03:49 R is equal to minus 3 divided by 0 .5 which will give us minus 6 we can also see that 18 divided by minus 3 that will also give us minus 6 and we can multiply the ratio with the term a 3 to get a 4 then we can multiply the ratio with a 4 to get a 5 the 4 the 4 can be written as multiplying 18 with minus 6 that will give us minus 108 the fifth term a 5 can be written as minus 108 multiplying with minus 6 which will give us 6 the 6 term a 6 can be written as 648 multiplying with minus 6 which will give minus 3 ,880 therefore the next 3 times of the given series will become minus 108 648 and minus 3 ,8888.

04:59 Now in the second part and third part of the question we have to find particular term in the given series.

05:05 In general form in a given series if the first term is given a and the common ratio it is given r then if we are taking the nf term tn is equal to a multiplied by r to d per in minus 1.

05:20 In the second part of the question you have been given that the series has a first term 10, therefore we are taking a is equal to 10, and the common ratio value is 3.

05:29 Therefore, we are taking r is equal to 3...

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