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# help

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A geometric series $$b_1+b_2+b_3+\cdots+b_{10}$$ has a sum of 180. Assuming that the common ratio of that series is 7/4, find the sum of the series $$b_2+b_4+b_6+b_8+b_{10}.$$

In a geometric sequence of real numbers, the sum of the first six terms is 9 times the sum of the first three terms. If the first term is 5, what is the third term?

The sum of the first three terms of a geometric sequence of positive integers is equal to seven times the first term, and the sum of the first four terms is 45. What is the first term of the sequence?

Feb 4, 2019

#1
+107348
+1

In a geometric sequence of real numbers, the sum of the first six terms is 9 times the sum of the first three terms. If the first term is 5, what is the third term?

We have that

9* ( 5 [ 1 - r^3]   =  5 [ 1 - r^6 ]

9 [ 1 - r^3]  = 1 - r^6

9 - 9r^3 = 1 - r^6     rearrange as

r^6 - 9r^3 + 8 = 0           factor

(r^3 - 8) (r^3 - 1) = 0

r^3 - 1 = 0   has a positive solution of r = 1.....but r cannot take on this value

r^3 - 8 = 0    has a positive solution of  r = 2

So....the third term is  5(2)^2 = 20

Feb 4, 2019
#2
+107348
+1

The sum of the first three terms of a geometric sequence of positive integers is equal to seven times the first term, and the sum of the first four terms is 45. What is the first term of the sequence?

We have that

7a1 =  a1 [ 1 - r^3] / [ 1 - r]

7 [ 1 - r ] = 1 - r^3       factor the right as a differece of cubes

7 [ 1 - r] = [ 1 - r ] [ r^2 + r  +1]            divide out [ 1 - r ]

7 =   r^2 + r + 1    rearrange

r^2 + r - 6  = 0     factor

(r + 3) ( r - 2) = 0

Set each factor to 0 and solve for r  and we get that  r = - 3  or r = 2

Either

45 = a1 [ 1 - (2)^4 ] / [ 1 - (2)]

45 = a1 [ -15]/ (-1)

-45 = a1 [-15]      divide by -15

3 = a1           this must be the correct value of r since all terms are positive integers

Feb 4, 2019
#3
0

1) - 180 =F * (1 - (7/4)^10) / (1 - 7/4), solve for F
F =47185920/93808891
b2+b4+b6+b8+b10
sum(82575360 / 93808891,  252887040 / 93808891,  774466560 / 93808891, 2371803840 / 93808891, 7263649260 / 93808891=1260 / 11

Feb 4, 2019
#4
+7763
+2

1.

Let a be the first term, r be the common ratio(which is 7/4).

$$b_1+b_2+\cdots+b_{10} = \dfrac{a(r^{10}-1)}{r-1}=180\\ a\cdot\dfrac{\left(\dfrac{7}{4}\right)^{10} -1}{\dfrac{7}{4}-1} = 180\\ a = \dfrac{47185920}{93808891}$$

So b1 = 47185920/93808891.

b2 = b1 * 7 / 4 = 82575360/93808891.

$$b_2 + b_4 + b_6 + b_8+b_{10} = \dfrac{82575360}{93808891} \cdot \left(\dfrac{\left(\dfrac{7}{4}\right)^{10}-1}{\left(\dfrac{7}{4}\right)^2-1}\right)=\dfrac{1260}{11}$$

Therefore the required answer is 1260/11.

Feb 4, 2019