SOLUTION: the 5th term in a geometric progression is 9 times the 3rd term and the sum of the 6th and 7th terms is 1944. determine the common ratio, the first term the sum of the 4th to 10th

Algebra ->  Sequences-and-series -> SOLUTION: the 5th term in a geometric progression is 9 times the 3rd term and the sum of the 6th and 7th terms is 1944. determine the common ratio, the first term the sum of the 4th to 10th       Log On


   

Question 1170436: the 5th term in a geometric progression is 9 times the 3rd term and the sum of the 6th and 7th terms is 1944. determine the common ratio, the first term the sum of the 4th to 10th terms inclusive
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
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the common ratio is 3 since between the 3rd and 5th is a 9-fold increase
a1 is unknown
a2=3a1
a3=9a1
a5=81a1
a6=243a1
a7=729a1
the last two add to 1944=972a1
so a1=2, and r=3
-
sum of first 10 terms is 2(1-3^10)/-2=59048
sum of first four terms is 2(1-3^4)/-2=80
the sum of the 4th-10th terms inclusive is 58968
the series is 2/6/18/54/162/486/1458/4374/13122/39366

Answer by ikleyn(52781) About Me  (Show Source):
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.


            There are  two different progressions  and  two different answers.


The common ratio is   +/- sqrt%289%29 = +/- 3.


Case 1.   Common ratio is  3 

            Then 


            a1 is unknown
            a2 = 3a1
            a3 = 9a1
            a5 = 81a1
            a6 = 243a1
            a7 = 729a1
            the last two terms add to 243a1 + 729a1 = 972a1 = 1944
            so a1 = 2, and r = 3
            -
            sum of first 10 terms is   2*(1-3^10)/(-2) = 59048
            sum of first  3 terms is   2*(1-3^3)/(-2)  =    26
            the sum of the 4th to 10th terms inclusive is  59048  - 26 = 59022
            the series is 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366.


Case 2.   Common ratio is  -3 

            Then 


            a1 is unknown
            a2 =   -3a1
            a3 =    9a1
            a5 =  -81a1
            a6 =  243a1
            a7 = -729a1
            the last two terms add to -729a1 + 243a1 = -486a1 = 1944
            so a1 = -4, and r = -3
            -
            sum of first 10 terms is   (-4)*(1-(-3)^10)/4 = 59048
            sum of first  3 terms is   (-4)*(1-(-3)^3)/4  =   -28
            the sum of the 4th to 10th terms inclusive is  59048 - (-28) = 59076
            the series is -4, 12, -36, 108, -324, 972, -2916, , 8748, -26244.

Solved.


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