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if sum of four terms of GP is 60 and am between first and last term is 18. then find a and r.
See this solution and answer me if you are satisfied. Let the numbers be a , ar , ar^2 , ar^3 . Now, sum=60. => a + ar + ar^2 +ar^3 =60. ...........(i). Also, a + ar^3 = 18*2 = 36. ...........(ii). => a ( r + r^3 ) = 36. => a = 36 / ( r + r^3). Now putting this in eq.(i) , we get 24(r^3) - 36(r^2) - 36r +24 = 0. => r = 1/2 , -1 , 2. r=1 doesn't satisfy the condition. So r=2 or r=1/2 When r=2 , a=4 , and when r=1/2 , a=32. Hence, solved.
See this solution and answer me if you are satisfied.
Let the numbers be a , ar , ar^2 , ar^3 .
Now, sum=60.
=> a + ar + ar^2 +ar^3 =60. ...........(i).
Also, a + ar^3 = 18*2 = 36. ...........(ii).
=> a ( r + r^3 ) = 36.
=> a = 36 / ( r + r^3).
Now putting this in eq.(i) , we get 24(r^3) - 36(r^2) - 36r +24 = 0.
=> r = 1/2 , -1 , 2.
r=1 doesn't satisfy the condition.
So r=2 or r=1/2
When r=2 , a=4 , and when r=1/2 , a=32. Hence, solved.
Dear Student,Please find below the solution to your problem.Let the numbers be a , ar , ar^2 , ar^3 .Now, sum=60.=> a + ar + ar^2 +ar^3 =60. ...........(i).Also, a + ar^3 = 18*2 = 36. ...........(ii).=> a ( r + r^3 ) = 36.=> a = 36 / ( r + r^3).Now putting this in eq.(i) , we get 24(r^3) - 36(r^2) - 36r +24 = 0.=> r = 1/2 , -1 , 2.r=1 doesn't satisfy the condition.So r=2 or r=1/2When r=2 , a=4 , and when r=1/2 , a=32.Thanks and Regards
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