Question Number 11256 by 786786AM last updated on 18/Mar/17
![In the arithmetic progression, u_(1 ) =1.Given that u_(7 ) , u_(11) and u_(17) are in geometric progression, find the value of each.](https://www.tinkutara.com/question/Q11256.png)
$$\mathrm{In}\:\mathrm{the}\:\mathrm{arithmetic}\:\mathrm{progression},\:\mathrm{u}_{\mathrm{1}\:} =\mathrm{1}.\mathrm{Given}\:\mathrm{that}\:\mathrm{u}_{\mathrm{7}\:} ,\:\mathrm{u}_{\mathrm{11}} \mathrm{and}\:\mathrm{u}_{\mathrm{17}} \:\mathrm{are}\:\mathrm{in}\:\mathrm{geometric}\: \\ $$$$\mathrm{progression},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{each}. \\ $$
Answered by ajfour last updated on 18/Mar/17
![n_7 =1+6d n_(11) =1+10d n_(17) =1+16d as (n_(11) )^2 =n_7 n_(17) (1+10d )^2 =(1+6d)(1+16d) 1+20d+100d^2 =1+22d+96d^2 if d≠0, then 4d =2 d=(1/2) n_7 = 1+6d =4 n_(11) = 1+10d =6 n_(17) = 1+16d =9](https://www.tinkutara.com/question/Q11259.png)
$$\mathrm{n}_{\mathrm{7}} =\mathrm{1}+\mathrm{6d} \\ $$$$\mathrm{n}_{\mathrm{11}} =\mathrm{1}+\mathrm{10d} \\ $$$$\mathrm{n}_{\mathrm{17}} =\mathrm{1}+\mathrm{16d} \\ $$$$\mathrm{as}\:\left(\mathrm{n}_{\mathrm{11}} \right)^{\mathrm{2}} \:=\mathrm{n}_{\mathrm{7}} \mathrm{n}_{\mathrm{17}} \\ $$$$\left(\mathrm{1}+\mathrm{10d}\:\right)^{\mathrm{2}} \:=\left(\mathrm{1}+\mathrm{6d}\right)\left(\mathrm{1}+\mathrm{16d}\right) \\ $$$$\mathrm{1}+\mathrm{20d}+\mathrm{100d}^{\mathrm{2}} \:=\mathrm{1}+\mathrm{22d}+\mathrm{96d}^{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{d}\neq\mathrm{0},\:\mathrm{then}\:\mathrm{4d}\:=\mathrm{2} \\ $$$$\mathrm{d}=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\mathrm{n}_{\mathrm{7}} \:=\:\mathrm{1}+\mathrm{6d}\:=\mathrm{4} \\ $$$$\mathrm{n}_{\mathrm{11}} \:=\:\mathrm{1}+\mathrm{10d}\:=\mathrm{6} \\ $$$$\mathrm{n}_{\mathrm{17}} \:=\:\mathrm{1}+\mathrm{16d}\:=\mathrm{9} \\ $$