Question

Известно что в геометрической прогрессии разность шестого и четвертого членов равна 56,а разность третьего и пятого членов равна 14. Найдите первый член данной прогрессии и знаменатель.​​ ​

Answer 1

$(b_n)$ - геометрическая прогрессия;

$b_6-b_4=56$

$b_3-b_5=14$

$b_1=?$

$q=?$

Решение.

1)    $b_n=b_1q^{n-1}$

   $\left \{ {{b_6-b_4=56} \atop {b_3-b_5=14}} \right.=> \left \{ {{b_1q^{5} -b_1q^{3} =56} \atop {b_1q^{2} -b_1q^{4} =14}} \right.=> \left \{ {{b_1q^{3}(q^{2}-1) =56} \atop {b_1q^{2}(1 -q^{2}) =14}} \right.=> \left \{ {{b_1q^{2}(q^{2}-1) =\frac{56}{q} } \atop {b_1q^{2}(q^{2}-1) =-14}} \right.=>$

$=>\frac{56}{q} =-14$

     $q=\frac{56}{-14}$

     $q=-4$

2)   $b_1q^{2}(q^{2}-1) =-14$    

     $b_1=\frac{-14}{q^2(q^2-1)}$

     $b_1=\frac{-14}{(-4)^2((-4)^2-1)}=\frac{-14}{16*15}=-\frac{7}{120}$

    $b_1=-\frac{7}{120}$

Ответ:      $b_1=-\frac{7}{120}$

                $q=-4$

Answer 2

Calculating the terms of the geometric progression

According to the information given, we know: The difference between the 6th and 4th terms is 56. The difference between the 3rd and 5th terms is 14.

Since this is a geometric progression, the ratio between consecutive terms is constant. Let's call this common ratio r.

Then: - The 4th term is the initial term a multiplied by r^3 - The 6th term is a multiplied by r^5

Using the first condition: a * r^3 - a * r = 56 a(r^3 - r) = 56 a = 56/(r^2 - r)

Similarly, from the second condition: a * r^2 - a * r^4 = 14 a(r^2 - r^4) = 14 a = 14/(r - r^3)

Equating the two expressions for a: 56/(r^2 - r) = 14/(r - r^3)

Solving this equation gives r = 2. Therefore, the first term a = 14.

So the first term of the geometric progression is 14, and the common ratio is 2.