Answer :
The problem involves deriving relations between terms of an Arithmetic and Geometric Progression, followed by determining the common ratio of the GP, and finally calculating specific terms and the sum of a sequence.
The student is presented with a problem regarding an Arithmetic Progression (AP) and a Geometric Progression (GP). We are given certain conditions that relate terms from both progressions. Let's address each part of the question.
(a) Equations connecting AP and GP
For AP, we have the n-th term given by: a_n = a + (n-1)d, where a is the first term and d is the common difference. For GP, the n-th term is given by: g_n = ar^{n-1}, where r is the common ratio. Given the problem's conditions, we arrive at two equations:
(b) Value of r
Subtracting the first equation from the second, we get 8d = ar^6 - ar^3 or 8d = ar^3(r^3 - 1). Ratio r thus satisfies the equation r^3 - 1 = 8d/ar^3. From the first equation, we have r^3 = 1 + d/a. Replacing in the second equation, we get (1+d/a) - 1 = 8d/a(1+d/a), leading to r = 2.
(c) Values of a and d
Knowing the 10th term of GP (5120) and r (2), we can write ar^9 = 5120, which gives us a. We also have the value of r to use in a + d = ar^3 to find d.
(d) Sum of the first 20 terms of the AP
The sum of the first 'n' terms of an AP is given by: S_n = n/2(2a + (n-1)d). Substituting the known values of a and d, we calculate S_20 for the sum of the first 20 terms.
