If each term of a geometric progression a_{1},a_{2},a_{3}, \ with a_{1} = \frac{1}{8} and a_{2} eq a_{1}, is the arithmetic mean of the next two terms and S_{n} = a_{1} + a_{2} + + a_{n}, then S_{20} - S_{18} is equal to\

Question

If each term of a geometric progression a_1,a_2,a_3, with a_1 = (1 / 8) and a_2 ≠ a_1, is the arithmetic mean of the next two terms and S_n = a_1 + a_2 + ... ... + a_n, then S_20 - S_18 is equal to

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Step 1: Identify the terms of the G.P. Let the common ratio be r . Then the terms of the G.P. are:\ a_1 = (1 / 8), a_2 = (1 / 8)r, a_3 = (1 / 8)r^2, ... \ bfStep 2: Use the condition of arithmetic mean Given that every term is the arithmetic mean of its two successive terms:\ a_n = fraca_n-1 + a_n+12 For n = 2 :\ a_2 = (a_1 + a_3 / 2) Substituting values:\ (1 / 8)r = frac(1 / 8) + (1 / 8)r^22 Multiplying both sides by 2 :\ (1 / 4)r = (1 / 8) + (1 / 8)r^2 Multiplying through by 8 :\ 2r = 1 + r^2 Rearranging:\ r^2 - 2r + 1 = 0 \ (r - 1)^2 = 0 => r = 1 Since a_1 eq a_2 , we consider another condition. \ bfStep 3: Use the condition for n = 3 \ a_3 = (a_2 + a_4 / 2) Substituting values:\ (1 / 8)r^2 = frac(1 / 8)r + (1 / 8)r^32 Multiplying both sides by 2 :\ (1 / 4)r^2 = (1 / 8)r + (1 / 8)r^3 Multiplying through by 8 :\ 2r^2 = r + r^3 Rearranging:\ r^3 - 2r^2 + r = 0 Factoring:\ r(r^2 - 2r + 1) = 0 \ r = 0, r = 1 Since r eq 1 , we take:\ r = -2 \ bfStep 4: Find the sum S_n The sum of the first n terms of a G.P. is:\ S_n = (a_1(1 - r^n) / 1 - r) \ S_18 = (1 / 8) * frac1 - (-2)^181 - (-2)= (1 / 8) * (1 - 262144 / 3) \ S_20 = (1 / 8) * frac1 - (-2)^201 - (-2)= (1 / 8) * (1 - 1048576 / 3) \ bfStep 5: Calculate S_20 - S_18 \ S_20 - S_18= (1 / 8) * (1 / 3)[(1 - 1048576) - (1 - 262144)] \ = (1 / 24)(-1048576 + 262144) \ = -32768 \ boxedS_20 - S_18 = -32768

Explanation

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