Q: A geometric progression of real numbers is such that the sum of its first four terms is equal to 30 and the sum of the squares of the first four terms is 340. Then

(A) two such G.P. are possible

(B) it must be a decreasing G.P.

(C) the common ratio is always rational

(D) the first term is always an even integer

Ans: (A), (C), (D)

Let the G.P be a, ar, ar^{2}, ….

We have a + ar + ar^{2} + ar^{3} = 30

a^{2} + a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6} = 340

$\large \frac{(1 + r + r^2 + r^3)^2}{1 + r^2 + r^4 + r^6} = \frac{900}{340} = \frac{45}{17}$

$\large \frac{(r^4 -1)^2 (r^2 -1)}{(r-1)^2 (r^8 -1)} = \frac{45}{17} $ ; (since , r ≠ 1)

$\large \frac{(r^4 -1) (r+1)}{(r-1) (r^4 +1)} = \frac{45}{17} $

$\large \frac{r^4 + 2r^3 + 2r^2 + 2r + 1}{r^4 +1} = \frac{45}{17} $

14 r^{4} – 17 r^{3} – 17 r^{2} – 17 r + 14 = 0

Dividing by r^{2 }, we get 14 – 17 – 17 = 0

$\large 14 (r^2 + \frac{1}{r^2}) – 17(r + \frac{1}{r}) -17 = 0$

$\large 14 [(r + \frac{1}{r})^2 -2 ] – 17(r + \frac{1}{r}) -17 = 0$

Putting , $r + \frac{1}{r} = y $

14y^{2} – 17 y – 45 = 0

$\large y = \frac{17 \pm 53}{28}$

=> y = 5/2 , – 9/7

=> r + 1/r = 5/2 => r = 2, 1/2 and r + 1/r = –9/7 has no real roots.

Putting r = 2 in (1), we get a = 2 and putting r = 1/2 in (1), we get a = 16

Then the progression is 2, 4, 8, 16, 32, 64, …….. or, 16, 8, 4, 2, 1, 1/2 , ….

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