Q9 - Arithmetic progression problem

Q9:

If an integer is added to each of the numbers 36, 300, 596, we get the square of three consecutive terms of an AP. Find K/25.
Answer:37
Solution: Let the three consecutive terms in AP be \( a - d, a, a + d \). $$ (a - d)^2 = 36 + k a^2 = 300 + k . (a + d)^2 = 596 + k . $$ Subtracting (I) from (III): \[ (a + d)^2 - (a - d)^2 = (596 + k) - (36 + k) \Rightarrow 4ad = 560 \] \[ ad = 140 \] Subtracting (II) from (III): \[ (a + d)^2 - a^2 = (596 + k) - (300 + k) \Rightarrow d^2 + 2ad = 296 \] Substitute \( ad = 140 \): \[ d^2 + 280 = 296 \Rightarrow d^2 = 16 \Rightarrow d = \pm 4 \] So a = 35.
k = 35*35 - 300 => K/25 = 49-12 = 37.

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