5. In an A.P, the sum of three consecutive terms is 24. When 1 is subtracted to the first term, 2 to the
second, then the terms form a G.P. Find the terms of the A.P.

Now a-d-1, a-2, a+d form a GP. It means, the common ratios should be equal or the middle term should be the Geometric Mean of the first and third terms, or

(a-d-1)(a+d) = (a-2)^2 or

(8-d-1)(8+d) = (8-2)^2, or

(7-d)(8+d) = 36, or

56 -8d +7d -d^2 = 36, or

56 -8d +7d -d^2 - 36 = 0, or

d^2 +d -20 = 0

(d+5)(d-4) = 0

Thus d = -5 or 4

So the AP is 13,8,3 or 4,8,12

Check: 13,8,3 in AP becomes (13–1), (8–2), 3 = 12, 6, 3 which is a GP with r = (1/2)

4,8,12 in AP becomes (4–1), (8–2), 12 = 3, 6, 12 which is a GP with r =2

5. In an A.P, the sum of three consecutive terms is 24. When 1 is subtracted to the first term, 2 to thesecond, then the terms form a G.P. Find the terms of the A.P.​

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5. In an A.P, the sum of three consecutive terms is 24. When 1 is subtracted to the first term, 2 to the
second, then the terms form a G.P. Find the terms of the A.P.