Question

Find the numbers in AP whose sum is 12 and if 2 is added to these third term they form GP

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Let the 3 terms of the AP: a - d, a, a + d => (a - d)+(a)+(a + d) = 3a = 12 => a = 4.

The AP: 4-d, 4, 4+ d. When 1, 2, 6 are added respectively the series becomes (5-d), 6, (10+d) which is in GP => 6^2 = (5-d)(10+d). 36= 50- 5d - d^2 = d^2 + 5d - 14 = 0.

d^2 + 5d - 14 = 0 => (d - 2)(d + 7) = 0 => d =

2 or -7. The AP series 2, 4, 6 or 11, 4, -3.

Resulting GP: 3, 6, 12 or 12, 6, 3.