Question

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Answer 1

hey mate
here's the solution

Answer 2

Let the required numbers be (a-d),a and (a+d).

Then,

$(a - d) + a + (a + d) = 15 \\ \\ 3a = 15 \\ \\ a = \frac{15}{3} \\ \\ a = 5$

So, the numbers are (5-d) , 5 and (5+d) .

Adding 1, 3, 9 respectively to these numbers, we will get the numbers as (6-d) , 8 and (14 + d).

These numbers are in G.P.

By using the condition of G.P.

${8}^{2} = (6 - d)(14 + d) \\ \\ 64 = 84 + 8d - {d}^{2} \\ \\ {d}^{2} + 8d - 20 = 0 \\ \\ {d}^{2} + 10d - 2d - 20 = 0 \\ \\ d(d - 10) - 2(d - 10) = 0 \\ \\ (d + 10)(d - 2) = 0 \\ \\ d = - 10 \: \: or \: \: 2.$
So , the required numbers are (3,5,7) or (15,5,-5).