Question -
Answer -
Let the first term ofan A.P. be┬атАШaтАЩ┬аand its common difference beтАШdтАЩ.
a1┬а+ a2┬а+a3┬а= 15
Where, the threenumber are: a,┬аa + d, and a + 2d
So,
a + a + d + a + 2d =15
3a + 3d = 15 or a + d= 5
d = 5 тАУ a тАж (i)
Now, according to thequestion:
a + 1, a + d + 3, anda + 2d + 9
they are in GP, thatis:
(a+d+3)/(a+1) =(a+2d+9)/(a+d+3)
(a + d + 3)2┬а=┬а(a+ 2d + 9) (a + 1)
a2┬а+ d2┬а+9 + 2ad + 6d + 6a = a2┬а+ a + 2da + 2d + 9a + 9
(5 тАУ a)2┬атАУ4a + 4(5 тАУ a) = 0
25 + a2┬атАУ10a тАУ 4a + 20 тАУ 4a = 0
a2┬атАУ18a + 45 = 0
a2┬атАУ15a тАУ 3a + 45 = 0
a(a тАУ 15) тАУ 3(a тАУ 15)= 0
a = 3 or a = 15
d = 5 тАУ a
d = 5 тАУ 3 or d = 5 тАУ15
d = 2 or тАУ 10
Then,
For a = 3 and d = 2,the A.P is 3, 5, 7
For a = 15 and d =-10, the A.P is 15, 5, -5
тИ┤┬аThe numbers are3, 5, 7 or 15, 5, тАУ 5