Question

The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.

Answer 1

Let the first term of the A.P. be a and the common difference be d.

$\therefore$ a = a, b = a + d and c = a + 2d

a + b + c = 18

$\Rightarrow a+(a+d)+(a+2d)=18$

$\Rightarrow 3a+3d=18$

$\Rightarrow a+d=6$ ........(i)

Now, according to the question, a + 4, a + d + 4 and a + 2d + 36 are in G.P.

$\therefore (a+d+4{)}^2=(a+4\left)\right(a+2d+36)$

$\Rightarrow (6-d+d+4{)}^2=(6-d+4\left)\right(6-d+2d+36)$

$\Rightarrow (6-d+d+4{)}^2=(6-d+4\left)\right(6-d+2d+36)$

$\Rightarrow (10{)}^2=(10-d\left)\right(42+d)$

$\Rightarrow 100=420+10d-42d-d^2$

$\Rightarrow d^2+32d-320=0$

$\Rightarrow (d+40)(d-8)=0$

$\Rightarrow d=8,-40$

Now, putting d = 8, -40 in equation (i), we get, a = -2, 46, respectively.

For a = - 2, and d = 8, we have:

a = -2, b = 6, c = 14

And, for a = 46 and d = - 40, we have;

a = 46, b = 6, c = -34