Question

Three number are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. Find the numbers.

Answer 1

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

$a_1+a_2+a_3=15$

$\Rightarrow a+(a+d)+(a+2d)=15$

$\Rightarrow 3a+3d=15$

$\Rightarrow a+d=5$ ......(i)

Now, according to the question :

$a+1,a+d+3anda+2d+9$ are in G.P.

$\Rightarrow (a+d+3{)}^2=(a+1\left)\right(a+2d+9)$

$\Rightarrow (5-d+d+3{)}^2=(5-d+1\left)\right(5-d+2d+9)$ [From (i)]

$\Rightarrow (8{)}^2=(6-d\left)\right(14+d)$

$\Rightarrow 64=84+6d-14d-d^2$

$\Rightarrow d^2+8d-20=0$

$\Rightarrow (d-2)(d+10)=0$

$\Rightarrow d=2,-10$

Now, putting d = 2, -10 in equation (i),

we get, a = 3, 15, respectively.

Thus, for a = 3 and d = 2, the A.P. is 3, 5, 7.

And, for a = 15 and d = - 10, the A.P. is 15, 5, -5.