Question

The sum of three numbers which are consecutive terms an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

Answer 1

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

$\therefore a_1+a_2+a_3=21$

$\Rightarrow a+(a+d)+(a+2d)=21$

$\Rightarrow 3a+3d=21$

$\Rightarrow a+d=7$ ..........(i)

Now, according to the questiong :

$a,a+d-1$ and $a+2d+1$ are in G.P.

$\Rightarrow (a+d-1{)}^2=a(a+2d+1)$

$\Rightarrow (7+a-a-1{)}^2=a[a+2(7-a)+1]$

$\Rightarrow (6{)}^2=a(15-a)$

$\Rightarrow 36=15a-a^2$

$\Rightarrow a^2-15a+36=0$

$\Rightarrow (a-3)(a-12)=0$

$\Rightarrow a=3,12$

Now, putting a = 2, 12 in equation (i), we get d = 5, -5, respectively.

Thus, for a = 2 and d = 5, the numbers are 2, 7 and 12.

And, for a = 12 and d = -5, the numbers are 12, 7 and 2.