Question

If the the number 1.4 and 3 subtracted from three consecutive term of an A.P..the resulting numbers are in G.P. find the number if their sum us 21

Answer 1

Answer:

Let the three consecutive geometric numbers be x , y , z .

Since, these numbers in geometric series, hence y2=xz ……..(1)

Since their sum is 13 , hence x+y+z=13 ……..(2)

Also, it is given that, x+1 , y+4 , z+3 are in arithmetic sequence, hence difference between consecutive terms must be constant.

Hence, (y+4)−(x+1)=(z+3)−(y+4)

y+4−x−1=z+3−y−4

x−2y+z=4 ……..(3)

Subtracting Eq.(2) from E(1) , we get, 3y=9 hence, y=3

Hence, Eq.(2) & (3) becomes, x+z=10 ……..(4)

From Eq.(1), we have y2=xz

Using values of ‘ y ’ and Eq.(4), above relation becomes, 32=x(10−x)

Hence, 9=10x−x2

So, x2−10x+9=0

i.e. (x−1)(x−9)=0

Hence, x=1 or x=9

From Eq.(4), we get z=9 when x=1

And also, we get z=1 when x=9

Hence, 1 , 3 , 9 or 9 , 3 , 1 is the sequence of the required numbers in geometric series.

Note here that:

1 , 3 , 9 is geometric series with first terms as ‘ 1 ’ and common ration of ‘ 3 ’ .

9 , 3 , 1 is geometric series with first terms as ‘ 9 ’ and common ration of ‘ 13 ’