Three positive number whose sum is 21 are in arithmetic progression, If 2, 2,14 are added to them respectively then resulting number are in geometric progression .Then which of the following is not among the three number ?
Step-by-step Solution:
Concept:
If a, b, c are in GP then \( \dfrac{b}{a} = \dfrac{c}{b} \)
Calculations:
Consider, three positive numbers \( a - d \), \( a \), and \( a + d \) are in AP.
These three positive numbers have a sum of 21:
Hence, \( a - d + a + a + d = 21 \)
Therefore, \( a = 7 \).
Hence, the three positive numbers are \( 7 - d \), \( 7 \), and \( 7 + d \).
It is given that if 2, 2, and 14 are added to them respectively, the resulting numbers are in geometric progression.
Thus, \( 9 - d \), \( 9 \), and \( 21 + d \) are in GP.
We can write the equation as:
\( \dfrac{9}{9 - d} = \dfrac{21 + d}{9} \)
Hence, \( 81 = (21 + d)(9 - d) \)
Which simplifies to:
\( d^2 + 12d - 108 = 0 \)
Factoring the quadratic equation:
\( (d + 18)(d - 6) = 0 \)
Therefore, \( d = -18 \) or \( d = 6 \).
When \( d = -18 \), the three numbers are 25, 7, and -11.
When \( d = 6 \), the three numbers are 1, 7, and 13.