Question

The number which should be added to the numbers 2, 14, 62 so that the resulting numbers may be in G.P., is
A.1
B.2
C.3
D.4

Answer 1

Hint: Here in this question, we have to find the number which should be added to the given number to become a geometric progression (G.P). For this, first we need to consider the unknown number be \[x\] and next consider the formula of the geometric mean of GP i.e., \[{b^2} = ac\] . To further simplify by using an algebraic identity and arithmetic operations to get the required solution.

Complete step-by-step answer:
Geometric Progression (G.P.) is a type of sequence where each succeeding term of the sequence is produced by multiplying each preceding term by a fixed number, which is called a common ratio.
In general, the geometric progression is represented by:
  \[a\] , \[ar\] , \[a{r^2}\] , \[a{r^3}\] , and so on.
Where ‘ \[a\] ’ is the first term and ‘ \[r\] ’ is the common ratio.
If a, b and c are three quantities in GP, then b is the geometric mean of a and c. Mathematically this can be written as: \[{b^2} = ac\]
Consider the given question:
Let us take \[x\] be any number added to 2, 14, 62.
 So, that the resulting numbers may be in GP is
 \[2 + x\] , \[14 + x\] , \[62 + x\] -----(1)
The geometric mean of the above G.P is
 \[ \Rightarrow {\left( {14 + x} \right)^2} = \left( {2 + x} \right)\left( {62 + x} \right)\]
Apply a algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] in LHS and multiply the binomials in RHS, then we have
 \[ \Rightarrow {14^2} + {x^2} + 2\left( {14} \right)\left( x \right) = 124 + 2x + 62x + {x^2}\]
 \[ \Rightarrow 196 + {x^2} + 28x = 124 + 64x + {x^2}\]
Take variable \[x\] and its coefficient term to the LHS and constant term to the RHS, then
 \[ \Rightarrow {x^2} + 28x - {x^2} - 64x = 124 - 196\]
 \[ \Rightarrow - 36x = - 72\]
Divide both side by \[ - 36\] , then we get
 \[ \Rightarrow x = \dfrac{{ - 72}}{{ - 36}}\]
 \[\therefore x = 2\]
The required resultant G.P is:
 \[ \Rightarrow 2 + 2\] , \[14 + 2\] , \[62 + 2\]
 \[\therefore 4\] , \[16\] , \[64\]
Because, here the common ratio \[r = 4\] is multiplied with two successive terms.
Hence, the required number is \[x = 2\]
Therefore, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: We must know about the geometric progression arrangement and it is based on the first term and common ratio. The common ratio \[r\] is multiplied uniformly with successive terms and If three quantities are in GP, then the middle one is called the geometric mean of the other two terms.