Question

In a G.P. The sum of three numbers is $14$, if $1$ is added to the first two numbers and the third number is decreased by $1$, the series becomes AP, find the geometric sequence.
A) $2,4,8$
B) $8,4,2$
C) $6,18,54$
D) $8,16,32$

Answer 1

In this question, we have to find the required geometric sequence.
First we need to find out the equations from the given sum by using the geometrical progression sum formula, then applying the second condition using the sum formula for A.P, we will get another equation .Solving these two equations we get the solution.

Formula used: The sum of the G.P. series is represented by $a + ar + a{r^2} + a{r^3} + ...... + a{r^{n - 1}}$
Where ‘a’ is the first term, and r is the factor between the terms called the "common ratio" and n is the number of terms in the G.P.
Property of A.P.:
${a_n} = \dfrac{1}{2}\left( {{a_{n - 1}} + {a_{n + 1}}} \right)$

Complete step-by-step answer:
It is given that, in a G.P. the sum of three numbers is $14$.
We need to find out the geometric sequence.
Using the sum formula for G.P. we get,
$a + ar + a{r^2} = 14$……..(i)
The above equation is found using the first condition. Since there are only three members in the sequence we get the above equation.
Also it is given that, when we add $1$ to the first two numbers and decrease the third number by $1$, the series becomes AP.
Since we increase 1 in the first two terms we get,
$\Rightarrow a + 1,ar + 1$
Also decrease of the third term will give us ${\rm{ }}a{r^2} - 1$
Using the A.P formula we get,
\Rightarrow$$$ar + 1 = \dfrac{1}{2}\left\\{ {\left( {a + 1} \right) + \left( {a{r^2} - 1} \right)} \right\\}$$ Let us multiply by 2 on both sides we get, \Rightarrow2ar + 2 = a + a{r^2}$$ Let us add $$ar$$ in both sides of the equation, $\Rightarrow3ar + 2 = a + ar + a{r^2} Let us now substitute (i) in the above equation we get, $\Rightarrow$$$3ar + 2 = 14
Let us now subtract by 2 on both sides and divide by three on both sides we get,
\Rightarrow$$$ar = \dfrac{{14 - 2}}{3} = \dfrac{{12}}{3} = 4$$ Let us now divide by r on both sides of the equation, we get, \Rightarrowa = \dfrac{4}{r}$$ Let us now substitute ‘a’ in (i) we get, $\Rightarrow\dfrac{4}{r} + \dfrac{4}{r} \times r + \dfrac{4}{r} \times {r^2} = 14 By cancelling the fraction we get, $\Rightarrow$$$\dfrac{4}{r} + 4 + 4r = 14
By solving the above equation we get,
\Rightarrow$$$\dfrac{{4 + 4r + 4{r^2}}}{r} = 14$$ Let us multiply by r on both sides we get, \Rightarrow4 + 4r + 4{r^2} = 14r$$ Let us subtract by 14r on both sides we get, $\Rightarrow4{r^2} - 10r + 4 = 0 Let us now factorize the above equation to get the value of r, $\Rightarrow$$$4{r^2} - 8r - 2r + 4 = 0
Let us take 4r in common from first two terms and -2 from next two terms we get,
\Rightarrow$$$4r\left( {r - 2} \right) - 2\left( {r - 2} \right) = 0$$ On solving the above equation we get, \Rightarrow\left( {r - 2} \right)\left( {4r - 2} \right) = 0$$ $\Rightarrowr = 2,r = \dfrac{2}{4} = \dfrac{1}{2}$Thus$r = 2{\rm{or}}\dfrac{1}{2}$Therefore, if$r = \dfrac{1}{2} we get, $\Rightarrow$$$a = \dfrac{4}{{\dfrac{1}{2}}} = 8
Let us now substitute the values of ‘a’ and ‘r’ in the sequence form we get,
The Sequence is $8,8 \times \dfrac{1}{2} = 4,8 \times \dfrac{1}{{{2^2}}} = 2$
The required G.P is $8,4,2$
If,$r = 2$ we get,
$\Rightarrow$$$a = \dfrac{4}{2} = 2$Let us now substitute the values of ‘a’ and ‘r’ in the sequence form we get, The Sequence is$2,2 \times 2 = 4,2 \times {2^2} = 8$The required G.P is$2,4,8$$.

Thus (A) and (B) are the correct options.

Additional Information: A geometric progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called a common ratio.
In General we write a Geometric Sequence like this:\left\\{ {a,ar,a{r^2},a{r^3},.......} \right\\} ,Where a is the first term, and r is the factor between the terms (called the "common ratio").

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. In General we write a Arithmetic Sequence like this:\left\\{ {a,a + d,a + 2d,a + 3d,.......} \right\\} ,Where a is the first term, and d is the difference between the terms (called the "common difference").

Note: While finding the value of $r$ the factorization can be done by using sridhar acharya's formula,
$r = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ Where $a = 4,b = - 10,c = 4$
We get, $r = \dfrac{{ - ( - 10) \pm \sqrt {{{( - 10)}^2} - 4(4)(4)} }}{{2(4)}} = \dfrac{{10 \pm \sqrt {100 - 64} }}{8} = \dfrac{{10 \pm 6}}{8}$
From this we get either $r = 2$ or $\dfrac{1}{2}$