Question

The sum of three consecutive terms in a geometric progression is $14$ . If 1 is added to the first and the second terms and $1$ is subtracted from the third, the resulting new terms are in arithmetic progression. Then the lowest of original term is
A) $1$
B) $2$
C) $4$
D) $8$

Answer 1

GP stands for Geometric Progression. $a,ar,a{{r}^{2}},a{{r}^{3}},.....$are said to be in GP where first term is $a$ and common ratio is $r$ .The ${{n}^{th}}$ term is given by
${{n}^{th}}term=a{{r}^{n-1}}$
The sum of $n$ terms is given by$\dfrac{a(1-{{r}^{n}})}{(1-r)}$, when $r<1$ and$\dfrac{a({{r}^{n}}-1)}{(r-1)}$, when $r>1$.
If three numbers $a,b,c$ in order are in A.P. Then, $2b=a+c$
If $a,b,c$ are in A.P., then $b$ is called the arithmetic mean (AM) between $a$ and $c$, that is $b=\dfrac{a+c}{2}$
The sum ${{S}_{n}}$ of$n$ terms of an A.P. with first term and common difference is
{{S}_{n}}=\dfrac{n}{2}\left\\{ 2a+(n-1)d \right\\}

Complete step-by-step solution:
Let $a,ar,a{{r}^{2}}$ be the three consecutive terms in Geometric Progression.
According to the question
$a+ar+a{{r}^{2}}=14$
Taking $a$common we get
$a(1+r+{{r}^{2}})=14$
Also $a+1,ar+1,a{{r}^{2}}-1$ is in arithmetic progression
So we get
$2(ar+1)=a+1+a{{r}^{2}}-1$
$2(ar+1)=a+a{{r}^{2}}$
From the above equations we get
$2(ar+1)=14-ar$
$3ar=12$
Further solving we get
$r=\dfrac{4}{a}$
Substituting this we get
$a\left( 1+\dfrac{4}{a}+\dfrac{16}{{{a}^{2}}} \right)=14$
${{a}^{2}}+4a+16=14a$
Further solving we get
${{a}^{2}}-10a+16=0$
Factorising we get
$(a-8)(a-2)=0$
$a=2,8$
When $a=2$
$r=2$
The series is $2,4,8$
When $a=8$
$r=\dfrac{1}{2}$
The series is $8,4,2$.
Therefore, the lowest term is $2$.
Hence, option $2$is the correct answer.

Note: Geometric progression problems require knowledge of exponent properties. A sequence or a series is an arrangement of numbers in a definite border according to some pattern. These types of questions require the knowledge of Arithmetic progression.