
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
480.6k+ views
Hint: 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.
\[{{n}^{th}}term=a{{r}^{n-1}}\]
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.
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