
The first term of a G.P. is \[7\] , the last term is \[448\] , and the sum of all the terms is \[889\] , then the common ratio is
A. \[5\]
B. \[4\]
C. \[3\]
D. \[2\]
Answer
492.6k+ 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 \[r>1\] .
Geometric progression is the series of non-zero numbers in which each term after the first is found by multiplying the previous number by a fixed non-zero number called the common ratio.
If all the terms of G.P. are multiplied or divided by the same non-zero constant, then it remains a G.P. with the same common ratio. The geometric mean of \[a\] and \[b\] is given by \[\sqrt{ab}\] . The reciprocals of the terms of a given G.P. form a G.P. If each term of a G.P. is raised to the same power the resulting sequence also forms a G.P.
Complete step-by-step answer:
According to the question we observe that
First term, \[a=7\]
The last term, \[{{a}_{{{n}_{{}}}}}=a{{r}^{n-1}}\]
Also given that, \[{{a}_{{{n}_{{}}}}}=448\]
The sum of all term, \[{{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}\]
Sum of all terms given is
\[{{S}_{n}}=889\]
Equating the equations we get
\[\dfrac{a({{r}^{n}}-1)}{r-1}=889\]
Solving further we get
\[\dfrac{a{{r}^{n-1}}r-a}{r-1}=889\]
Substituting the value we get
\[448r-7=889r-889\]
Transposing the values from both the sides we get
\[889r-448r=889-7\]
Further solving we get
\[441r=882\]
Dividing both side by \[441\]
\[r=2\]
Thus, the common ratio is \[2\] .
Therefore, option (D) is the correct answer.
So, the correct answer is “Option D”.
Note: Geometric progression problems require knowledge of exponent properties. Exponent properties are also called laws of indices. Exponent is the power of the base value. Power is the expression that represents repeated multiplication of the same number. Exponent is the quantity representing the power to which the number is raised.
\[{{n}^{th}}term=a{{r}^{n-1}}\]
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 \[r>1\] .
Geometric progression is the series of non-zero numbers in which each term after the first is found by multiplying the previous number by a fixed non-zero number called the common ratio.
If all the terms of G.P. are multiplied or divided by the same non-zero constant, then it remains a G.P. with the same common ratio. The geometric mean of \[a\] and \[b\] is given by \[\sqrt{ab}\] . The reciprocals of the terms of a given G.P. form a G.P. If each term of a G.P. is raised to the same power the resulting sequence also forms a G.P.
Complete step-by-step answer:
According to the question we observe that
First term, \[a=7\]
The last term, \[{{a}_{{{n}_{{}}}}}=a{{r}^{n-1}}\]
Also given that, \[{{a}_{{{n}_{{}}}}}=448\]
The sum of all term, \[{{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}\]
Sum of all terms given is
\[{{S}_{n}}=889\]
Equating the equations we get
\[\dfrac{a({{r}^{n}}-1)}{r-1}=889\]
Solving further we get
\[\dfrac{a{{r}^{n-1}}r-a}{r-1}=889\]
Substituting the value we get
\[448r-7=889r-889\]
Transposing the values from both the sides we get
\[889r-448r=889-7\]
Further solving we get
\[441r=882\]
Dividing both side by \[441\]
\[r=2\]
Thus, the common ratio is \[2\] .
Therefore, option (D) is the correct answer.
So, the correct answer is “Option D”.
Note: Geometric progression problems require knowledge of exponent properties. Exponent properties are also called laws of indices. Exponent is the power of the base value. Power is the expression that represents repeated multiplication of the same number. Exponent is the quantity representing the power to which the number is raised.
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