Question

If in a certain G.P. \[\mathop S\nolimits_6 \]= 126, and \[\mathop S\nolimits_3 \]= 14, find a and r.

Answer 1

Hint: A sequence is said to be in a geometric progression or G.P., if the ratio of any term to its preceding term is the same throughout.
This constant factor is called the common ratio.
The first term of the G.P. is denoted by ‘a’ and its common ratio by ‘r’.
The sum \[\mathop S\nolimits_n \]of the first $n$ terms of G.P. is the key formula to solve these types of questions.
For solving two unknowns, minimum two equations are required.

Complete step by step solution:
Step 1
Given that:
Sum of first 3 numbers in G.P. = 14
Sum of first 6 numbers in G.P. = 126
Step 2
We have to find:
first number, ‘a’ and common ratio, ‘r’
Step 3: state the formula for :
Sum of first n terms in G.P. \[ = \dfrac{{a(\mathop r\nolimits^n - 1)}}{{(r - 1)}}\] $r \ne 1$
Where, a is the first term of the G.P. and r is the common ratio.
Step 4: substituting given value of \[\mathop S\nolimits_3 \]:
Therefore, the sum of the first 3 numbers in G.P. , \[\mathop S\nolimits_3 \]\[ = \dfrac{{a(\mathop r\nolimits^3 - 1)}}{{(r - 1)}}\] = 14 (given)
                                                                                                                                                       …… (1)
Step 5: substituting given value of \[\mathop S\nolimits_6 \]
Therefore, the sum of the first 6 numbers in G.P. , \[\mathop S\nolimits_6 \]\[ = \dfrac{{a(\mathop r\nolimits^6 - 1)}}{{(r - 1)}}\] = 126 (given)
                                                                                                                                                       …… (2)
Step 6:
On Dividing (1) and (2) ‘a’ and ‘(r – 1)’ gets cancelled, equation left only in term of ‘r’:
\[\dfrac{{\mathop S\nolimits_3 }}{{\mathop S\nolimits_6 }} = \] \[{\rm{ }}\dfrac{{\dfrac{{a(\mathop r\nolimits^3 - 1)}}{{(r - 1)}}}}{{\dfrac{{a(\mathop r\nolimits^6 - 1)}}{{(r - 1)}}}} = \dfrac{{14}}{{126}}\]
\[\begin{array}{l}
 \Rightarrow {\rm{ }}\dfrac{{\dfrac{{{a}(\mathop r\nolimits^3 - 1)}}{{{{(r - 1)}}}}}}{{\dfrac{{{a}(\mathop r\nolimits^6 - 1)}}{{{{(r - 1)}}}}}} = \dfrac{1}{9}\\
 \Rightarrow {\rm{ }}\dfrac{{(\mathop r\nolimits^3 - 1)}}{{(\mathop r\nolimits^6 - 1)}} = \dfrac{1}{9}\\
 \Rightarrow {\rm{ }}9(\mathop r\nolimits^3 - 1) = (\mathop r\nolimits^6 - 1)\\
 \Rightarrow {\rm{ }}9\mathop r\nolimits^3 - 9 = \mathop r\nolimits^6 - 1\\
 \Rightarrow \mathop r\nolimits^6 - 9\mathop r\nolimits^3 + 8 = 0
\end{array}\]
Take $\mathop r\nolimits^3 = t$ so that equation will be converted into quadratic equation of type: $a\mathop x\nolimits^2 + bx + c$
$\therefore {\rm{ }}\mathop t\nolimits^2 - 9t + 8 = 0$
Let’s solve this quadratic equation by middle term splitting method:
$\begin{array}{l}
\therefore {\rm{ }}\mathop t\nolimits^2 - 8t - t + 8 = 0\\
 \Rightarrow {\rm{ }}t(t - 8) - 1(t - 8) = 0\\
 \Rightarrow {\rm{ }}(t - 8)(t - 1) = 0
\end{array}$
$\begin{array}{l}
t - 8 = 0\\
 \Rightarrow t = 8\\
 \Rightarrow \mathop r\nolimits^3 = 8
\end{array}$
$ \Rightarrow r = 2$
$\begin{array}{l}
t - 1 = 0\\
 \Rightarrow t = 1\\
 \Rightarrow \mathop r\nolimits^3 = 1
\end{array}$
$ \Rightarrow r = 1$
Step 8
Put r = 2 in equation (1) to find a,
\[\mathop S\nolimits_3 \]\[ = \dfrac{{a(\mathop r\nolimits^3 - 1)}}{{(r - 1)}}\] = 14
\[\begin{array}{l}
 \Rightarrow \dfrac{{a(\mathop 2\nolimits^3 - 1)}}{{(2 - 1)}} = 14\\
 \Rightarrow {\rm{ }}\dfrac{{a(8 - 1)}}{1} = 14\\
 \Rightarrow {\rm{ }}a(7) = 14\\
 \Rightarrow {\rm{ }}a = \dfrac{{14}}{7}
\end{array}\]
Hence, a = 2.
Put r = 1 in equation (1) to find a,
\[\mathop S\nolimits_3 \]\[ = \dfrac{{a(\mathop r\nolimits^3 - 1)}}{{(r - 1)}}\] = 14
\[ \Rightarrow \dfrac{{a(\mathop 1\nolimits^3 - 1)}}{{(1 - 1)}} = 14\]
Denominators cannot be 0 (zero). Therefore, r = 1 is discarded.

The first term of the G.P. ‘a’ is equal to 2 and the common ratio ‘r’ is equal to 2.

Note:
The general term or the $\mathop n\nolimits^{th} $of the G.P. is given by $\mathop a\nolimits_n = a\mathop r\nolimits^{n - 1} .$
Where a is the first term, r is the common ratio.
The geometric mean, G.M. of any two positive numbers a and b is given by $\sqrt {ab} $ i.e., the sequence a, G, b is G.P.