Question

If the sequence \[\left\{ {{a_n}} \right\}\] is in GP, such that \[\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4}\] and \[{a_2} + {a_5} = 216\], then find \[{a_1}\] .
A.12 or \[\dfrac{{108}}{7}\]
B.10
C.7 or \[\dfrac{{54}}{7}\]
D. None of these.

Answer 1

Hint: Write the nth term formula of GP, then apply the given conditions \[\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4}\] and \[{a_2} + {a_5} = 216\], then obtain r as \[ \pm 2\] from the condition \[\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4}\] and put in the condition \[{a_2} + {a_5} = 216\] to obtain the values of a.

Formula Used: The nth term formula of a GP \[a,ar,a{r^2},...\] is \[{t_n} = a{r^{n - 1}}\] .

Complete step by step solution: It is given that \[\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4}\] and \[{a_2} + {a_5} = 216\].
Now,
\[{a_2} = ar,{a_4} = a{r^3},{a_5} = a{r^4},{a_6} = a{r^5}\] .
So, \[\dfrac{{a{r^3}}}{{a{r^5}}} = \dfrac{1}{4}\]
\[\dfrac{1}{{{r^2}}} = \dfrac{1}{4}\]
\[{r^2} = 4\]
\[r = \pm 2\]
Now,
\[\begin{array}{l}ar + a{r^4} = 216\\ar(1 + {r^3}) = 216 - - - - (1)\end{array}\]
Put \[r = 2\] in the equation (1) to obtain the value of a.
\[2a(1 + {2^3}) = 216\]
\[18a = 216\]
\[a = 12\]
Again, put \[r = - 2\] in the equation (1) to obtain another value of a.
 \[( - 2)a(1 + {( - 2)^3}) = 216\]
\[14a = 216\]
\[a = \dfrac{{108}}{7}\]

Option ‘A’ is correct

Additional Information: A geometric progression is also known as geometric sequence. The ratios of two consecutive terms in a geometric sequence is always constant. The constant is known as common ratio.


Note: Sometimes students only take the positive value of r but not the negative one, that does not give the correct answer, as in GP series the common ratio can be negative or positive so we have to take both the values of r.