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Question: If the sequence \(\\{ {a_n}\\} \) is in GP, such that \(\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4}\) an...

If the sequence an\\{ {a_n}\\} is in GP, such that a4a6=14\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4} and a2+a5=216{a_2} + {a_5} = 216 , then a1{a_1} is equal to
a.1212 or 1087\dfrac{{108}}{7}
b.  10\;10
c.77or 54/754/7
d.None of these

Explanation

Solution

In this question, we have been given that the sequence in geometric progression or GP. We know the general form of the geometric expression is a,ar,ar2,ar3...a,ar,a{r^2},a{r^3}... , where aa is the first term of the progression and rr is the common ratio. We should know that we can write a4{a_4} as a1r3{a_1}{r^3} . Similarly, we will write the other term also and then we will first find the common ratio of the given sequence. After this, we will simplify the expression.

Complete answer:
Here we have been given in the question a4a6=14\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4} .
We can write
a4{a_4} =a1r3 = {a_1}{r^3} .
Similarly, by applying the same method we can write a6=a1r5{a_6} = {a_1}{r^5} .
By substituting these in the equation we have:
a1r3a1r5=14\Rightarrow \dfrac{{{a_1}{r^3}}}{{{a_1}{r^5}}} = \dfrac{1}{4}
We will now simplify this, we can break down the value in the denominator using exponential powers:
r3r2×r3=14\Rightarrow \dfrac{{{r^3}}}{{{r^2} \times {r^3}}} = \dfrac{1}{4}
Now we have
1r2=14\Rightarrow \dfrac{1}{{{r^2}}} = \dfrac{1}{4}
By cross multiplying the terms, it gives us
r2=4\Rightarrow {r^2} = 4
r=4\Rightarrow r = \sqrt 4
So we have the value of common ratio i.e.
r=±2r = \pm 2
It is given that
a2+a5=216{a_2} + {a_5} = 216 .
We will again break down the terms and it can be written as
a1r+a1r4=216\Rightarrow {a_1}r + {a_1}{r^4} = 216
Let us take case one, where we take the value of r=2r = 2
By applying this in the expression, we have:
2a1+24a1=216\Rightarrow 2{a_1} + {2^4}{a_1} = 216
2a1+16a1=216\Rightarrow 2{a_1} + 16{a_1} = 216
On adding the terms it gives us
18a1=216\Rightarrow 18{a_1} = 216
Therefore we have the value of a1{a_1} as
21618=12\Rightarrow \dfrac{{216}}{{18}} = 12
Now let us solve for case two, where we have a negative value of common ratio i.e. r=2r = - 2
By applying this in the expression, we have:
2a1+(2)4a1=216\Rightarrow - 2{a_1} + {( - 2)^4}{a_1} = 216
2a1+16a1=216\Rightarrow - 2{a_1} + 16{a_1} = 216
On simplifying the terms it gives us
14a1=216\Rightarrow 14{a_1} = 216
Therefore we have the value of a1{a_1} as
21614=1087\Rightarrow \dfrac{{216}}{{14}} = \dfrac{{108}}{7}

Hence the correct option is (a) 1212 or 1087\dfrac{{108}}{7}

Note:
We should know that Geometric progression or GP is a type of sequence in which each succeeding term is multiplied by multiplying each preceding term with a mixed number, which is called a common ratio. We should note that the formula of the nthnth term of a GP is an=arn1{a_n} = a{r^{n - 1}} .
By applying this we can write
a4=ar41\Rightarrow {a_4} = a{r^{4 - 1}} .
So it can also be written as ar3a{r^3}