Question
Question: If the sequence \(\\{ {a_n}\\} \) is in GP, such that \(\dfrac{{{a_4}}}{{{a_6}}} = \dfrac{1}{4}\) an...
If the sequence an is in GP, such that a6a4=41 and a2+a5=216 , then a1 is equal to
a.12 or 7108
b.10
c.7or 54/7
d.None of these
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... , where a is the first term of the progression and r is the common ratio. We should know that we can write a4 as a1r3 . 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 a6a4=41 .
We can write
a4 =a1r3 .
Similarly, by applying the same method we can write a6=a1r5 .
By substituting these in the equation we have:
⇒a1r5a1r3=41
We will now simplify this, we can break down the value in the denominator using exponential powers:
⇒r2×r3r3=41
Now we have
⇒r21=41
By cross multiplying the terms, it gives us
⇒r2=4
⇒r=4
So we have the value of common ratio i.e.
r=±2
It is given that
a2+a5=216 .
We will again break down the terms and it can be written as
⇒a1r+a1r4=216
Let us take case one, where we take the value of r=2
By applying this in the expression, we have:
⇒2a1+24a1=216
⇒2a1+16a1=216
On adding the terms it gives us
⇒18a1=216
Therefore we have the value of a1 as
⇒18216=12
Now let us solve for case two, where we have a negative value of common ratio i.e. r=−2
By applying this in the expression, we have:
⇒−2a1+(−2)4a1=216
⇒−2a1+16a1=216
On simplifying the terms it gives us
⇒14a1=216
Therefore we have the value of a1 as
⇒14216=7108
Hence the correct option is (a) 12 or 7108
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 nth term of a GP is an=arn−1 .
By applying this we can write
⇒a4=ar4−1 .
So it can also be written as ar3