Question

if 1,4,a. are in GP then a is​

Answer 1

In geometric progression, the series have a common ratio.

let the series is a1 , a2 ,a3 is 1,4,a

the common ratio is a2÷a1 = 4÷1 = 4

so a3 ÷ a2 should also be equal to 4.

therefore ,

a÷4 = 4

therefore a = 16.

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Answer 2

Answer:

the $a$ in the geometric progression is $16$.

Step-by-step explanation:

As per the data in the question, a geometric progression is given.

The given geometric progression is $1, 4, a..$

We know that geometric progression (GP) is a sequence, in which each term after the first term is multiplying with a constant value. This constant value is known as common ratio ($r$).

Let us take the terms in the geometric progression as   $a_{1} ,a_{2} ,a_{3} ..$

Compare standard  geometric progression with the question.

Take first term term of GP, $a_{1} =1$

Second term of GP,  $a_{2} =4$

Third term of GP, $a_{3} =a$

We have to find the common ratio in geometric progression to find the third term.

Therefore,

The common ratio in geometric progression,  $r=\frac{a_{2}}{a_{1} }$

⇒ $r=\frac{4}{1}$

⇒ $r=4$

The common ratio of this geometric progression is $4$,

Therefore the third term is,

$a_{3} =a= a_{2}$ × $r$

Substitute the values,

⇒ $a= 4$×$4=16$

Hence the $a$ in the geometric progression is $16$.

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