Problem Solving

varun289 wrote:The 6th term of a geometric series falls between 4090 and 5010. What could possibly be the first term of the series if the first term and common ratio are equal?

A geometric progression with first term a and common ratio r is given as : a, ar, ar², ar³, ar�, ar�, ar�...

So, 6th term = ar�
Here, a = r
Hence, 6th term = a*(a�) = a�

So, 4090 < a� < 5010
--> 3600 < a� < 6400
--> 60 < a³ < 80

Check the options...

• A. a = 3 ---> a³ = 3³ = 27 < 60 --> NO
  B. a = 4 ---> a³ = 4³ = 64 ---> 60 < 64 < 80 --> YES
  C. a = 5 ---> a³ = 5³ = 125 > 80 --> NO
  No need to check the rest as they are greater than 5.

The correct answer is B.

hemant_rajput wrote:--> 3600 < a� < 6400

How you arrive at this ?

# Short answer:
As a� is greater than 4090, it'll be greater than 3600 too and as it is less than 5010, it'll be less than 6400 too.


# Long answer:
We know that 4090 < a� < 5010
Now, power of 6 is difficult to handle and numbers are pretty large.
So, we should try to manipulate this inequality in such a way that the numbers becomes smaller.

Now, power of a is 6.
This should be our clue that we can take either square root or cube root of a� easily. But that means we have to take the square roots or cube roots of 4090 and 5010 too. Taking square roots of these numbers will be easier than taking cube roots.

Now, taking square root of 4090 and 5010 is time consuming and calculation intensive process. Instead of that we can take the square roots of some known squares close to 4090 and 5010. Have a look at leftmost two digits of 4090 and 5010, they are 40 and 50. What are the perfect squares that are just less than 40 and just greater than 50? 36 and 64.

So, a� must be greater than 3600 and less than 6400.

Note that this is an approximation as 36 is the perfect square just less than 40, but 3600 is not the perfect square just less than 4090. Similarly, 6400 is not the perfect square just greater than 50. We are doing this to simplify the process of taking square root without doing much calculation. By doing this approximation we are expanding the range of a� and in turn the range of a. That's why we need to check one more option even after getting a possible value of a.

Hope that helps.