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CAT 2019 Question Paper | Quants Slot 1

CAT Previous Year Paper | CAT Quants Questions | Question 20

Here is a tough question from Arithmetic progressions. As we have already seen with CAT previous year paper, progressions questions are either extremely easy or super tough. This falls into the latter bucket. Try cracking it. The key lies in doing a small operation that will unlock the solution. With good practice you can crack these kind of progressions questions.

Question 20 : If a1, a2, ......... are in A.P , \\frac{1}{\sqrt{a_1} + \sqrt{a_2}}) + \\frac{1}{\sqrt{a_2} + \sqrt{a_3}}) + ......... + \\frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}) then , is equal to

  1. \\frac{n}{\sqrt{a_1} + \sqrt{a_{n+1}}})
  2. \\frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}})
  3. \\frac{n}{\sqrt{a_1} - \sqrt{a_{n+1}}})
  4. \\frac{n - 1}{\sqrt{a_1} + \sqrt{a_{n-1}}})

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Explanatory Answer

Consider the first term
Multiply numerator and denominator by -
We get. We know, a2 - a1 = d
Similarly, repeat the same for the entire series.
We get+ ..... + = (- )
We do not know the value of d.
We know an+1 = a1 + nd
an+1 - a1 = nd
d =
=
So, our expression = =


The question is "If a1, a2, ......... are in A.P , \\frac{1}{\sqrt{a_1} + \sqrt{a_2}}) + \\frac{1}{\sqrt{a_2} + \sqrt{a_3}}) + ......... + \\frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}) then , is equal to"

Hence, the answer is \\frac{n}{\sqrt{a_1} + \sqrt{a_{n+1}}})

Choice A is the correct answer.

 

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