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CAT 2019 Slot 1 — QA Question 24

Arithmetic ProgressionEasy

If a1, a2, ... are in A.P., then, 1a1+a2 + 1a2+a3 + ... + 1an+an+1 is equal to

Answer & solution

  • A

    na1-an+1

  • na1+an+1

  • C

    n-1a1+an-1

  • D

    n-1a1+an

Solution

Easy

Each term has the form 1ak+ak+1\dfrac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}. Rationalising the denominator turns every term into a difference of two square roots, so the whole sum telescopes. Use the common difference d=ak+1−akd=a_{k+1}-a_k to finish.

1

Rationalise a general term. Multiply numerator and denominator by ak+1−ak\sqrt{a_{k+1}}-\sqrt{a_k}.

1ak+ak+1=ak+1−akak+1−ak(conjugate)⇒ =ak+1−akd(d=common difference)\begin{aligned} &\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k}\quad\text{(conjugate)}\\ &\Rightarrow\ =\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}\quad\text{(}d=\text{common difference)} \end{aligned}
2

Add all terms — telescoping. Summing k=1k=1 to nn, intermediate roots cancel.

∑k=1nak+1−akd=an+1−a1d(telescope)\begin{aligned} &\sum_{k=1}^{n}\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}=\frac{\sqrt{a_{n+1}}-\sqrt{a_1}}{d}\quad\text{(telescope)} \end{aligned}
3

Express dd via a1a_1 and an+1a_{n+1}. In an AP, an+1=a1+nda_{n+1}=a_1+nd, so nd=an+1−a1nd=a_{n+1}-a_1.

an+1−a1d=n(an+1−a1)nd(multiply by n/n)⇒ =n(an+1−a1)an+1−a1(nd=an+1−a1)⇒ =n(an+1−a1)(an+1−a1)(an+1+a1)(factor an+1−a1)⇒ =na1+an+1(cancel)\begin{aligned} &\frac{\sqrt{a_{n+1}}-\sqrt{a_1}}{d}=\frac{n\left(\sqrt{a_{n+1}}-\sqrt{a_1}\right)}{nd}\quad\text{(multiply by }n/n)\\ &\Rightarrow\ =\frac{n\left(\sqrt{a_{n+1}}-\sqrt{a_1}\right)}{a_{n+1}-a_1}\quad\text{(}nd=a_{n+1}-a_1)\\ &\Rightarrow\ =\frac{n\left(\sqrt{a_{n+1}}-\sqrt{a_1}\right)}{\left(\sqrt{a_{n+1}}-\sqrt{a_1}\right)\left(\sqrt{a_{n+1}}+\sqrt{a_1}\right)}\quad\text{(factor }a_{n+1}-a_1)\\ &\Rightarrow\ =\frac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}\quad\text{(cancel)} \end{aligned}
na1+an+1\frac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}

Plug numbers: take a1=a2=a3=1a_1=a_2=a_3=1 (so n=2n=2). The sum is 12+12=1\tfrac12+\tfrac12=1. Only na1+an+1=21+1=1\dfrac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}=\dfrac{2}{1+1}=1 matches.

CAT 2019 Slot 1 QA Q24: If a 1 , a 2 , ... are in A.P., then, 1 a 1 + a 2 + 1 a 2 + a 3 + ... + 1 a n + a n + 1 is equal to — Solution | TheCATExam