If {a}₁, {a}₂, , {a}ₙ are in AP then {1}{≤ft({a₁}+{a₂})}+{1}{≤ft({a₂}+{a₃})}++{1}{≤ft({aₙ-1}}+{aₙ})} is equal to

Correct Answer is : (n−1)(a1+an){(n-1)}{≤ft({{a}₁}+{{a}_{{n}}})}(a1+an)(n−1)

{(n-1)}{≤ft({{a}₁}+{{a}_{{n}}})}

{n}{≤ft({{a}₁}+{{a}_{{n}+1}})}

{(n-1)}{≤ft({a₁}+{aₙ-1}})}

{n}{≤ft({{a}₁}-{{a}_{{n}+1}})}

Correct Answer is : (n−1)(a1+an){(n-1)}{≤ft({{a}₁}+{{a}_{{n}}})}(a1+an)(n−1)

Test Index

MAT Exam 2022 Paper

Question : 117 of 200

Marks: +1, -0

\mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_n

\frac{1}{\left(\sqrt{a_{1}}+\sqrt{a_{2}}\right)}+\frac{1}{\left(\sqrt{a_{2}}+\sqrt{a_{3}}\right)}+\ldots+\frac{1}{\left(\sqrt{a_{n-1}}+\sqrt{a_{n}}\right)}

is equal to

$\frac{(n-1)}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right)}$
$\frac{n}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}+1}}\right)}$
$\frac{(n-1)}{\left(\sqrt{a_{1}}+\sqrt{a_{n-1}}\right)}$
$\frac{n}{\left(\sqrt{\mathrm{a}_{1}}-\sqrt{\mathrm{a}_{\mathrm{n}+1}}\right)}$

Solution:

Concept:

Rationalise each term and use the common difference of an AP to create a telescoping sum.

Explanation:

Let the common difference of the AP be

d

Then,

a_{r+1}-a_r = d

Rationalising the general term:

\frac{1}{\sqrt{a_r}+\sqrt{a_{r+1}}} = \frac{\sqrt{a_{r+1}}-\sqrt{a_r}}{a_{r+1}-a_r} = \frac{\sqrt{a_{r+1}}-\sqrt{a_r}}{d}

So, each term becomes

\frac{\sqrt{a_{r+1}}-\sqrt{a_r}}{d}

Adding from

r=1

r=n-1

, all middle terms cancel:

S = \frac{\sqrt{a_n}-\sqrt{a_1}}{d}

Also,

a_n-a_1=(n-1)d

, hence

d=\frac{a_n-a_1}{n-1}

Substitute this value of

d

S = \frac{\sqrt{a_n}-\sqrt{a_1}}{a_n-a_1}(n-1)

Since

a_n-a_1 = (\sqrt{a_n}-\sqrt{a_1})(\sqrt{a_n}+\sqrt{a_1})

, we get:

S = \frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}

This matches option A.

Answer:

Option A:

\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}

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