If a1- a2- a3- - an are in A-P- where a1 - 0 - i then 1-a1-a2-1-a2-a3-1-an-1-an-k-a1-an- The value of k is

The correct option is C n 1 T _ 1 = 1 √( a _ 1 ) +√( a _ 2 ) #160; =√( a _ 2 ) √( a _ 1 ) #160; a _ 2 a _ 1 (Rationalizing) =√( a _ 2 ) ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

If a1, a2, ...

Question

If $a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}$ are in A.P. where $a_{1} > 0$ $\forall$ i then

$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{k}{\sqrt{a_{1}} + \sqrt{a_{n}}}$ . The value of $k$ is

1 - n

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n - 1

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n

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n + 1

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Solution

The correct option is C $n - 1$
$T_{1} = \frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} = \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}}$ (Rationalizing)

$= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{d}$

$T_{1} = \frac{\sqrt{a_{2}}}{d_{1}} - \frac{\sqrt{a_{1}}}{d}$

$T_{2} = \frac{\sqrt{a_{3}}}{d} - \frac{\sqrt{a_{2}}}{d}$

$T_{3} = \frac{\sqrt{a_{4}}}{d} - \frac{\sqrt{a_{3}}}{d}$
.
.
.
.
$T_{n} = \frac{\sqrt{a_{n}}}{d} - \frac{\sqrt{a_{n - 1}}}{d}$

$S_{n} = \frac{\sqrt{a_{n}}}{d} - \frac{\sqrt{a_{1}}}{d}$

$∴ S_{n} = \frac{\sqrt{a_{n}} - \sqrt{a_{1}}}{d}$

$= \frac{\sqrt{a_{n}} - \sqrt{a_{1}}}{\frac{a_{n} - a_{1}}{(n - 1)}}$

$= \frac{n - 1}{\sqrt{a_{n}} + \sqrt{a_{1}}}$

$⟹ K = n - 1$

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If a1,a2,a3,.......an are in A.P. where a1>0 ∀ i then 1√a1+√a2+1√a2+√a3+.....+1√an−1+√an=k√a1+√an. The value of k is

The correct option is C n−1T1=1√a1+√a2=√a2−√a1a2−a1 (Rationalizing)=√a2−√a1dT1=√a2d1−√a1dT2=√a3d−√a2dT3=√a4d−√a3d....Tn=√and−√an−1dSn=√and−√a1d ∴Sn=√an−√a1d =√an−√a1an−a1(n−1) =n−1√an+√a1⟹K=n−1

If $a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}$ are in A.P. where $a_{1} > 0$ $\forall$ i then

$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{k}{\sqrt{a_{1}} + \sqrt{a_{n}}}$ . The value of $k$ is