If a1- a2- an are in A-P- where ai - 0 for all ii- show that 1-a1-a2-1-a2-a3 -1-an-1-an-n-1-a1-an-

T_1=1√(a_1)+√(a_2)=√(a_1) √(a_2) d ∴ S_n=∑ T_1= 1d[√(a_1) √(a_n)] ∵ The terms will cancel diagonally ∴ S_n= 1da_1 a_n√(a_1)+√(a_n) = 1d ...

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Byju's Answer

Standard XII

Mathematics

nth Term of A.P

If a1, a2, ...

Question

If $a_{1}, a_{2},$ ________ $a_{n}$ are in A.P. where $a_{i} > 0$ for all ii, show that $\frac{1}{\sqrt{(a_{1})} + \sqrt{(a_{2})}} + \frac{1}{\sqrt{(a_{2})} + \sqrt{(a_{3})}}$ _______ $+ \frac{1}{\sqrt{(a_{n - 1})} + \sqrt{(a_{n})}} = \frac{n - 1}{\sqrt{(a_{1})} + \sqrt{(a_{n})}}$ .

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Solution

$T_{1} = \frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} = \frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{- d}$
$∴ S_{n} = \sum T_{1} = - \frac{1}{d} [\sqrt{a_{1}} - \sqrt{a_{n}}]$
$∵$ The terms will cancel diagonally
$∴ S_{n} = - \frac{1}{d} \frac{a_{1} - a_{n}}{\sqrt{a_{1}} + \sqrt{a_{n}}}$
$= - \frac{1}{d} [\frac{- (n - 1) d}{\sqrt{a_{1}} + \sqrt{a_{n}}}] = \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}$ .

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If a1,a2,________an are in A.P. where ai>0 for all ii, show that 1√(a1)+√(a2)+1√(a2)+√(a3)_______+1√(an−1)+√(an)=n−1√(a1)+√(an).

T1=1√a1+√a2=√a1−√a2−d∴Sn=∑T1=−1d[√a1−√an]∵ The terms will cancel diagonally∴Sn=−1da1−an√a1+√an=−1d[−(n−1)d√a1+√an]=n−1√a1+√an.

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