If a1-a2-a3-an are in A-P of non-zero terms- Prove that 1a1a2-1a2a3-1a3a4-1an-1an-n-1a1an

Let d be the common difference of the A.P ∴ d=a_2 a_1=a_3 a_2=a_4 a_3=...=a_n a_n 1 Given 1a_1a_2+1a_2a_3+1a_3a_4+...1a_n 1a_nn 1a_1a_n =1d( ...

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Sum of n Terms

If a1,a2,a3...

Question

If $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ are in A.P of non-zero terms, Prove that $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}$

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a_{r} > 0; \forall r, n \in N

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

a_{1}, a_{2},

a_{n}

a_{1} = 0

\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \frac{a_{5}}{a_{4}} +

+ \frac{a_{n}}{a_{n - 1}} - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n - 2}}) = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}

a_{r} > 0, r \in N

a_{1}, a_{2}, a_{3}, . . ., a_{2 n}

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

a_{1}, a_{2}, a_{3}, . . . . . . a_{n}

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} . . . . + a_{n - 1} a_{n} =

\frac{1}{a_{1}}, \frac{1}{a_{2}}, \dots \frac{1}{a_{n}}

\frac{a_{n}}{a_{1} + a_{2} + \dots + a_{n - 1}}, . . . . ., \frac{a_{2}}{a_{1} + a_{3} + \dots + a_{n}} + \frac{a_{1}}{a_{2} + a_{3} + \dots + a_{n}}

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