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Harmonic Progression

Let the seque...

Question

Let the sequence $a_{1}, a_{2},$ ________ $a_{n}$ form an A.P. and let $a_{1} = 0$ , prove that $\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \frac{a_{5}}{a_{4}} +$ ________ $+ \frac{a_{n}}{a_{n - 1}} - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n - 2}}) = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}$ .

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Solution

Let d be the common difference of the given A.P.
Then since $a_{1} = 0$ , we have $a_{2} = d$ ,
$a_{3} = 2 d$ ,.... $a_{n} = (n - 1) d$ .
Hence L.H.S. $= \frac{2 d}{d} + \frac{3 d}{2 d} + \frac{4 d}{3 d} + . . . + \frac{(n - 1) d}{(n - 2) d} - d (\frac{1}{d} + \frac{1}{2 d} + \frac{1}{3 d} + . . . + \frac{1}{(n - 3) d})$
$= (1 + 1) + (1 + \frac{1}{2}) + (1 + \frac{1}{3}) + . . . + (1 + \frac{1}{n - 3}) + (1 + \frac{1}{n - 2}) - (1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n - 3})$
$= 1 + 1 + 1 + . . .$ to $(n - 2)$ terms $+ \frac{1}{n - 2}$
$= (n - 2) + \frac{1}{n - 2} = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}$ .

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Similar questions

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}}

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

are first n terms of an A.P. with first term

a_{1} = 0

and common difference

d \neq 0

\frac{a_{3} - a_{2}}{a_{2}} + \frac{a_{4} - a_{2}}{a_{3}} + \frac{a_{5} - a_{2}}{a_{4}} . . . . . . . . . . . . \frac{a_{n} - a_{2}}{a_{n - 1}}

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

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

are in A.P, then

\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_{r} > 0, r \in N

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

are in A.P. then

\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}}}

x + 1

f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0

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