If a-a2-a3-an are first n terms of an AP with first term a-0 and common difference d- 0 Then value of a3-a2-a2-a4-a2-a3-a5-a2-a4-an-a2-an-1 is

The correct option is C n 2 Given, #160; a_3 a_2/a_2 + a_4 a_2/a_3 + a_5 a_2/a_4+ ............+a_n a_2/a_n 1 Since, a_n=a+(n 1)d #1 ...

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Arithmetic Progression

If a,a2,a3,...

Question

If $a, a_{2}, a_{3}, . . . ., a_{n}$ are first $n$ terms of an AP with first term $a = 0$ and common difference $d \neq 0$ Then value of $\frac{a_{3} - a_{2}}{a_{2}} + \frac{a_{4} - a_{2}}{a_{3}} + \frac{a_{5} - a_{2}}{a_{4}} . . . . + \frac{a_{n} - a_{2}}{a_{n - 1}}$ is

n

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\frac{(n - 1) n}{2}

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n - 2

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\frac{n (n + 1)}{2}

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Solution

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Similar questions

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

a_{1} = 0

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}

\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_{4}

a i \in N \forall i \in N

4 \sum i = 1 a_{i} = 4 (a_{3} - a_{2}) + 32

(a_{2} + \frac{a_{4}}{a_{3}})

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

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}}}

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