If a1- a2- a3-an are first n terms of an A-P- with first term a1 - 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 D n 2 Given: #160; #160; a_1=0 ∴ a_2=d, a_3=2d,..... ∴a_3 a_2a_2+a_4 a_2a_3+a_5 a_2a_4+......+a_n a_2a_n 1 = ...

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Sum of n Terms

If a1, a2, ...

Question

If $a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}$ are first n terms of an A.P. with first term $a_{1} = 0$ and common difference $d \neq 0$ 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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a, a_{2}, a_{3}, . . . ., a_{n}

n

a = 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_{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}

\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_{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}})

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