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Question: If \(a_{1},a_{2},......,a_{n + 1}\) are in A.P., then \(\frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} ...

If a1,a2,......,an+1a_{1},a_{2},......,a_{n + 1} are in A.P., then 1a1a2+1a2a3+.....+1anan+1\frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} + ..... + \frac{1}{a_{n}a_{n + 1}} is

A

n−1a1an+1\frac{n - 1}{a_{1}a_{n + 1}}

B

1a1an+1\frac{1}{a_{1}a_{n + 1}}

C

n+1a1an+1\frac{n + 1}{a_{1}a_{n + 1}}

D

n−1a1an+1\frac{n - 1}{a_{1}a_{n + 1}}

Answer

n−1a1an+1\frac{n - 1}{a_{1}a_{n + 1}}

Explanation

Solution

S=1a1a2+1a2a3+....+1anan+1=(1a1−1a2)(a2−a1)+(1a2−1a3)(a3−a2)+......+(1an−1an+1)(an+1−an)S = \frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} + .... + \frac{1}{a_{n}a_{n + 1}} = \frac{\left( \frac{1}{a_{1}} - \frac{1}{a_{2}} \right)}{(a_{2} - a_{1})} + \frac{\left( \frac{1}{a_{2}} - \frac{1}{a_{3}} \right)}{(a_{3} - a_{2})} + ...... + \frac{\left( \frac{1}{a_{n}} - \frac{1}{a_{n + 1}} \right)}{(a_{n + 1} - a_{n})}As a1,a2,a3,.......,an,an+1a_{1},a_{2},a_{3},.......,a_{n},a_{n + 1} are in A.P., i.e.

a2−a1=a3−a2=.........=an+1−an=da_{2} - a_{1} = a_{3} - a_{2} = ......... = a_{n + 1} - a_{n} = d (say)

∴ S=1d[(1a1−1a2)+(1a2−1a3)+......+(1an−1an+1)]=S = \frac{1}{d}\left\lbrack \left( \frac{1}{a_{1}} - \frac{1}{a_{2}} \right) + \left( \frac{1}{a_{2}} - \frac{1}{a_{3}} \right) + ...... + \left( \frac{1}{a_{n}} - \frac{1}{a_{n + 1}} \right) \right\rbrack = $$\frac{1}{d}\left\lbrack \frac{1}{a_{1}} - \frac{1}{a_{n + 1}} \right\rbrack = \frac{a_{n + 1} - a_{1}}{d.a_{1}.a_{n + 1}} = \frac{\lbrack a_{1} + (n + 1 - 1)d\rbrack - a_{1}}{d.a_{1}.a_{n + 1}}

S=ndda1an+1=na1an+1S = \frac{nd}{da_{1}a_{n + 1}} = \frac{n}{a_{1}a_{n + 1}}