If a 1 , a 2 , … , a n + 1 are in A.P., then 1 a 1 a 2 + 1 a 2 a 3 + ? + 1 a n a n + 1 is

Question

If $a_{1}, a_{2}, \dots, a_{n + 1}$ are in A.P., then $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots + \frac{1}{a_{n} a_{n + 1}}$ is

Smart Achievers · Accepted Answer

(4)

$Let d be the common difference of the A.P.$

$then d = a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{n + 1} - a_{n}$

$Now, \frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots + \frac{1}{a_{n} a_{n + 1}}$

$= \frac{1}{d} {\frac{d}{a_{1} a_{2}} + \frac{d}{a_{2} a_{3}} + \dots + \frac{d}{a_{n} a_{n + 1}}}$

$= \frac{1}{d} {\frac{a_{2} - a_{1}}{a_{1} a_{2}} + \frac{a_{3} - a_{2}}{a_{2} a_{3}} + \dots + \frac{a_{n + 1} - a_{n}}{a_{n} a_{n + 1}}}$

$= \frac{1}{d} {(\frac{1}{a_{1}} - \frac{1}{a_{2}}) + (\frac{1}{a_{2}} - \frac{1}{a_{3}}) + \dots + (\frac{1}{a_{n}} - \frac{1}{a_{n + 1}})}$

$= \frac{1}{d} {\frac{1}{a_{1}} - \frac{1}{a_{n + 1}}} = \frac{a_{n + 1} - a_{1}}{d \cdot a_{1} a_{n + 1}} = \frac{n d}{d \cdot a_{1} a_{n + 1}} = \frac{n}{a_{1} a_{n + 1}}$

Solution

Q.
If $a_{1}, a_{2}, \dots, a_{n + 1}$ are in A.P., then $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots + \frac{1}{a_{n} a_{n + 1}}$ is

1 $\frac{n - 1}{a_{1} a_{n + 1}}$

2 $\frac{1}{a_{1} a_{n + 1}}$

3 $\frac{n + 1}{a_{1} a_{n + 1}}$

4 $\frac{n}{a_{1} a_{n + 1}}$

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Solution

Q. If a1,a2,…,an+1 are in A.P., then 1a1a2+1a2a3+⋯+1anan+1 is

1 n-1a1an+1

2 1a1an+1

3 n+1a1an+1

4 na1an+1