There is this question in one of my math textbooks which I can't seem to figure out how to solve, it'd be awesome if you could help me :
If a1,a2,a3,...,an IS an arithmetic progression and an is NOT equal to 0 then prove the following statement :
1a1a2+1a2a3+1a3a4+...+1an−1an=n−1a1an
Thanks in advance...
Answer
Here's sketch, not complete proof.
Since A is an arithmetic progression, Let ai+1−ai=d
1aiai+1=dd1aiai+1=1dai+1−aiaiai+1since ai+1−ai=d=1d(1ai−1ai+1)
Therefore, 1a1a2+1a2a3+⋯+1an−1an=n−1∑i=11aiai+1=1dn−1∑i=11ai−1ai+1=1d(1a1−1a2+1a2−⋯−1an−1+1an−1−1an)=1d(1a1−1an)
Now, take LCM and use the fact that an=a1+(n−1)d
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