Question
If $a_{1}, a_{2}, a_{3}, a_{4}$ are in H.P., then $\frac{1}{a_{1} a_{4}} \sum_{r=1}^{3} a_{r} a_{r+1}$ is aroot of(A) $x^{2}+2 x+15=0$(B) $x^{2}+2 x-15=0$(C) $x^{2}-6 x-8=0$(D) $x^{2}-9 x+20=0$
Step 1
This means that their reciprocals are in arithmetic progression. So, we have $\frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}} = \frac{1}{a_{4}} - \frac{1}{a_{3}} = d$, where $d$ is the common difference. Show more…
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