If a1, a2, a3, a4, a5 are in H.P then a1a2 + a2a3 + a3a4 + a4a5 is equal to

1a2−1a1=1a3−1a2=1a4−1a3=1a5−1a4=da1a2+a2a3+a3a4+a4a5=a1−a5dand1a5=1a1+4d⇒a1−a5=4da1a5⇒a1−a5d=4a1a5

If a1, a2, a3, a4, a5 are in H.P then a1a2 + a2a3 + a3a4 + a4a5 is equal to

1a2−1a1=1a3−1a2=1a4−1a3=1a5−1a4=da1a2+a2a3+a3a4+a4a5=a1−a5dand1a5=1a1+4d⇒a1−a5=4da1a5⇒a1−a5d=4a1a5

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If a₁, a₂, a₃, a₄, a₅ are in H.P then a₁a₂ + a₂a₃ + a₃a₄ + a₄a₅ is equal to

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2a₁a₅

3a₁a₅

4a₁a₅

–4

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}} = \frac{1}{a_{4}} - \frac{1}{a_{3}} = \frac{1}{a_{5}} - \frac{1}{a_{4}} = d \\ a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + a_{4} a_{5} = \frac{a_{1} - a_{5}}{d} \\ a n d \frac{1}{a_{5}} = \frac{1}{a_{1}} + 4 d \\ \Rightarrow a_{1} - a_{5} = 4 {da}_{1} a_{5} \Rightarrow \frac{a_{1} - a_{5}}{d} = 4 a_{1} a_{5} \end{array}$