If a1,a2,...,an are in H.P. and d is the common difference of the corresponding A.P., then the expression a1a2+a2a3+.....+an−1an is equal to
A
a1−and
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B
(n−1)(a1−an)
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C
n(a1−an)
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D
(n−1)a1an
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Solution
The correct options are Aa1−and D(n−1)a1an a1a2=a1a2dd=a1a2d(1a2−1a1)=a1−a2d Similarly, a2a3=a2−a3d,a3a4=a3−a4d,...,an−1an=an−1−and Now, a1a2+a2a3+a3a4+.....+an−1an =1d[a1−a2+a2−a3+.....+an−1−an] =a1−and.....(i) We have, 1an=1a1+(n−1)d ⇒1an−1a1n−1=d Substituting the value of d in (i), we get, a1a2+a2a3+.....+an−1an=(n−1)a1an. Hence, options A and D are correct.