If a_1, a_2, , a_n are in H.P., then the expression a_1a_2 + a_2a_3 + + a_n-1a_n is equal to: - Mathematics MCQ AIEEE 2006 2026 | ExamTest.live

mediumMCQAIEEE 20062026Mathematics

1 attempts0% success rate1 mark

If a_{1}, a_{2}, \dots, a_{n} are in H.P., then the expression a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n} is equal to:

Solution & Step-by-step Explanation

Since a_{1}, a_{2}, \dots, a_{n} are in H.P., \frac{1}{a _{1}}, \frac{1}{a _{2}}, \dots, \frac{1}{a _{n}} are in A.P.Let d be the common difference. \frac{1}{a _{2}} - \frac{1}{a _{1}} = d ⟹ \frac{a _{1} - a _{2}}{a _{1} a _{2}} = d ⟹ a_{1} a_{2} = \frac{a _{1} - a _{2}}{d} Similarly, a_{2} a_{3} = \frac{a _{2} - a _{3}}{d}, \dots, a_{n - 1} a_{n} = \frac{a _{n - 1} - a _{n}}{d} .Sum S = \frac{1}{d} [(a_{1} - a_{2}) + (a_{2} - a_{3}) + \dots + (a_{n - 1} - a_{n})] = \frac{a _{1} - a _{n}}{d} \dots (1) Also, \frac{1}{a _{n}} = \frac{1}{a _{1}} + (n - 1) d ⟹ \frac{a _{1} - a _{n}}{a _{1} a _{n}} = (n - 1) d ⟹ d = \frac{a _{1} - a _{n}}{( n - 1 ) a _{1} a _{n}} .Substituting d into (1): S = \frac{a _{1} - a _{n}}{\frac{a _{1} - a _{n}}{( n - 1 ) a _{1} a _{n}}} = (n - 1) a_{1} a_{n} .

Try it yourself before checking the explanation above.

If a_{1}, a_{2}, \dots, a_{n} are in H.P., then the expression a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n} is equal to:

n (a_{1} - a_{n})

(n - 1) (a_{1} - a_{n})

n a_{1} a_{n}

(n - 1) a_{1} a_{n}

Practice with timed mock tests and track your performance across Mathematics.