A sequence has a 1 = 19 , a 9 = 99 , and for n ≥ 3 , a n is the arithmetic mean of the first n − 1 terms. Find a 2 .

Since a 3 = ( a 1 + a 2 ) /2 , the next mean remains a 3 , so a 3 = ⋯ = a 9 = 99 . Therefore ( 19 + a 2 ) /2 = 99 and a 2 = 179 .

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MathematicsSequence & SeriesA.P.Medium2 minQB

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A sequence has $a_{1} = 19, a_{9} = 99$ , and for $n \geq 3, a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_{2}$ .

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Solution:

Since $a_{3} = (a_{1} + a_{2}) /2$ , the next mean remains $a_{3}$ , so $a_{3} = \dots = a_{9} = 99$ . Therefore $(19 + a_{2}) /2 = 99$ and $a_{2} = 179$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:A.P.

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