﻿ Let a_n be the n-th term of an arithmetic sequence, where the common difference is

#### Question

Let a_n be the n-th term of an arithmetic sequence, where the common difference is a non-zero constant. It is given that a_1 + a_3 + a_5 =\dfrac{72}{5} and a^2_1= a^2_{11}

Express a_n in terms of n.

Find the first negative term of the sequence.

Elaine claims that there exists a positive integer h such that 5a_h = h^2. Do you agree? Explain your answer.

Collected in the board: Arithmetic sequence

Steven Zheng posted 5 months ago

Since a_n is an arithmetic sequence, let d be the common difference of the sequence

a_n=a_1+(n-1)d

a_3=a_1+2d

a_5=a_1+4d

a_{11}=a_1+10d

\because a_1+a_3+a_5=\dfrac{72}{5}

\therefore 3a_1+6d=\dfrac{72}{5}
(1)
\because a^2_1= a^2_{11}

a^2_1= (a_1+10d)^2

a_1+5d =0
(2)

Solve the function system (1) and (2)

d = -\dfrac{8}{5} and a_1 =8

Therefore

a_n = 8-(n-1)\dfrac{8}{5}

=\dfrac{40-8n+8}{5}

=\dfrac{48}{5}-\dfrac{8}{5}n

If a_n <0

\dfrac{48}{5}-\dfrac{8}{5}n< 0

n>6

So the first negative term happens when n=7

Plug in the formula for general terms

a_7 = \dfrac{48}{5}-\dfrac{8}{5}\times 7

=\dfrac{48-56}{5}

=-\dfrac{8}{5}

Let h be a positive integer such that 5a_h = h^2, that is

5( \dfrac{48}{5}-\dfrac{8}{5}h ) = h^2

h^2+8h-48 = 0

(h+12)(h-4) = 0

h = 4

So yes, there exists a positive integer such that 5a_h = h^2

Steven Zheng posted 4 months ago

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