Answer:

\begin{gathered}\huge\blue{\underbrace{\overbrace{\mathbb{ANSWER}}}\end{gathered}

→ \boxed{\bold{d=-16,\:a_n=-16n+68}}

d=−16,a

n

=−16n+68

←

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{\mathrm{Find\:the\:sequence\:for\:the\:terms\:}52,\:36,\:20:\quad d=-16,\:a_n=-16n+68}Findthesequencefortheterms52,36,20:d=−16,a

n

=−16n+68

\begin{gathered}\tt{Arithmetic\:Progression\:\left(Arithmetic\:Sequence\right)\\\\\end{gathered}

\mathrm{An\:arithmetic\:sequence\:has\:a\:constant\:dfferencce\:}Anarithmeticsequencehasaconstantdfferencce dd \mathrm{and}and is defined by a_{n} = a_{1} + (n-1)da

n

=a

1

+(n−1)d

Check whether the difference is constant: d= -16

\begin{gathered}52,\:36,\:20\\\\\mathrm{Compute\:the\:difference\:of\:all\:adjacent\:terms:}\: d =a_{n+1} - a_n\end{gathered}

52,36,20

Computethedifferenceofalladjacentterms:d=a

n+1

−a

n

\begin{gathered}36-52=-16,\:\quad \:20-36=-16\\\\\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}\\d=-16\\\\\end{gathered}

36−52=−16,20−36=−16

Thedifferencebetweenalloftheadjacenttermsisthesameandequalto

d=−16

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\begin{gathered}\mathrm{The\:first\:element\:of\:the\:sequence\:is}\\\\a_1=52\\\\a_n=a_1+\left(n-1\right)d\\\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:\\\\Refine\\d=-16,\:a_n=-16n+68\\\end{gathered}

Thefirstelementofthesequenceis

a

1

=52

a

n

=a

1

+(n−1)d

Therefore,thenthtermiscomputedby

Refine

d=−16,a

n

=−16n+68

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