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Question

# Which term of the AP: $121,117,113,...$ is its first negative term?

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Solution

## Determine the first negative term.Use the ${n}^{\mathrm{th}}$ term formula: ${a}_{n}=a+\left(n-1\right)d$The given AP is $121,117,113,...$.The first term is $a=121$.The common difference is:$\begin{array}{rcl}d& =& 117-121\\ & ⇒& d=-4\end{array}$Now,${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}⇒{a}_{n}=121+\left(n-1\right)\left(-4\right)\phantom{\rule{0ex}{0ex}}⇒{a}_{n}=121-4n+4\phantom{\rule{0ex}{0ex}}⇒{a}_{n}=125-4n$Let ${a}_{n}$ be the first negative term of the given AP so${a}_{n}<0\phantom{\rule{0ex}{0ex}}⇒125-4n<0\phantom{\rule{0ex}{0ex}}⇒125<4n\phantom{\rule{0ex}{0ex}}⇒n>\frac{125}{4}\phantom{\rule{0ex}{0ex}}\therefore n>31.25$So we can take $n=32$Hence, the ${32}^{nd}$ term will be the first negative term of this AP.

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