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Which term of the list of numbers

    \[\mathbf{20},\text{ }\mathbf{19}{\scriptscriptstyle 1\!/\!{ }_4},\text{ }\mathbf{18}{\scriptscriptstyle 1\!/\!{ }_2},\text{ }\mathbf{17}{\scriptscriptstyle 3\!/\!{ }_4},\text{ }\ldots \]

is the first negative term?
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
Which term of the list of numbers

    \[\mathbf{20},\text{ }\mathbf{19}{\scriptscriptstyle 1\!/\!{ }_4},\text{ }\mathbf{18}{\scriptscriptstyle 1\!/\!{ }_2},\text{ }\mathbf{17}{\scriptscriptstyle 3\!/\!{ }_4},\text{ }\ldots \]

is the first negative term?

From the question it is given that,

First term a =

    \[20\]

Common difference d =

    \[19{\scriptscriptstyle 1\!/\!{ }_4}\text{ }\text{ }20\text{ }=\text{ }77/4\text{ }\text{ }20\text{ }=\text{ }\left( 77\text{ }\text{ }80 \right)/4\text{ }=\text{ }-{\scriptscriptstyle 3\!/\!{ }_4}\]

We know that, an = a + (n – 1)d

    \[\begin{array}{*{35}{l}} {{a}_{n}}~=\text{ }20\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\text{ }\left( -{\scriptscriptstyle 3\!/\!{ }_4} \right) \\ {{a}_{n}}~=\text{ }20\text{ }\text{ }{\scriptscriptstyle 3\!/\!{ }_4}n\text{ }+\text{ }{\scriptscriptstyle 3\!/\!{ }_4} \\ {{a}_{n}}~=\text{ }20\text{ }+\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\text{ }\text{ }{\scriptscriptstyle 3\!/\!{ }_4}n \\ {{a}_{n}}~=\text{ }\left( 80\text{ }+\text{ }3 \right)/4\text{ }\text{ }{\scriptscriptstyle 3\!/\!{ }_4}n \\ {{a}_{n}}~=\text{ }83/4\text{ }\text{ }{\scriptscriptstyle 3\!/\!{ }_4}n\text{ }<\text{ }0 \\ 83/4\text{ }<\text{ }{\scriptscriptstyle 3\!/\!{ }_4}n \\ 83\text{ }<\text{ }3n \\ 83/3\text{ }<\text{ }n \\ 28\text{ }<\text{ }n \\ \end{array}\]

Therefore,

    \[{{28}^{th}}\]

 is the first negative term.

i

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