Find first negative term 121\’117\’113\’109
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Here the given AP is :\xa0121,117,113……..a=121 and d=117-121 = -4The general term of an AP is given byan\xa0= a+(n-1)dLet nth term is negative{tex}\\Rightarrow{/tex}\xa0a+(n-1)d<0{tex}\\Rightarrow{/tex}\xa0121+(n-1)(-4)<0{tex} \\Rightarrow {/tex}\xa0121-4n+4<0{tex} \\Rightarrow {/tex}\xa0-4n+125 < 0{tex} \\Rightarrow {/tex}\xa0-4n< -125Hence, 4n>125{tex} \\Rightarrow {/tex}\xa0n >{tex}\\frac{{125}}{4}{/tex}So, n > 31.25Hence, the 32nd term is first negative term.