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Which term of the progression $20, 19 \frac{1}{4},18 \frac{1}{2}, 17 \frac{3}{4},...$ is the first negative term ?

Answers (1)

Given : $a=20$

$d=19\frac{1}{4}-20=\frac{-3}{4}$

$a_n=a+(n-1)d$

First negative term $\Rightarrow a_n<0$

$\Rightarrow 20+(n-1) \left ( \frac{-3}{4}\right )<0$

$\Rightarrow (n-1)\times \left ( \frac{3}{4}\right )>20$

$\Rightarrow (n-1) > \frac{20 \times 4}{3}$

$\Rightarrow n > \frac{20 \times 4}{3}+1$

$\Rightarrow n > \frac{80}{3}$

$\Rightarrow n > 27$

Let $n=28\; \; (\because n>27)$

$a_{28}=20+(28-1)\left ( \frac{-3}{4} \right )$ $=20+ \frac{27 \times (-3)}{4}$

$=20- \frac{81}{4}$

$=- \frac{1}{4}$

As $a_{28}$ is negative hence the first negative term is $\left ( -\frac{1}{4} \right )$

Posted by

Ravindra Pindel

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