Question

which term of sequence 17,81/5,77/5,73/5 is first negative term

Answer 1

here is the solution. hope it helps.

Answer 2

$Given \: sequence \: 17, \frac{81}{5}, \frac{77}{5}, \frac{73}{5},.$

$First \:term (a) = 17$

$i )a_{2} - a_{1} \\= \frac{81}{5} - 17 \\= \frac{81-85}{5} \\= \frac{-4}{5} \: --(1)$

$ii)a_{3} - a_{2} \\=\frac{77}{5} - \frac{81}{5} \\= \frac{77-81}{5} \\= \frac{-4}{5} \: --(2)$

/* From (1) and (2) */

$\therefore a_{2} - a_{1} = a_{3} - a_{2} = \frac{-4}{5}$

$\blue { Given \: sequence \:is \: in \: A.P}$

$\therefore Common \:difference (d) = \frac{-4}{5}$

/* According to the problem given */

$n^{th} \: term < 0$

$\implies a + (n-1)d < 0$

$\implies 17 + (n-1)\Big(\frac{-4}{5}\Big) <0$

$\implies (n-1)\Big(\frac{-4}{5}\Big) < -17$

$\implies (n-1)\Big(\frac{4}{5}\Big) > 17$

$\implies (n-1)>17 \times \frac{5}{4}$

$\implies n > \frac{85}{4} + 1$

$\implies n > \frac{85+4}{4}$

$\implies n > \frac{89}{4}$

$\implies n > 22\frac{1}{4}$

$\implies n = 23^{rd} \:term$

Therefore.,

$\red { First \: Negative \:term \: in \:the \: sequence} \\\green { \:is \ : 23^{rd} \:term}$

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