Question

find first -ve tern of series 292,289,286...................................................................................................

Answer 1

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please refer to the attachment above....

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Answer 2

a=292
d=289-292=-3

Let nth term of the AP is first negative term.

${n}^{th} \: term < 0 \\ a + (n - 1)d < 0 \\ 292 + (n - 1)( - 3) < 0 \\ (n - 1)( - 3) < - 292 \\ (n - 1) > \frac{ - 292}{ - 3} \\ n - 1 > 97.33 \\ n > 98.33 \\ so \: the \: value \: of \: n \: is \: 99.$
99 th term is the first negative term.

99th term =292+98×(-3)
=292-294
=-2