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# Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]

Given the AP series is 121, 117, 113, . . .,
Thus, first term, a = 121
Common difference, d = 117-121= -4

By the $$n^{th}$$ term formula,
$$a_n = a+(n -1)d$$
Therefore,
$$a_n = 121+(n-1)(-4)$$
= 121-4n+4
=125-4n

To find the first negative term of the series, $$a_n < 0$$
Therefore,

$$\Rightarrow$$ 125-4n < 0
$$\Rightarrow$$ 125 < 4n
$$\Rightarrow$$ n>125/4
$$\Rightarrow$$ n>31.25

Therefore, the first negative term of the series is $$32^{nd}$$ term.