Question

Find first negative term from following A.P.
122, 116, 110, ... (Note: find smallest n such that t_n < 0 )

Answer 1

In the given A.P., we have:
a = 122 and d = t₂ – t₁ = 116 – 122 = – 6
We know that the n^th term of an A.P. is t_n = a + (n – 1)d.
To find the first negative term in the given A.P., we need to find the smallest integer using the inequality as t_n < 0 $\Rightarrow$a + (n – 1)d < 0.
Thus, we have:
122 + (n – 1)(–6) < 0
$\Rightarrow$122 – 6n + 6 < 0

$\Rightarrow$–6n < – 128

$\Rightarrow$n >$\frac{128}{6}=\frac{64}{3}=21\frac{1}{3}$

$\Rightarrow$n > $21\frac{1}{3}$

Thus, the smallest integer value of n is 22 and the first negative term in the given A.P. is t₂₂.
To find the 22^nd term of the given A.P., we need to put n = 22 in t_n = a + (n – 1)d.
Thus, we get:
t₂₂ = 122 + (22 – 1)(–6)
= 122 – 21 × 6
= 122 – 126
= –4