Write the first negative term of the sequence20, 19\(\frac{1}{4}\), 18

1.	Write the first negative term of the sequence20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\) ….
Answer» Given sequence is 20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\)..... Or 20\(\frac{77}{4}\), \(\frac{37}{2}\), \(\frac{71}{4}\)...... Here, a = 20, d = \(\frac{77}{4}\) − 20 = \(\frac{37}{2}\) − \(\frac{77}{4}\) = − \(\frac{3}{4}\) (it is an A.P) Let n^th term be its first negative term, i.e. T_n = a + (n − 1)d < 0 ⇒ 20 + (n − 1)(\(-\frac{3}{4}\)) < 0 ⇒ 80 − 3n + 3 < 0 ⇒ 83 − 3n < 0 ⇒ n > \(\frac{83}{3}\) ⇒ n > \(27\frac{2}{3}\) Hence, n = 28

Write the first negative term of the sequence20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\) ….

Answer»

Given sequence is

20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\).....

Or 20\(\frac{77}{4}\), \(\frac{37}{2}\), \(\frac{71}{4}\)......

Here, a = 20, d = \(\frac{77}{4}\) − 20 = \(\frac{37}{2}\) − \(\frac{77}{4}\) = − \(\frac{3}{4}\) (it is an A.P)

Let n^th term be its first negative term, i.e.

T_n = a + (n − 1)d < 0

⇒ 20 + (n − 1)(\(-\frac{3}{4}\)) < 0

⇒ 80 − 3n + 3 < 0

⇒ 83 − 3n < 0

⇒ n > \(\frac{83}{3}\)

⇒ n > \(27\frac{2}{3}\)

Hence, n = 28