Which term of the progression 19, 18(1/5) , 17(2/5), ..... is the first negative term?

Question

Which term of the progression 19, \(18\frac{1}{5}\) , \(17\frac{2}{5}\),..... is the first negative term?

Answer 1

We must find the common difference of the A.P:-

Common difference (d) = first term − second term

\(= 18 \frac 15 - 19\)

\(= \frac{91}5 - \frac{95}5\)

\(= \frac{-4}5\)

Hence,

the common difference is \(\frac{-4}5\)​ or − 0.8

Now to estimate which term of this A.P will be negative let us assume

the nth term of A.P is 0.

Formula for the nth term of an A.P is = a + (n − 1)d

then,

0 = 19 + (n − 1) × −0.8

0 = 19 − 0.8n + 0.8

0.8n = 19.8

n = \(\frac{19.8}{0.8}\)

n = 24.75​

If the 24.75th term is 0. Then surly 25th term of the A.P will be negative.

Let us make sure, just in this case:

25th term of an A.P = a + (n − 1)d

= 19 + (25 − 1) × −0.8

= 19 + (24 × (−0.8))

= 19 + (−19.2)

= 19 − 19.2

= −0.2​

Thus, the 25th term of the progression is first negative term.

Answer 2

Here a = 19, d = \(18\frac{1}{5} -19\) = - \(\frac{4}{5}\)

Let the nth term be the first negative term. Then,

Tn < 0 ⇒ a + (n – 1) d < 0 

⇒ 19 + (n – 1) (−4/5) < 0

⇒ 19 – \(\frac{4}{5}\)n + \(\frac{4}{5}\) < 0 

⇒ \(\frac{99}{5}\) - \(\frac{4}{5}\)n < 0 

⇒ n > \(\frac{99}{5}\) × \(\frac{5}{4}\) 

⇒ n > \(\frac{99}{4}\)  = \(24\frac{3}{4}\)

⇒ n = 25.