Question

Which Progression is the first negative term ??

ANSWER PLEASE !!​

Answer 1

$19 \frac{1}{4} = \frac{77}{4} \\ 18 \frac{1}{2} = \frac{37}{2} = \frac{74}{4} \\ 17 \frac{3}{4} = \frac{71}{4}$

this sequence forms an A.P.

common difference, d = 74/4 - 77/4 = -3/4

first term, a = 77/4

for first negative term,

$t_{n} < 0 \\ a + (n - 1)d < 0 \\ \frac{77}{4} + (n - 1)( - \frac{3}{4} ) < 0 \\ \frac{77}{4} - \frac{3n}{4} + \frac{3}{4} < 0 \\ \frac{80}{4} - \frac{3n}{4} < 0 \\ \\ multiplying \: throughout \: by \: 4 \\ \\ {80 - 3n} < 0 \\ 80 < 3n \\ \frac{80}{3} < n \\ n > 26.6666... \\ \\ n \: is \: an \: integer \\ hence \: value \: of \: n \: closest \: to \: greater \: than \: 26 .6666... \\ \\ n = 27$

Hence, the first negative term of the given progression will be the 27th term.