Question

the first negative term of the sequence 45, 42 1/4, 39 1/2, 36 3/4,........ is (A) 16th (B) 17th (C) 18th (D) 19th. pls dont spam. give answer with explanation or attachment ^_^

Answer 1

Answer:

d)19th is the first negative sequence

Answer 2

$Given \: sequence \: 45 ,42\frac{1}{4},39\frac{1}{2} , 36\frac{3}{4},\ldots$

$Rewrite \: the \: sequence \: 45,42.25,39.5$

$36.75 ,\ldots$

$First \:term (a) = a_{1} = 45 ,$

$i )a_{2} - a_{1}$

$= 42.25 - 45$

$= -2.75$

$ii )a_{3} - a_{2}$

$= 39.5 - 42.25$

$= -2.75$

$\pink {\therefore a_{2} - a_{1} = a_{3} - a_{2}}$

$Given \: sequence \: is \: an \: A.P$

$Common \: difference (d) = -2.75$

$We \:know \: that ,$

$\blue{ n^{th} \:term \:( a_{n}) = a + (n-1)d }$

$Let \: a_{n} = 0$

$\implies 45 + ( n - 1 )( -2.75) = 0$

$\implies (n-1)(-2.75) = - 45$

$\implies n-1 = \frac{ - 45}{-2.75}$

$\implies n = \frac{4500}{275} + 1$

$\implies n = \frac{180}{11} +1$

$\implies n = \frac{180+11}{11}$

$\implies n = \frac{191}{11}$

$\implies n = 17.\bar{36}$

$Now , Let \: n = 18$

$a_{18} = a + 17d$

$= 45 + 17(-2.75)$

$= 45 - 46.75$

$= - 1.75$

$\lt 0$

Therefore.,

$\red{ First \: negative \: term \: in \: the }$

$\red{ sequence } \green { = 18^{th} \:term }$

$Option \: \pink { (C ) } \:is \: correct.$

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