Question

Consider the A. P 171, 167, 163,........ Is 0 a term of the sequence. How many positive terms are in this sequence.

Answer 1

$Given \:A.P : 171,167,163,\ldots$

$First \:term = a = a_{1} = 171$

$Common \: difference (d) = a_{2} - a_{1}$

$= 167 - 171$

$= -4$

/* According to the problem given */

$Let \:n^{th} \: term = 0$

$a_{n} = 0$

$\implies a + (n-1)d = 0$

$\implies 171 + (n-1)(-4) = 0$

$\implies (n-1)(-4) = -171$

$\implies n - 1 = \frac{-171}{-4}$

$\implies n = \frac{171}{4} + 1$

$\implies n = \frac{171+ 4}{4}$

$\implies n = \frac{175}{4}$

/* But n should not be a fraction */

$\blue {zero \: is \: not \: a \:term \: in \: given}$

$\blue{ sequence .}$

$Now , \pink { Number \:of \: positive\:terms }$

$\pink{sequence}$

$n = 42$

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