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Hint- In order to solve such a question consider 0 to be a term, then with the help of formula of nth term of an A.P. find the value of n or the term number. If the value of n be an integer then our consideration will be right otherwise wrong.

Complete step-by-step solution -

Given A.P. is $40,37,34,31,....$

For a general A.P. with $a$ as first term and $d$ be its common difference.

Nth term of the general A.P. is

${a_n} = a + \left( {n - 1} \right)d$

For the given A.P.

$

a = 40 \\

d = {a_2} - {a_1} = 37 - 40 = - 3 \\

$

Let us consider 0 is the nth term of the A.P.

$ \Rightarrow {a_n} = 0$

Also we have

$

\Rightarrow {a_n} = a + \left( {n - 1} \right)d = 0 \\

\Rightarrow 40 + \left( {n - 1} \right)\left( { - 3} \right) = 0 \\

$

Solving the equation for the value of n

\[

\Rightarrow 40 + \left( {n - 1} \right)\left( { - 3} \right) = 0 \\

\Rightarrow \left( {n - 1} \right)\left( { - 3} \right) = - 40 \\

\Rightarrow \left( {n - 1} \right)\left( 3 \right) = 40 \\

\Rightarrow \left( {n - 1} \right) = \dfrac{{40}}{3} \\

\Rightarrow n = \dfrac{{40}}{3} + 1 \\

\Rightarrow n = \dfrac{{40 + 3}}{3} \\

\Rightarrow n = \dfrac{{43}}{3} \\

\]

Since the term number in an A.P. cannot be a decimal number. It can only be an integer. So our consideration is false.

Hence, 0 is not a term of the given A.P.

Note- An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 2, 4, 6, 8 ... is an arithmetic progression with common difference 2. Always remember the formula of the nth term of an A.P. and the sum of an A.P.

Complete step-by-step solution -

Given A.P. is $40,37,34,31,....$

For a general A.P. with $a$ as first term and $d$ be its common difference.

Nth term of the general A.P. is

${a_n} = a + \left( {n - 1} \right)d$

For the given A.P.

$

a = 40 \\

d = {a_2} - {a_1} = 37 - 40 = - 3 \\

$

Let us consider 0 is the nth term of the A.P.

$ \Rightarrow {a_n} = 0$

Also we have

$

\Rightarrow {a_n} = a + \left( {n - 1} \right)d = 0 \\

\Rightarrow 40 + \left( {n - 1} \right)\left( { - 3} \right) = 0 \\

$

Solving the equation for the value of n

\[

\Rightarrow 40 + \left( {n - 1} \right)\left( { - 3} \right) = 0 \\

\Rightarrow \left( {n - 1} \right)\left( { - 3} \right) = - 40 \\

\Rightarrow \left( {n - 1} \right)\left( 3 \right) = 40 \\

\Rightarrow \left( {n - 1} \right) = \dfrac{{40}}{3} \\

\Rightarrow n = \dfrac{{40}}{3} + 1 \\

\Rightarrow n = \dfrac{{40 + 3}}{3} \\

\Rightarrow n = \dfrac{{43}}{3} \\

\]

Since the term number in an A.P. cannot be a decimal number. It can only be an integer. So our consideration is false.

Hence, 0 is not a term of the given A.P.

Note- An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 2, 4, 6, 8 ... is an arithmetic progression with common difference 2. Always remember the formula of the nth term of an A.P. and the sum of an A.P.

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