Answer

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Hint: If -150 lies in the series then we will have a integer value for n when we substitute -150 in place of ${{a}_{n}}$ in the formula ${{a}_{n}}=a+(n-1)d$ where a is the first term of the series and d is the common difference of the series of the A.P.

Complete step-by-step answer:

An A.P. is a series of the form a, a+d , a+2d , …………… and so on. Therefore the terms are increasing or decreasing by a single value that is d.

This sequence is an A.P. as we can see the terms are decreasing by 3 when we go from left to right in the series.

Now to find out if -150 is a term of the series or not we will need to first find out the first term and common difference of the series.

In the sequence we can see that the first term is 11 i.e. a=11.

And to find the common difference we need to subtract the subsequent terms since, $d={{a}_{n}}-{{a}_{n-1}}$

When we take n=2, we have

$\begin{align}

& d={{a}_{2}}-{{a}_{1}} \\

& =8-11 \\

& =-3 \\

\end{align}$

Therefore, d=-3

The nth term of the series is given by ${{a}_{n}}=a+(n-1)d$

Now we will substitute ${{a}_{n}}$ by -150, a by 11 and d by -3 in the formula.

$\begin{align}

& {{a}_{n}}=a+(n-1)d \\

& \Rightarrow -150=11+(n-1)(-3) \\

& \Rightarrow -150=11-3(n-1) \\

\end{align}$

Taking -150 to RHS and 3(n-1) to LHS we have,

$\begin{align}

& \Rightarrow 3(n-1)=150+11 \\

& \Rightarrow 3(n-1)=161 \\

\end{align}$

Dividing 3 both sides we have,

$\Rightarrow n-1=\dfrac{161}{3}$

Adding 1 both sides we have,

$\begin{align}

& \Rightarrow n=\dfrac{161}{3}+1 \\

& \Rightarrow n=\dfrac{164}{3} \\

\end{align}$

Now if 164 is divisible by 3 then n will be an integer.

To check if the given number is an integer we have a divisibility rule of 3 which says that if the sum of digits of the number is divisible by 3 then the number is divisible by 3.

For 164 we have,

1+6+4=11

Since, 11 is not divisible by 3 then the given number is not divisible by 3. Therefore, n is not an integer. And since n is not an integer we can say that -150 is not a term of the series.

Hence, -150 is not a term of the given A.P.

Note: One can get confused when they see d=-3 and the term which is being asked in the question is -150. By appearance without checking they may answer that the given term lies in the series just because -150 is divisible by -3.

Complete step-by-step answer:

An A.P. is a series of the form a, a+d , a+2d , …………… and so on. Therefore the terms are increasing or decreasing by a single value that is d.

This sequence is an A.P. as we can see the terms are decreasing by 3 when we go from left to right in the series.

Now to find out if -150 is a term of the series or not we will need to first find out the first term and common difference of the series.

In the sequence we can see that the first term is 11 i.e. a=11.

And to find the common difference we need to subtract the subsequent terms since, $d={{a}_{n}}-{{a}_{n-1}}$

When we take n=2, we have

$\begin{align}

& d={{a}_{2}}-{{a}_{1}} \\

& =8-11 \\

& =-3 \\

\end{align}$

Therefore, d=-3

The nth term of the series is given by ${{a}_{n}}=a+(n-1)d$

Now we will substitute ${{a}_{n}}$ by -150, a by 11 and d by -3 in the formula.

$\begin{align}

& {{a}_{n}}=a+(n-1)d \\

& \Rightarrow -150=11+(n-1)(-3) \\

& \Rightarrow -150=11-3(n-1) \\

\end{align}$

Taking -150 to RHS and 3(n-1) to LHS we have,

$\begin{align}

& \Rightarrow 3(n-1)=150+11 \\

& \Rightarrow 3(n-1)=161 \\

\end{align}$

Dividing 3 both sides we have,

$\Rightarrow n-1=\dfrac{161}{3}$

Adding 1 both sides we have,

$\begin{align}

& \Rightarrow n=\dfrac{161}{3}+1 \\

& \Rightarrow n=\dfrac{164}{3} \\

\end{align}$

Now if 164 is divisible by 3 then n will be an integer.

To check if the given number is an integer we have a divisibility rule of 3 which says that if the sum of digits of the number is divisible by 3 then the number is divisible by 3.

For 164 we have,

1+6+4=11

Since, 11 is not divisible by 3 then the given number is not divisible by 3. Therefore, n is not an integer. And since n is not an integer we can say that -150 is not a term of the series.

Hence, -150 is not a term of the given A.P.

Note: One can get confused when they see d=-3 and the term which is being asked in the question is -150. By appearance without checking they may answer that the given term lies in the series just because -150 is divisible by -3.

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