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Make $'d'$ the subject of the formula:
$s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Hence, find $'d'$ , if $s=144,a=1$ and $n=12$.

Last updated date: 24th Jul 2024
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Hint: To find the value of $'d'$ , we need to use the given formula. Since, we have given values for some variables. We will substitute them in the formula accordingly and then will simplify the equation with various mathematical operations to obtain the required answer.

Complete step by step answer:
It is given in the question:
  & \Rightarrow s=144 \\
 & \Rightarrow a=1 \\
 & \Rightarrow n=12
And we had given the formula also that is:
$\Rightarrow s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Now, we will substitute the values for variables as $144$ for $s$ , $1$ for $a$ and $12$ for $n$ in the given formula $s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ as:
$\Rightarrow 144=\dfrac{12}{2}\left[ 2\times 1+\left( 12-1 \right)d \right]$
Here, we multiply by $2$each side as:
$\Rightarrow 2\times 144=2\times \dfrac{12}{2}\left[ 2\times 1+\left( 12-1 \right)d \right]$
Now, we will use multiplication wherever it is needed and cancel out the equal like terms as:
$\Rightarrow 288=12\left[ 2+\left( 12-1 \right)d \right]$
After solving the operation within the small bracket that is the subtraction of $1$ from $12$ , we will have the above equation as:
$\Rightarrow 288=12\left[ 2+11d \right]$
Now, we will open the bracket as:
$\Rightarrow 288=12\times 2+12\times 11d$
Here, we will complete the multiplication. We will have $24$ by multiplying $12$ and $2$ and will get $132$ by multiplying $12$ and $11$. Now, the above equation will be:
$\Rightarrow 288=24+132d$
Now, we will subtract $24$ both sides in the above step as:
$\Rightarrow 288-24=24+132d-24$
We will have $264$ in the left side of the equal with the use of subtraction and in the right side, we will cancel out the equal like term as:
$\Rightarrow 264=132d$
Here, we will divide by $132$ both sides in the above step as:
$\Rightarrow \dfrac{264}{132}=\dfrac{132d}{132}$
After dividing $264$ by $132$ , we will have $2$ and we will cancel out the equal like term also. Then, we have:
$\Rightarrow 2=d$
Hence, the required answer is $d=2$ .

Note: Now, we will check the solution by applying the obtained value of $d$ and $a=1$ and $n=12$in the given formula with $s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ and will find $s$ that is already given in the question.
Since, the given formula is:
$\Rightarrow s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Substituting the values of $a,d,n$ in the above formula as:
$\Rightarrow s=\dfrac{12}{2}\left[ 2\times 1+\left( 12-1 \right)\times 2 \right]$
Now, we will simply the equation doing necessary calculations as:
  & \Rightarrow s=\dfrac{12}{2}\left[ 2+11\times 2 \right] \\
 & \Rightarrow s=\dfrac{12}{2}\left[ 2+22 \right] \\
 & \Rightarrow s=\dfrac{12}{2}\left[ 24 \right] \\
Here, we will open the bracket and will simplify as:
  & \Rightarrow s=\dfrac{12}{2}\times 24 \\
 & \Rightarrow s=12\times 12 \\
 & \Rightarrow s=144 \\
So, we got the given value of $s$. Hence, the solution is correct.