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Question: Make \('d'\) the subject of the formula: \(s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\) ...

Make d'd' the subject of the formula:
s=n2[2a+(n1)d]s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]
Hence, find d'd' , if s=144,a=1s=144,a=1 and n=12n=12.

Explanation

Solution

To find the value of d'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:
s=144 a=1 n=12\begin{aligned} & \Rightarrow s=144 \\\ & \Rightarrow a=1 \\\ & \Rightarrow n=12 \end{aligned}
And we had given the formula also that is:
s=n2[2a+(n1)d]\Rightarrow s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]
Now, we will substitute the values for variables as 144144 for ss , 11 for aa and 1212 for nn in the given formula s=n2[2a+(n1)d]s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] as:
144=122[2×1+(121)d]\Rightarrow 144=\dfrac{12}{2}\left[ 2\times 1+\left( 12-1 \right)d \right]
Here, we multiply by 22each side as:
2×144=2×122[2×1+(121)d]\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:
288=12[2+(121)d]\Rightarrow 288=12\left[ 2+\left( 12-1 \right)d \right]
After solving the operation within the small bracket that is the subtraction of 11 from 1212 , we will have the above equation as:
288=12[2+11d]\Rightarrow 288=12\left[ 2+11d \right]
Now, we will open the bracket as:
288=12×2+12×11d\Rightarrow 288=12\times 2+12\times 11d
Here, we will complete the multiplication. We will have 2424 by multiplying 1212 and 22 and will get 132132 by multiplying 1212 and 1111. Now, the above equation will be:
288=24+132d\Rightarrow 288=24+132d
Now, we will subtract 2424 both sides in the above step as:
28824=24+132d24\Rightarrow 288-24=24+132d-24
We will have 264264 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:
264=132d\Rightarrow 264=132d
Here, we will divide by 132132 both sides in the above step as:
264132=132d132\Rightarrow \dfrac{264}{132}=\dfrac{132d}{132}
After dividing 264264 by 132132 , we will have 22 and we will cancel out the equal like term also. Then, we have:
2=d\Rightarrow 2=d
Hence, the required answer is d=2d=2 .

Note: Now, we will check the solution by applying the obtained value of dd and a=1a=1 and n=12n=12in the given formula with s=n2[2a+(n1)d]s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] and will find ss that is already given in the question.
Since, the given formula is:
s=n2[2a+(n1)d]\Rightarrow s=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]
Substituting the values of a,d,na,d,n in the above formula as:
s=122[2×1+(121)×2]\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:
s=122[2+11×2] s=122[2+22] s=122[24] \begin{aligned} & \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] \\\ \end{aligned}
Here, we will open the bracket and will simplify as:
s=122×24 s=12×12 s=144 \begin{aligned} & \Rightarrow s=\dfrac{12}{2}\times 24 \\\ & \Rightarrow s=12\times 12 \\\ & \Rightarrow s=144 \\\ \end{aligned}
So, we got the given value of ss. Hence, the solution is correct.