Appropriate Question:

Make d the subject of the formula : l = a+ (n - 1)d

And find d when l = 10, a = 1 and n = 4

Answer:

[tex]\sf\:\boxed{\begin{aligned}& \qquad \:\bf \: d = \dfrac{l - a}{n - 1} \qquad \: \\ \\& \qquad \:\bf \: d = 3 \end{aligned}}[/tex]

Step-by-step explanation:

Given expression is

[tex]\sf\: l = a + (n - 1)d \\ [/tex]

[tex]\sf\: l - a = (n - 1)d \\ [/tex]

[tex]\implies\sf\:d = \dfrac{l - a}{n - 1} \\ [/tex]

On substituting the values l = 10, a = 1 and n = 4, we get

[tex]\sf\: d = \dfrac{10 - 1}{4 - 1} \\ [/tex]

[tex]\sf\: d = \dfrac{9}{3} \\ [/tex]

[tex]\implies\sf\:d = 3 \\ [/tex]

Hence,

[tex]\implies\sf\:\sf\:\boxed{\begin{aligned}& \qquad \:\bf \: d = \dfrac{l - a}{n - 1} \qquad \: \\ \\& \qquad \:\bf \: d = 3 \end{aligned}}[/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information:

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} =  {x}^{2}  + 2xy +  {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2}  =  {x}^{2} - 2xy +  {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} -  {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2}  -  {(x - y)}^{2}  = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2}  +  {(x - y)}^{2}  = 2( {x}^{2}  +  {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} =  {x}^{3} +  {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} =  {x}^{3} -  {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3}  +  {y}^{3} = (x + y)( {x}^{2}  - xy +  {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]