Answer:

Common difference of an AP is 3

Step-by-step explanation:

Let assume that first term and common difference of an AP be a and d respectively.

Given that, 23rd term of an AP is 54.

[tex] \sf \: a_{23} = 54 \\ [/tex]

[tex] \sf \: a + ({23} - 1)d = 54 \\ [/tex]

[tex] \sf \: a + 22d = 54 \\ [/tex]

[tex] \implies\sf\: a = 54 - 22d - - - (1) \\ [/tex]

Further given that, 37th term of an AP is 96.

[tex] \sf \: a_{37} = 96 \\ [/tex]

[tex] \sf \: a + ({37} - 1)d = 96 \\ [/tex]

[tex] \sf \: a + 36d = 96 \\ [/tex]

On substituting the value of a from equation (1), we get

[tex] \sf \: 54 - 22d + 36d = 96 \\ [/tex]

[tex] \sf \: 54 + 14d = 96 \\ [/tex]

[tex] \sf \: 14d = 96 - 54 \\ [/tex]

[tex] \sf \: 14d = 42 \\ [/tex]

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

Hence, Common difference of an AP is 3

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

Additional Information:

↝ Sum of n  terms of an arithmetic progression is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \end{gathered}[/tex]

↝ nᵗʰ term of an arithmetic progression is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \end{gathered}[/tex]

↝ nᵗʰ term of an arithmetic progression from end is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:l\:-\:(n\:-\:1)\:d}}}}}} \end{gathered}[/tex]

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

aₙ is the nᵗʰ term