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Hint: We can use sum of n terms of an arithmetic progression i.e. \[{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\]

Where ${{S}_{n}}$ is the sum of n terms of an A.P

n is the number of terms

a is the first term of A.P

d is a common difference of two consecutive terms.

__Complete step-by-step solution -__

Explanation:

We know that sum of n terms of an arithmetic progression whose term is ‘a’ and common difference ‘d’ is given by

\[{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\]

In given question \[a=3,n=8,{{S}_{8}}=192\]

So we can write

\[\Rightarrow 192=\dfrac{8}{2}[2\times 3+(8-1)d]\]

\[\Rightarrow 192=4[6+7d]\]

\[\Rightarrow \dfrac{192}{4}=6+7d\]

\[\Rightarrow 48=6+7d\]

\[\Rightarrow 48-6=7d\]

\[\Rightarrow 42=7d\]

\[\Rightarrow d=\dfrac{42}{7}=6\]

So, \[d=6\].

Hence the value of common difference ‘d’ is 6.

Note: We need to be careful about the formula. In arithmetic, we have two formulas for finding sum as below

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]=\dfrac{n}{2}(a+l)$

So in this, if we will use the formula for the last term, here we will have to find the value of last terms due to which the solution will become lengthy and complicated.

Where ${{S}_{n}}$ is the sum of n terms of an A.P

n is the number of terms

a is the first term of A.P

d is a common difference of two consecutive terms.

Explanation:

We know that sum of n terms of an arithmetic progression whose term is ‘a’ and common difference ‘d’ is given by

\[{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\]

In given question \[a=3,n=8,{{S}_{8}}=192\]

So we can write

\[\Rightarrow 192=\dfrac{8}{2}[2\times 3+(8-1)d]\]

\[\Rightarrow 192=4[6+7d]\]

\[\Rightarrow \dfrac{192}{4}=6+7d\]

\[\Rightarrow 48=6+7d\]

\[\Rightarrow 48-6=7d\]

\[\Rightarrow 42=7d\]

\[\Rightarrow d=\dfrac{42}{7}=6\]

So, \[d=6\].

Hence the value of common difference ‘d’ is 6.

Note: We need to be careful about the formula. In arithmetic, we have two formulas for finding sum as below

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]=\dfrac{n}{2}(a+l)$

So in this, if we will use the formula for the last term, here we will have to find the value of last terms due to which the solution will become lengthy and complicated.

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