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# In an A.P. given $a=3,n=8,s=192$. Find d.  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.
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