For a given A.P.,the 5^{th} term is 30 and th | vedclass

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(D) Let the first term be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.Given $a_5 = 30$,so $a +

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$2475$

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(D) Let the first term be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.Given $a_5 = 30$,so $a + 4d = 30$ (Equation $1$).Given $a_{12} = 65$,so $a + 11d = 65$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(a + 11d) - (a + 4d) = 65 - 30$.$7d = 35$,which gives $d = 5$.Substituting $d = 5$ into Equation $1$: $a + 4(5) = 30 \implies a + 20 = 30 \implies a = 10$.The sum of the first $n$ terms is $S_n = \frac{n}{2} [2a + (n-1)d]$.For $n = 30$: $S_{30} = \frac{30}{2} [2(10) + (30-1)5]$.$S_{30} = 15 [20 + 29 	imes 5] = 15 [20 + 145] = 15 	imes 165 = 2475$.

For a given $A.P.$,the $5^{th}$ term is $30$ and the $12^{th}$ term is $65$. Find the sum of first $30$ terms of this $A.P.$

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