For an A.P.,the 5^{th} term is 30 and the 12^ | vedclass

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(B) Let the first term be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.For the $5^{th}$ term:

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$1700$

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(B) Let the first term be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.For the $5^{th}$ term: $a + 4d = 30$  --- $(1)$For the $12^{th}$ term: $a + 11d = 100$ --- $(2)$Subtracting equation $(1)$ from $(2)$:$(a + 11d) - (a + 4d) = 100 - 30$$7d = 70$$d = 10$Substituting $d = 10$ in equation $(1)$:$a + 4(10) = 30$$a + 40 = 30$$a = -10$The sum of the first $n$ terms is $S_n = \frac{n}{2} [2a + (n - 1)d]$.For $n = 20$:$S_{20} = \frac{20}{2} [2(-10) + (20 - 1)10]$$S_{20} = 10 [-20 + 190]$$S_{20} = 10 [170]$$S_{20} = 1700$.

For an $A.P.$,the $5^{th}$ term is $30$ and the $12^{th}$ term is $100$. Find the sum of the first $20$ terms of the $A.P.$

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