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(C) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$The second term is $a + d = 12$ .....$(1)$The sum of the first nine

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$20$

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(C) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$The second term is $a + d = 12$ .....$(1)$The sum of the first nine terms is given by:${S_9} = \frac{9}{2}(2a + 8d) = 9(a + 4d)$Given that $200 $200 Substitute $a = 12 - d$ from equation $(1)$ into the inequality:$200 $200 $200 Subtract $108$ from all parts:$92 Since the terms are positive integers,$d$ must be an integer. Testing values for $d$:If $d = 3$,$27 	imes 3 = 81$ (too small).If $d = 4$,$27 	imes 4 = 108$ (which satisfies $92 If $d = 5$,$27 	imes 5 = 135$ (too large).Thus,$d = 4$.From equation $(1)$,$a + 4 = 12$,so $a = 8$.The $4^{th}$ term is $a + 3d = 8 + 3(4) = 8 + 12 = 20$.

Given an $A.P.$ whose terms are all positive integers. The sum of its first nine terms is greater than $200$ and less than $220$. If the second term is $12$,then its $4^{th}$ term is:

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