For an A.P.,the 12^{th} term is 4 and the 20^ | vedclass

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Question

(A) Let the first term be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.For the $12^{th}$ term:

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$T_n = -3n + 40$

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(A) Let the first term be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.For the $12^{th}$ term: $a + 11d = 4$  --- $(1)$For the $20^{th}$ term: $a + 19d = -20$ --- $(2)$Subtracting equation $(1)$ from equation $(2)$:$(a + 19d) - (a + 11d) = -20 - 4$$8d = -24$$d = -3$Substitute $d = -3$ into equation $(1)$:$a + 11(-3) = 4$$a - 33 = 4$$a = 37$The $n^{th}$ term is $T_n = a + (n - 1)d = 37 + (n - 1)(-3) = 37 - 3n + 3 = -3n + 40$.

For an $A.P.$,the $12^{th}$ term is $4$ and the $20^{th}$ term is $-20$. Find the $n^{th}$ term of the $A.P.$

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