For the finite A.P. 12, 21, 30, ldots, 363, f | vedclass

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(B) The given $A.P.$ is $12, 21, 30, \ldots, 363$.Here,the first term $a = 12$ and the common difference $d = 21 - 12 = 9$.The last term $l = 363$.To

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$264$

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(B) The given $A.P.$ is $12, 21, 30, \ldots, 363$.Here,the first term $a = 12$ and the common difference $d = 21 - 12 = 9$.The last term $l = 363$.To find the $n^{th}$ term from the end of an $A.P.$,we use the formula: $n^{th}$ term from end $= l - (n - 1)d$.Here,$n = 12$,$l = 363$,and $d = 9$.Substituting the values: $12^{th}$ term from end $= 363 - (12 - 1) 	imes 9$.$= 363 - (11 	imes 9) = 363 - 99 = 264$.Thus,the $12^{th}$ term from the end is $264$.

Part $I$	Part $II$
$1$. Sixth term of the $A.P.$ $1, 3, 5, 7, \ldots$	$a$. $105$
$2$. Eleventh term of the $A.P.$ $3, 6, 9, 12, \ldots$	$b$. $11$
$3$. Sixteenth term of the $A.P.$ $4, 6, 8, 10, \ldots$	$c$. $33$
$4$. Twenty-first term of the $A.P.$ $5, 10, 15, 20, \ldots$	$d$. $34$

For the finite $A.P.$ $12, 21, 30, \ldots, 363,$ find the $12^{th}$ term from the end.

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