For the finite A.P. 40, 35, 30, ldots, -200, | vedclass

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(C) To find the $n^{th}$ term from the end of an $A.P.$,we can reverse the sequence or use the formula: $l - (n - 1)d$,where $l$ is the last term,$n$

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$-155$

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(C) To find the $n^{th}$ term from the end of an $A.P.$,we can reverse the sequence or use the formula: $l - (n - 1)d$,where $l$ is the last term,$n$ is the position from the end,and $d$ is the common difference of the original sequence.Given $A.P.$ is $40, 35, 30, \ldots, -200$.Here,the first term $a = 40$,common difference $d = 35 - 40 = -5$,and the last term $l = -200$.We need to find the $10^{th}$ term from the end.Using the formula: $a_n (	ext{from end}) = l - (n - 1)d$.Substituting the values: $a_{10} = -200 - (10 - 1)(-5)$.$a_{10} = -200 - (9)(-5)$.$a_{10} = -200 + 45$.$a_{10} = -155$.Thus,the $10^{th}$ term from the end is $-155$.

For the finite $A.P.$ $40, 35, 30, \ldots, -200,$ find the $10^{th}$ term from the end.

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