Which term of the A.P. 5, 10, 15, ldots excee | vedclass

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(D) Given $A.P.$ is $5, 10, 15, \ldots$Here,first term $a = 5$ and common difference $d = 10 - 5 = 5$.The $n^{th}$ term of an $A.P.$ is given by $a_n

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(D) Given $A.P.$ is $5, 10, 15, \ldots$Here,first term $a = 5$ and common difference $d = 10 - 5 = 5$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.The $31^{st}$ term is $a_{31} = 5 + (31 - 1)5 = 5 + 30 	imes 5 = 5 + 150 = 155$.Let the $n^{th}$ term exceed the $31^{st}$ term by $130$,so $a_n = a_{31} + 130$.$a_n = 155 + 130 = 285$.Using the formula $a_n = a + (n - 1)d$:$285 = 5 + (n - 1)5$.$280 = (n - 1)5$.$n - 1 = 280 / 5 = 56$.$n = 56 + 1 = 57$.Therefore,the $57^{th}$ term exceeds the $31^{st}$ term by $130$.

Which term of the $A.P.$ $5, 10, 15, \ldots$ exceeds its $31^{st}$ term by $130$ (in $^{th}$)?

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