Which term of the A.P. 5, 15, 25, ldots excee | vedclass

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(D) Given $A.P.$ is $5, 15, 25, \ldots$Here,the first term $a = 5$ and the common difference $d = 15 - 5 = 10$.The $n^{th}$ term of an $A.P.$ is given

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(D) Given $A.P.$ is $5, 15, 25, \ldots$Here,the first term $a = 5$ and the common difference $d = 15 - 5 = 10$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.The $31^{st}$ term is $a_{31} = a + 30d = 5 + 30(10) = 5 + 300 = 305$.Let the $n^{th}$ term exceed the $31^{st}$ term by $130$.So,$a_n = a_{31} + 130$.$a_n = 305 + 130 = 435$.Using the formula $a_n = a + (n - 1)d$:$435 = 5 + (n - 1)10$.$430 = (n - 1)10$.$43 = n - 1$.$n = 44$.Therefore,the $44^{th}$ term exceeds the $31^{st}$ term by $130$.

Which term of the $A.P.$ $5, 15, 25, \ldots$ exceeds its $31^{st}$ term by $130$ (in $^{th}$)?

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