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

  • A
    $40$
  • B
    $96$
  • C
    $60$
  • D
    $57$

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Find the sum of all the $11$ terms of an $AP$ whose middle most term is $30$.

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The first term of an $AP$ is $-5$ and the last term is $45$. If the sum of the terms of the $AP$ is $120$,then find the number of terms and the common difference.

The first term of a finite $A.P.$ is $5$ and its last term is $95$. If the common difference of the $A.P.$ is $5$,then there are $\ldots \ldots \ldots$ terms in the $A.P.$

Two APs have the same common difference. The first term of one $AP$ is $2$ and that of the other is $7$. The difference between their $10^{\text{th}}$ terms is the same as the difference between their $21^{\text{st}}$ terms,which is the same as the difference between any two corresponding terms. Why?

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The $10^{\text{th}}$ term of the $AP: 5, 8, 11, 14, \ldots$ is

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