The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is
$5$
$10$
$12$
$14$
A man deposited $Rs$ $10000$ in a bank at the rate of $5 \%$ simple interest annually. Find the amount in $15^{\text {th }}$ year since he deposited the amount and also calculate the total amount after $20$ years.
How many terms of the $A.P.$ $-6,-\frac{11}{2},-5, \ldots \ldots$ are needed to give the sum $-25 ?$
Let the sum of the first $n$ terms of a non-constant $A.P., a_1, a_2, a_3, ……$ be $50\,n\, + \,\frac{{n\,(n\, - 7)}}{2}A,$ where $A$ is a constant. If $d$ is the common difference of this $A.P.,$ then the ordered pair $(d,a_{50})$ is equal to
If the ratio of the sum of $n$ terms of two $A.P.'s$ be $(7n + 1):(4n + 27)$, then the ratio of their ${11^{th}}$ terms will be