The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is

  • A

    $5$

  • B

    $10$

  • C

    $12$

  • D

    $14$

Similar Questions

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.

Let $S _{ n }=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots$ upto n terms. If the sum of the first six terms of an $A.P.$ with first term $- p$ and common difference $p$ is $\sqrt{2026 S_{2025}}$, then the absolute difference between $20^{\text {th }}$ and $15^{\text {th }}$ terms of the $A.P.$ is

  • [JEE MAIN 2025]

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

  • [JEE MAIN 2019]

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