Number of terms of A.P. 3,7,11,15... to be taken so that sum is 406 is

English

Gujarati

Hindi

8. Sequences and Series

easy

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

$14$

Solution

(d) $S = \frac{n}{2}[2a + (n – 1)d]$

==> $406 = \frac{n}{2}\left[ {6 + (n – 1)4} \right]$

==> $812 = n\,[6 + 4n – 4]$

==> $812 = 2n + 4{n^2}$

==> $406 = 2{n^2} + n$

==> $2{n^2} + n – 406 = 0$

$ \Rightarrow $ $ = \frac{{ – 1 \pm \sqrt {1 + 4.2.406} }}{2.2}$

$ = \frac{{ – 1 \pm \sqrt {3249} }}{4}$ $=\frac{{-1 \pm 57}}{{4}}$

Taking $(+)$ sign, $n = 14$.

Standard 11

Mathematics

Share

Similar Questions

If the ${p^{th}}$ term of an $A.P.$ be $q$ and ${q^{th}}$ term be $p$, then its ${r^{th}}$ term will be

easy

If the sum of the series $2 + 5 + 8 + 11…………$ is $60100$, then the number of terms are

easy

Let $S_n$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20}=790$ and $S_{10}=145$, then $S_{15}-$ $S_5$ is:

medium

(JEE MAIN-2024)

Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is

hard

(JEE MAIN-2013)

Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$ If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals

hard

(AIEEE-2006)