Question

the number of terms of the AP 3,7,11,15,… to be taken so that the sum is 406 is​

Answer 1

Answer:

The sum of the A.P.(s) = $\frac{n}{2}[2a+(n-1)d]$

We have to determine the number of terms = n

$406= \frac{n}{2}[6+(n-1)4]$

$406 = \frac{n}{2}[6+4n-4]$

$406 = \frac{n}{2}[4n+2]$

$406 = \frac{4n^{2}+2n }{2}$

$406 = \frac{2n(4n+2)}{2}$

$406 = n(2n+1)$

$406=2n^2+n$

$2n^2+n-406=0$

$n=14$

PLEASE MARK ME BRAINLIEST

Step-by-step explanation: