Question

The number of terms of AP:3,7,11,15,.... to taken so that sum is 406

Answer 1

ANSWER : n = 14

Sn = 406
a = 3
d = 7 - 3 = 4
n = ?

Sn = n/2 ( 2a + (n-1) d)
406 = n/2 ( 2 x 3 + (n-1) 4)
406 x 2 = n ( 6 + 4n -4)
812 = n ( 2 + 4n)
812 = 2n + 4n sq.

4n sq. + 2n - 812 = 0
2n sq. + n - 406 = 0. .....[ taking 2 common ]
2n sq. - 28n + 29n - 406 = 0
2n ( n - 14) + 29 ( n - 14) = 0
(2n + 29) (n-14) = 0

2n + 29 = 0
n = -29/2

n - 14 = 0
n = 14

SINCE VALUE OF n CANNOT BE IN FRACTION, THEREFORE n = 14

HOPE IT HELPS
☜☆☞

Answer 2

$\underline{\bold{Given:-}}$

3, 7, 11, 15,. ....... forms an A.P.

$\underline{\bold{Solution:-}}$

First term (a) = 3

It forms an A.P. so there is the common difference in between them.

Common difference (d) = 7-3 = 4

$S_{n} = 406 \\ \\ \frac{n}{2}[ 2a + (n - 1)d ]= 406 \\ \\ n[2(3) + (n - 1)(4)] = 406 \times 2 \\ \\ n[6 + 4n - 4 ]= 812 \\ \\ n(4n + 2) = 812$
$4 {n}^{2} + 2n - 812 = 0 \\ \\ On \: dividing \: both \: sides \: by \: 2 \\ \\ 2 {n}^{2} + n - 406 = 0$
$2 {n}^{2} -28n + 29n - 406 = 0$
$2n(n - 14) + 29(n - 14) = 0 \\ \\ (3n + 29)(n - 14) = 0 \\ \\ if \: 2n + 29 = 0 \\ \\ 2n = - 29 \\ \\ n = \frac{ - 29}{2} \: \: \: (not \: possible) \\ \\ if \: n - 14 = 0 \\ \\ n = 14$

So, there will be 14 terms.