Question

How many terms of the series 3+7+11+....must be taken to get a sum of 1176

Answer 1

↑ Here is your answer ↓
_____________________________________________________________
_____________________________________________________________

$a = 3$

$d = 4$

$S_n = 1176$

$S_n = n/2 (2a+(n-1)d)$

$= n/2(6+(n-1)4)$

$= n/2(6+4n-4)$

$= n/2(2+4n)$

$1176 = (2n+4n^2)/2$

$2352 = 2n + 4n^2$

$4n^2 + 2n - 2352 = 0$

By solving qudratic equation,

$n = -49/2, n = 24$

Reject the negative term,

$n = 24$

Answer 2

Given AP : 3 + 7 + 11+....

a= 3, d= 7-3=4, Sn= 1176

We know that 

Sn= n/2( 2a + (n-1)d)
⇒1176= n/2 ( 2 * 3 + (n-1) 4)
⇒1176 = n/2 ( 6 + 4n -4)
⇒1176= n/2 ( 2 + 4n)
⇒2352 = n( 2 + 4n)
⇒2352 = 2n + 4n²
⇒ 4n² + 2n - 2352= 0

So now we have to middle term split 4n² + 2n - 2352=0

We get n= -49/2 or 24

Therefore, n=24