Question

In the ap 3,7,11...... Sum of 1st 'n' terms is 253. Find the value of n

Answer 1

sn = 253 = n/2 [2a + [n-1]d]

253 = n/2[6+[n-1]4]

506= n[6+4n-4]

506= 2n +4n^2

4n^2 + 2n -506=0

2n^2 +n-253 = 0

on factorisation we have,

2n^2 -23+22-253 = 0

2n[n+11] -23[n+11]

thus, n = -11 or 23/2

Answer 2

Here ,

a = 3
d = 4
Sⁿ = 253
n =?

Sⁿ = n/2 ( 2a + (n-1) d)

➡ 253 = n/2 ( 6 + (n-1) 4)

➡ n( 6+(n-1)4) = 506

➡ n ( 6 + 4n - 4 ) = 506

➡ 4n² + 6n - 4n - 506 = 0

➡ 4n² +2n - 506 = 0

➡ 2n² + n - 253 = 0

➡ 2n² -22n + 23 n - 253 = 0

➡ 2n ( n - 22) +23 ( n - 22)

➡ (2n +23) + (n-22)

➡ n = 22

Hope it will help u !!