In the ap 3,7,11...... Sum of 1st 'n' terms is 253. Find the value of n
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Answered by
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
Answered by
0
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 !!
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 !!
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