Answer :
It is an arithmetic series.
a = -4 d = +3
-4 - 1 + 2 + 5 + .... + (x-3) + x = 437
Sum of the series = [ 2 a + (n-1)d ]n /2 = 437
[ - 8 + 3(n-1) ] n / 2 = 437
- 11n + 3n² = 874
3 n² -11n - 874 = 0
n = (11 +- 103 )/6 = 114/6 = 19 as n is +ve only.
So x = a + (n-1)d = -4 + 3 (19-1) = 50
a = -4 d = +3
-4 - 1 + 2 + 5 + .... + (x-3) + x = 437
Sum of the series = [ 2 a + (n-1)d ]n /2 = 437
[ - 8 + 3(n-1) ] n / 2 = 437
- 11n + 3n² = 874
3 n² -11n - 874 = 0
n = (11 +- 103 )/6 = 114/6 = 19 as n is +ve only.
So x = a + (n-1)d = -4 + 3 (19-1) = 50
first term(a) = - 4
common difference(d) = 3
- 4 - 1 + 2 + 5 + .... + x = 437
so now using formula for sum of n terms
[tex] S_{n}=\frac{n}{2}[2a+(n-1)d] [/tex]
⇒437= n/2[ -8 + (n-1)3]
⇒n[ -8 + 3(n-1) ] = 874
⇒ -11x=n + 3n² = 874
⇒3n² -11n - 874 = 0
Using sidharacharya
no. of terms cannot be negative the we take the positive value
n = 19
So
x = a + (n-1)d = -4 + 3 (19-1) = 50
common difference(d) = 3
- 4 - 1 + 2 + 5 + .... + x = 437
so now using formula for sum of n terms
[tex] S_{n}=\frac{n}{2}[2a+(n-1)d] [/tex]
⇒437= n/2[ -8 + (n-1)3]
⇒n[ -8 + 3(n-1) ] = 874
⇒ -11x=n + 3n² = 874
⇒3n² -11n - 874 = 0
Using sidharacharya
no. of terms cannot be negative the we take the positive value
n = 19
So
x = a + (n-1)d = -4 + 3 (19-1) = 50