if 1+2+3+ +n = 666, find the - Maths - Arithmetic Progressions

Question

if 1+2+3+ +n = 666, find the value of n

Jeel · Accepted Answer

a=1
d=2-1
=1
Sn=666
(n/2)(2a+(n-1)d)=666
(n/2)(2+n-1)=666
(n^(2)+n)=1332
n^(2)+n-1332=0
n^(2)+37n-36n-1332=0
n(n+37)-36(n+7)=0
(n-36)(n+37)=0
n=36 or -37
n cannot be negative
So n=36
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if 1+2+3+..............+n = 666, find the value of n.