Find positive integer x, if 1+2+3 +...+x=120.

Step-by-step explanation:

Given the expression  1+2+3 +...+x=120, this means that the sum of the arithmetic sequence  1+2+3 +...+x is 120.

First we need to find the xth term of the arithmetic sequence 1+2+3 +...+x

Sum of an arithmetic seq Sx = x/2(2a+(x-1)d)

a = first term

d = common difference

x is the number of terms

from the sequence, a = 1 and d = 1

Sx = x/2(2(1)+(x-1)*1)

Sx = x/2(2+x-1)

Sx = x/2(1+x)

Sx = x²+x/2

Substitutig thid into the question to calculate x;

x²+x/2 = 120

x²+x = 240

x²+x-240 = 0

x = 1±√1²-4(-240)/2

x = 1±√961/2

x = 1±31/2

x = 1+31/2 or 1-31/2

x = 32/2 or -30/2

x = 16 or -15

Since we are to look for the positive integer of x, then x is 16

the answer is simply just 6n add the two together as they have an n