Question

Consider the sequence of consecutive odd numbers starting with 1 and ending with n. If 1+3+5+…+n=49, what is the value of n?

Answer 1

Answer:

$\sqrt{48}$

Step-by-step explanation:

1+3+5...+n = 49

Sum = [n/2][2a+(n-1)d]

49 = [n/2 ] {2(1) + (n-1) 2}

98 = n {2+2n-2}

98 = 2$n^{2}$

$n^{2}$ = 48

n = $\sqrt{48}$

Hope Helped! If Yes mark me as Brainiest!

Answer 2

Answer:

Answer is 13

I honsetly dont know how to explain it, very sorry if your reading this.