A question?

Question

The other day at college my math tutor gave me this question, and it did it, and to my tutor's surprise i had got it right. She said that this was an extremely difficult question and that the average person couldn't do it.

I need proof before i can believe her, so below is the question she gave me.. Try it out and see if you get it right.

An arithmetic progression is such that the eighth term is twice the second term and the 11th term is 18.

Find the smallest term of progression whose value exceeds 126.

{Common difference = 1.2 and first term= 6 the 26th term= 546}

Answer 1

an = a1 + (n - m)d;
Knowns: d = 1.2, a1 = 6, a26 = 546, a8 = 2a2, a11 = 18;

a11 = 18 = a1 + (18 - 1)×d;
a8 = a1 + (8 - 1)×d = 2a2 = 2(a1 + (2 - 1)×d);
a1 = 5d, d = 18/15 = 1.2, a1 = 6;

an= 6 + (n - 1)×d

n = 101, an = 126;
n = 102, an = 127.2;

Answer: n = 102... This is a question every high school student is expected to know (arithmetic progression is tested on the SAT II math 2C), so it depends what your teacher means by "extremely difficult", and "average person." In any case, I'd say the majority of the world can't solve it.

Answer 2

I got the same values for the common difference and the first term, but I got the 102nd term for the smallest that exceeds 126.

The general expression for an arithmetic progresison is:
{a + kd} for k = 0 to n

So the 8th term is a + 7d, the second term is a + 1*d, and the 11th term is a + 10d. (Remember that the first term in the progression is when k=0, not k=1.)

If the 8th term is twice the second term, and the 11th term is 18, then:

a + 7d = 2(a+d)
a + 10d = 18

Solve for 2 equations with 2 unknowns:
a + 7d = 2a + 2d; 5d = a for the first equation. Plug this into the second to get 5d + 10 = 18; 15d=18, so d = 18/15 = 6/5. If that's the value of d, then a = (6/5)*5 = 6.

So the kth term of the progression is 6 + (k-1)(6/5). To find out when this is greater than 126, set up the inequality:

6 + (k-1)(6/5) > 126
(k-1)(6/5) > 120
(k-1) > 120*(5/6)
(k-1) > 20*(5)
k > 101

So if k has to be greater than 101, then the first term that exceeds 126 is for when k=102.

Notice that 6 + (101-1)(6/5) = 6 + (20)(6) = 6 + 120 = 126 exactly, so the next term (k=102) must be the first to exceed this.

Answer 3

Is your self esteem that low ? Take the compliment and run with it. Congratulations.

Answer 4

I'm probably wrong, but is it 21?

Anyway, congrats on the compliment.