MHB Show that the number of questions he answered is 14, Arithmetic progressions

[mathlearn]

A competitor participating in a programme organised by a certain telelvision channel, wins by answering 15 questions correctly. The price money of 50 for the first question, 75 for second question , 100 for third question etc ... given for correct answers , are in an arithmetic progressionA competitor has to leave the programme if a wrong answer is given . In this case , the competitor's prize money is half the amount allocated for all the questions he has answered correctly thus far. If a certain competitor had to leave the competition with 1300 due to not giving the correct answer to a certain question , show that the number of questions he answered is 14.

Thoughts

The original prize money he earned is 2600

This is an sum of an arithmetic progression right?

Formula preferences

I would think that two formulas are possible in this case

$\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$

or $\displaystyle S_n=\frac{n}{2}\left(a_1+l\right)$ since we know the last term

My attempt

I tried substituting the values into the formula

a=50, n = ? , d= 25

$\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$

$\displaystyle 2600=\frac{n}{2}\left(50+(n-1)25\right)$

$\displaystyle 2600=\frac{n}{2}\left(50+25n-25\right)$

$\displaystyle 5200={n}\left(50+25n-25\right)$

$\displaystyle 5200=50n+25n^2-25n$

$\displaystyle 5200=25n+25n^2$

$\displaystyle 208=n+n^2$

Assuming that my method of solving this method was true what should be done further to obtain the answer

Many Thanks :)

 

[MarkFL]

You substituted 50 for $2a_1$, when you need to use 100...and then your resulting equation would be:

$$208=3n+n^2$$

or

$$n^2+3n-208=0$$

Now, we may observe that:

$$(16)(-13)=-208$$ and $$(16)+(-13)=3$$

Thus, the equation may be written as:

$$(n+16)(n-13)=0$$

Discarding the negative root, we are left with:

$$n=13$$

And this is the number of questions answered correctly, and so the error must have occurred on the 14th question.

 

[kaliprasad]

A competitor participating in a programme organised by a certain telelvision channel, wins by answering 15 questions correctly. The price money of 50 for the first question, 75 for second question , 100 for third question etc ... given for correct answers , are in an arithmetic progressionA competitor has to leave the programme if a wrong answer is given . In this case , the competitor's prize money is half the amount allocated for all the questions he has answered correctly thus far. If a certain competitor had to leave the competition with 1300 due to not giving the correct answer to a certain question , show that the number of questions he answered is 14.

Thoughts

The original prize money he earned is 2600

This is an sum of an arithmetic progression right?

Formula preferences

I would think that two formulas are possible in this case

$\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$

or $\displaystyle S_n=\frac{n}{2}\left(a_1+l\right)$ since we know the last term

My attempt

I tried substituting the values into the formula

a=50, n = ? , d= 25

$\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$

$\displaystyle 2600=\frac{n}{2}\left(50+(n-1)25\right)$

$\displaystyle 2600=\frac{n}{2}\left(50+25n-25\right)$

$\displaystyle 5200={n}\left(50+25n-25\right)$

$\displaystyle 5200=50n+25n^2-25n$

$\displaystyle 5200=25n+25n^2$

$\displaystyle 208=n+n^2$

Assuming that my method of solving this method was true what should be done further to obtain the answer

Many Thanks :)

The problematic point is
$\displaystyle 2600=\frac{n}{2}\left(50+25n-25\right)$
which should be
$\displaystyle 2600=\frac{n}{2}\left(2* 50+25n-25\right)$

proceeding you shall get

$\displaystyle 208=3n+n^2$

this is quadratic in n and we have
$(n^2+3n- 208 = 0$ factoring we get $(n-13)(n+16) = -$ so n = 13 or - 16
but n is positive so n = 13
one incorrect ans giving 14 as the answer