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I am stuck on this AP questions. I know I need to find the difference, first time and such. But I am not sure how.

The sum of the first 15 terms of an Arithmetic Progression is 100 and its 10th term is 5. Determine the 5th term and the sum of the first 50 terms.

Any help will be really helpful.

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closed as off-topic by user21820, kingW3, user8795, zoli, Daniel W. Farlow Apr 2 '17 at 23:54

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  • $\begingroup$ Have you gotten any formulas to use? $\endgroup$ – mathreadler Feb 26 '17 at 23:14
  • $\begingroup$ I wasn't give any for the question but I guess I have to use one of these S = n/2[2a + (n - 1)d] S = n/2[a + l] an = a+(n−1)d but i am not sure which one to use and how. $\endgroup$ – anon420 Feb 26 '17 at 23:19
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The classic formula for the sum of $n$ terms of an AP is the average of the first and last multiplied by the number of terms:$$ s_n=n\cdot \frac{a_1+a_n}{2}$$

Any two terms equidistant from first and last should have the same average, which for the given example must be $\frac{\large 100}{\large 15}$. So since $a_8$ is equidistant from $a_1$ and $a_{15}$, we have $$a_8 = \frac{100}{15} = \frac{20}{3}$$

Since $a_{10}=5=\frac{15}{3}$, we can see that the step between terms $d$ is $ -\frac{5}{6}$. Thus we have

$$a_5 = a_8-3d = \frac{20}{3} + 3\cdot \frac 56 = \frac{55}{6} $$

This allows you to determine $a_1$ and $a_{50}$ easily and complete the question.

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  • $\begingroup$ I believe your common step value is wrong $\endgroup$ – mrnovice Feb 26 '17 at 23:36
  • $\begingroup$ I believe d = $-\frac{5}{6}$ $\endgroup$ – mrnovice Feb 26 '17 at 23:39
  • $\begingroup$ Well I just tried your values for the information given in the question, and it fails, so I'm pretty sure it's incorrect. $\endgroup$ – mrnovice Feb 26 '17 at 23:41
  • $\begingroup$ Ok it's possible I'm just having a brainfart, so what would your value of $a_{1}$ and $a_{15}$ be? $\endgroup$ – mrnovice Feb 26 '17 at 23:43
  • $\begingroup$ @mrnovice thanks for checking, fixed $\endgroup$ – Joffan Feb 26 '17 at 23:51
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In general the nth term in an AP can be written as $u_{n} = u_{1} + (n-1)d$, where $d$ is the common difference between terms. Let the sum of the AP to $n$ terms be $S_{n}$

We are given that $S_{15} = 100$ and $u_{10} = 5$.

The general formula for the sum of an AP can be derived as follows:

$S_{n} = u_{1} + u_{1} + d + ... + u_{1} + (n-1)d$

$S_{n} = u_{n} + u_{n}-d + ... + u_{n} -(n-1)d$

$2S_{n} = n(u_{1} + u_{n})$

$S_{n} = \frac{n}{2}(u_{1} +u_{n})$

We can now work out $u_{1} + 9d = 5 = u_{10} \Rightarrow u_{1} = 5-9d$

Note that $u_{15} = u_{10} + 5d = 5+5d$

Then we have $S_{15} = \frac{15}{2}(5-9d + 5+ 5d) = 100$

$\frac{40}{3}= 10-4d$

$\Rightarrow d=-\frac{5}{6}$

$\Rightarrow u_{1} = 5-9\cdot (-\frac{5}{6}) = \frac{25}{2}$

$\Rightarrow u_{50} = u_{1} + 49d = \frac{-85}{3}$

$\Rightarrow S_{50} = 25(\frac{25}{2} -\frac{85}{3}) = -\frac{2375}{6}$

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The question can be answered without the use of "formulas", just by noting that if you have an odd number of consecutive terms in an arithmetic progression, their average equals the middle term.

If you have $15$ terms of an arithmetic progression $a_1,\dots,a_{15}$ adding up to $100$, then their average $\frac{100}{15}=\frac{20}{3}$ equals the middle term, i.e. the $8^{th}$ term. The difference between consecutive terms is half the difference between the $10^{th}$ and the $8^{th}$, $\frac{5-20/3}{2}=-\frac{5}{6}$.

The $5^{th}$ term is then equal to the $10^{th}$ minus $5$ times the difference, $5-5\cdot(-\frac{5}{6})=\frac{55}{6}$.

Similarly, the $33^{rd}$ term equals the $10^{th}$ plus $23$ times the difference, $5+23\cdot(-\frac{5}{6})=-\frac{85}{6}$, and since it's the middle term of the $35$ terms from the $16^{th}$ to the $50^{th}$, the sum of those terms is $35\cdot(-\frac{85}{6})=-\frac{2975}{6}$. Adding the sum of the first $15$ terms, $100=\frac{600}{6}$, we obtain the sum of the first $50$ terms of the sequence, $-\frac{2375}{6}$.

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    $\begingroup$ Despite the fact that you're saying it can be done without the use of formulas, you are basically using those formulas without quoting them. $\endgroup$ – mrnovice Feb 26 '17 at 23:50
  • $\begingroup$ @mrnovice I believe I am only using two facts: that in an arithmetic progression with an odd number of terms, their average equals the middle term, and the basic definition of arithmetic progression (every time you "advance" one term, you add the same constant quantity). Call them "formulas" if you wish! $\endgroup$ – Anonymous Feb 27 '17 at 0:28
  • $\begingroup$ Ok now maybe, but before editing it wasn't ^^ $\endgroup$ – mrnovice Feb 27 '17 at 0:29
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Let the progression be,

$$x,x+y,x+2y,x+3y,x+4y,\cdots$$

Then the sum of the first $15$ terms is $100$, in other words:

$$\sum_{n=0}^{14} (x+ny)=100$$

$$15x+105y=100$$

And we also have the tenth term is $5$:

$$x+9y=5$$

This is a system which can be solved.

$$x=\frac{25}{2}$$

$$y=-\frac{5}{6}$$

Now you can get whatever you want.

The fifth term would be,

$$\frac{25}{2}-4(\frac{5}{6})=\frac{55}{6}$$

The sum of the first $50$ terms would be,

$$\sum_{n=0}^{49} (\frac{25}{2}-\frac{5}{6}n)=-\frac{2375}{6}$$

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  • $\begingroup$ What was the reason for a downvote? $\endgroup$ – Ahmed S. Attaalla Apr 2 '17 at 16:58

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