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 September 17th, 2015, 08:35 AM #1 Newbie   Joined: Sep 2015 From: Kenya Posts: 6 Thanks: 0 Arithmetical and Geometrical Progression The 2nd, 5th and 14th term in an arithmetical progression form the 1st three terms in a Geometrical Progression. The 10th term in the Arithmetical progression is 57. Find the : a) 1st term and common difference in the AP b) common ratio in the GP HELP PLEASE
September 17th, 2015, 01:19 PM   #2
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Quote:
 Originally Posted by Worryinglamb08 The 2nd, 5th and 14th term in an arithmetical progression form the 1st three terms in a Geometrical Progression. The 10th term in the Arithmetical progression is 57. Find the : a) 1st term and common difference in the AP b) common ratio in the GP HELP PLEASE
Let a = AP first term, i = AP increment
Let c = GP first term, m = GP multiplier

a+i=c
a+4i=cm
$\displaystyle a+13i=cm^2$
a+9i=57

You now have 4 equations in 4 unknowns.

 September 17th, 2015, 01:49 PM #3 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 We have the AP: $\displaystyle a_1,\ a_2,\ a_3, ...$ If the common difference is d, then $\displaystyle a_2=a_1+d \\\;\\ a_5=a_1+4d \\\;\\ a_{14}=a_1+13d$ We have the GP: $\displaystyle a_1+d,\ a_1+4d,\ a_1+13d,\ ...$ These three consecutive terms will satisfy the equation: $\displaystyle (a_1+4d)^2=(a_1+d)(a_1+13d) \ \Longrightarrow\ ...\ \Longrightarrow\ d = 2a_1\ \ \ (*)$ In the AP, we know that: $\displaystyle a_{10} = 57 \ \Longrightarrow\ a_1+9d=57 \ \stackrel{(*)}{\Longrightarrow}\ a_1+9\cdot2a_1=57 \ \Longrightarrow\ 19a_1=57 \ \Longrightarrow\ a_1=3. \\\;\\ a_1=3\ \stackrel{(*)}{\Longrightarrow} d= 6.$
 September 17th, 2015, 03:12 PM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 As 2 + 2 * 14 = 3 * 10, $\displaystyle a_2 + a_{14} + a_{14} = 3 \cdot a_{10} = 171$, 171 = a2 * (2r^2 + 1) where r is the common ratio. (14 - 5) / (5 - 2) = 3 so r = 3 giving $\displaystyle a_2 = 9$ so $\displaystyle a_{14} = 81$ giving d = (81 - 9) / (14 - 2) = 6.
 September 17th, 2015, 03:55 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond $\displaystyle a_1=57-9d$ $\displaystyle \dfrac{a_1+4d}{a_1+d}=\dfrac{a_1+13d}{a_1+4d}$ Substitute for $\displaystyle a_1$ and solve for $\displaystyle d$. Thanks from Hoempa and Worryinglamb08
 September 18th, 2015, 02:26 AM #6 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Greg, that shows two solutions! Thanks from greg1313 and Worryinglamb08
 September 18th, 2015, 07:35 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Yes, one of which is always $\displaystyle d=0$. Thanks from Worryinglamb08
September 18th, 2015, 11:17 AM   #8
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Quote:
 Originally Posted by greg1313 Yes, one of which is always $\displaystyle d=0$.

If d = 0, we have a beautiful and gentle banality:

$\displaystyle a_1=a_2=a_3= ... =a_{10}= ...= a_n = 57$

So, we all celebrate another reality.

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