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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 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. Tags arithmetical, geometrical, progression Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post raul21 Number Theory 7 June 10th, 2014 01:41 PM jiasyuen Algebra 1 May 1st, 2014 03:40 AM PKSpark Computer Science 7 February 28th, 2014 02:16 AM mirror Number Theory 3 September 7th, 2013 08:21 AM teodork Number Theory 1 December 10th, 2007 02:24 PM

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