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In an arithmetic progression the difference between the any [#permalink]
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19 Jul 2011, 05:36
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In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94? A. 47 B. 63 C. 55 D. 94 E. It can't be determined
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Re: Arithmetic Progression [#permalink]
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19 Jul 2011, 05:51
kannn wrote: In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?
A. 47 B. 63 C. 55 D. 94 E. It can't be determined a10 = a1 + 9d
a14 = a1 + 13d
=> a10 + a14 = 2 * a1 + 22d = 94
Now, a1 + a23 = a1 + [a1 + (231)d]
=> a1 + a23 = a1 + a1 + 22d
=> a1 + a23 = 2 * a1 + 22d = 94
So AM from a1 to a23 = 94/2 = 47 (Because it's an AP)
Answer  A
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Re: Arithmetic Progression [#permalink]
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19 Jul 2011, 05:52
10th term: t10 = a1 + (101)d t10 = a1 + 9d 14th term: t14 = a1 +13d t10+t14=94 2a1+22d=94 Sum of 23 terms of sequence= 23/2 [2a1+(231)d] =23/2 [2a1+22d] =23/2 * 94 =1081 Arithmetic mean = sum of 23 terms / 23 = 1081/23 =47
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Re: Arithmetic Progression [#permalink]
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19 Jul 2011, 09:27
mean = (first term + last term)/2 first term = a last term = 23rd = a + (n1)d where d is the common difference. thus 23rd term = a + 22d now sum given is a+9d + a+ 13d = 2a + 22d = 94 thus mean = [a + (a+(n1)d) ]/2 = (2a+ 22d)/2 = 94/2 = 47
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Re: Arithmetic Progression [#permalink]
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19 Jul 2011, 09:44
kannn wrote: In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?
A. 47 B. 63 C. 55 D. 94 E. It can't be determined Arithmetic mean from 1st to 23rd term would be the number in 12th term: (23+1)/2=12 We just need to find the 12th term. 10,11,12,13,14 We know that the 12th term is midway from 10th and 14th. And 12th term would be (10th+14th)/2=94/2=47 Ans: "A"
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Re: Arithmetic Progression [#permalink]
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20 Aug 2012, 05:23
The problem focuses upon concepts rather than formula i.e. There is no need to use any formula for this question. In order to solve this , we must know one property of AP which is 1) Any term is equal to half the sum of the terms which are equidistant from it. or 2) The sum of terms equidistant from the beginning & end is always same and is equal to the sum of the first & last term. Now back to the problem: As the series contain 23 terms, the mean of the series will the middle term of the series which is 12th term. 12th term is equal distance from both 10th & 12th term Thus 12th term = average of 10th & 14th term = 94/2 = 47
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Re: In an arithmetic progression the difference between the any [#permalink]
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11 Dec 2013, 11:17
kannn wrote: In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?
A. 47 B. 63 C. 55 D. 94 E. It can't be determined OK this is how you solve this question in under a minute. We are asked what is (s1+s23)/2 s23 = s1+22k, where k is a constant So we need to find (2s1+22k)/2 or s1+11k We are given s10+s14=94 So s1+9k+s1+13k=94 2s1+22k=94 s1+11k=47 So 47 is our answer Hence, (A) Hope it helps Cheers! J



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Re: In an arithmetic progression the difference between the any [#permalink]
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04 Jul 2014, 09:30
jlgdr wrote: kannn wrote: In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?
A. 47 B. 63 C. 55 D. 94 E. It can't be determined OK this is how you solve this question in under a minute. We are asked what is (s1+s23)/2 s23 = s1+22k, where k is a constant So we need to find (2s1+22k)/2 or s1+11k We are given s10+s14=94 So s1+9k+s1+13k=94 2s1+22k=94 s1+11k=47 So 47 is our answer Hence, (A) Hope it helps Cheers! J QUESTION : HOW DO WE KNOW WE NEED TO FIND (2S1 + 22K ) / 2 , what manipulation was done to figure this out?



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In an arithmetic progression the difference between the any [#permalink]
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07 Oct 2014, 18:35
We can directly predicate it as question mention difference is constant so mean = median let see adding the value of first and last term/2 ,second first and second last/2.......  we can get mean. so Values of any  1+23/2, 2+22/2, 3+21/2, 4+20/2, 5+19/2, 6+18/2, 7+17/2, 8+16/2, 9+15/2, 10+14 = 94/2(given)=47(mean) mean =median =47 so answer is 47.



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Re: In an arithmetic progression the difference between the any [#permalink]
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08 Oct 2014, 00:28
Let the numbers be as follows: x, x+y, x+2y, x+3y....................................................... x+22y 1st, 2nd ..............................................................................23rd 10th number = x+9y 14th number = x+13y Given that x+9y+x+13y = 94 2x + 22y = 94 x + 11y = 47Note that x+11y is the middle term of the series which will also be the mean of the series. Answer = A
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Re: In an arithmetic progression the difference between the any [#permalink]
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09 Mar 2016, 10:10
Most of the previous analyses seem way too complicated.
Here's mine:  we're given that members of the set are evenly spaced, so it's an evenlyspaced set.  in an evenlyspaced (ordered) set the mean of any two members that are equidistant from the center is equal to the mean value of all members of the set.  the 10th and 14th members are equidistant from the center, which is the 12th member.  the mean of member 10 and member 14 = 47 = mean of the set. > A.



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Re: In an arithmetic progression the difference between the any [#permalink]
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09 Mar 2016, 11:09
Very simple one to explain. Sum of 10 th term and 14 th term is a+9d + a+13d= 2a+ 22 d = 94. Therefore a + 11d = t(12 term)=47. Mean of arithmetic progression is sum of first and last term /2. Therefore sum of 1st and 23 rd /2= 12 term = 47.



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Re: In an arithmetic progression the difference between the any [#permalink]
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06 Aug 2016, 00:42
Superb Question Here we need to use the basic principle involving mean and AP series Mean of any AP series = Median Here N=23 Median = 12th term = A+11D fro first term being A and D being the common Difference Now Given A10 + A14 =94 => A+9D+A+13D = 2A+22D=94 => A+11D=47 Hence Median = 47 So the mean = 47 Smash that A
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Re: In an arithmetic progression the difference between the any [#permalink]
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07 Aug 2017, 00:53
Average of 23 terms should be the 13th term in the series. Now to find the 13th term we can either add t1+t23 or as given in problem statement t10+t14 are also equi distant to LHS & RHS respectively of the t13. Hence avg of t10+t14=47 which should also be the 13th term and hence the avg of Sum of 1st 23 terms in the series. Option A.



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In an arithmetic progression the difference between the any [#permalink]
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07 Aug 2017, 13:00
kannn wrote: In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?
A. 47 B. 63 C. 55 D. 94 E. It can't be determined the sum of any two terms in an arithmetic progression will equal twice the term exactly between them thus the 12th term=94/2=47=mean in a 23 term progression A




In an arithmetic progression the difference between the any
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