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If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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30 Jul 2015, 11:11
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If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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30 Jul 2015, 11:18
Bunuel wrote: If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?
A. 60 B. 120 C. 160 D. 240 E. 840
Kudos for a correct solution. nth Term of an Arithmetic Progression, \(T_n = a + (n1)d\)
Sum of n terms, \(S_n = (n/2)*[2a + (n1)*d\)
Where, a = first term of Progression, d = common difference (Second term  first term or Third  second term etc.)4th Term, \(T_4 = a + (41)*d = a+3d\) 12th Term, \(T_{12} = a + (121)*d = a+11d\) Given : (a+11d) + (a+3d) = 8i.e. 2a + 14d = 8 Question : Sum of 15 terms, \(S_{15} = (15/2)*[2a + (151)*d = (15/2)*[2a+14d] = (15/2)*8 = 60\) Answer: Option A
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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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30 Jul 2015, 15:03
Bunuel wrote: If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?
A. 60 B. 120 C. 160 D. 240 E. 840
Kudos for a correct solution. IMO: A nth term of an AP = a+(n1)dLet 1st term of AP = a+d 12th term of AP = a+11d Sum of 1st & 12th = 8 a+d+a+11d =8 2a+14d=8 a+7d = 4 (i)Sum of 1st 15 terms = (a+d)+(a+2d)+(a+3d)+.....(a+14d) =15a+(d+2d+3d+4d+..14d) = 15a+d(1+2+3+..14) Sum of 1st n natural numbers = \(\frac{n(n+1)}{2}\) = 15a +d (\(\frac{14*15}{2}\)) = 15a +d(7*15) =15(a+7d) (sub eq(i)) =15*4=60
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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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31 Jul 2015, 02:08
Ans = A
4th term + 12th term = 8 i.e., (a+3d)+(a+11d) = 8 2a+14d = 8  (1)
Now, Sum of first 15 terms = (15/2) * [2a + (151)d] = (15/2) * [2a + 14d] = (15/2) * 8  From (1) = 60



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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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31 Jul 2015, 02:35
Bunuel wrote: If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?
A. 60 B. 120 C. 160 D. 240 E. 840
Kudos for a correct solution. First term=a, common difference=d 4th term=a+3d 12th term=a+11d Sum=2a+14d=8 Now , sum upto first 15 terms= 15/2(2a+14d)=15/2(8)=60 Answer A



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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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31 Jul 2015, 07:18
Let a = first term, d = common difference. 4th term = a+3d 12th term = a+11d a+3d + a+11d = 8 2a + 14d = 8 a + 7d = 4. Sum of the first 15 terms = 15/2 (2a + 14d) = 15 (a+7d) = 15 (4) = 60. Ans (A).
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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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05 Aug 2015, 23:49
Bunuel wrote: If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?
A. 60 B. 120 C. 160 D. 240 E. 840 Ans: A solution: given that, lets say first term is a and common difference is d then, a+3d+a+11d=8 = a+7d =4 ===this is the 8th term, which will be the median of the set of 15 terms in AP so the sum of all the terms in AP= number of terms*median = 15*4 = 60
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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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06 Aug 2016, 00:47



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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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06 Aug 2016, 14:54
t4+t12=8 in AP 1,2,3,4,5.., where a and d=1, t4+t12=2*8 if we assume a and d=1/2, then ∑15=15/2*(1+14*1/2)=60



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If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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24 Aug 2017, 12:20
Bunuel wrote: If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?
A. 60 B. 120 C. 160 D. 240 the 8th term is the middle term between the 4th and 12th terms the middle term is half the sum of equidistant terms on either side the 8th term=4 because the 8th term is the middle term of a 15 term progression, it is also the median 4*15=60 A



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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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29 Aug 2017, 16:15
Bunuel wrote: If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?
A. 60 B. 120 C. 160 D. 240 E. 840 An arithmetic progression (sequence) is characterized by having each term separated from the next term by a common difference. We can let d = the common difference (i.e., the difference between each pair of consecutive terms) and let the first term = a. Thus, the first term is a, second term is a + d, and third term is a +2d, so the 4th term is a + 3d and the 12th term is a + 11d. Thus: (a + 3d) + (a + 11d) = 8 2a + 14d = 8 a + 7d = 4 We are asked to find the sum of the first 15 terms. Since this is an arithmetic progression, we can use the formula sum = quantity x average. The ‘quantity’ is 15 since there are 15 terms. The ‘average’ of a finite arithmetic progression is also the median, which in this case is the 8th term. The 8th term in terms of a and d is a + 7d, and we have found that to be 4. Thus: Sum = 15 x 4 = 60 Answer: A
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Re: If the sum of the 4th term and the 12th term of an arithmetic progress [#permalink]
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07 Sep 2017, 10:18
We can use the concept of corresponding pairs if the series is in AP. here is what i mean suppose there are 15 terms in AP the sequence is given below is in AP 2 , 4, 6, 8, 10,12,14,16,18,20,22,24,26,28,30 Then the corresponding pairs are (2,30),(4,28), (6,26), (8,24), (10,22), (12,20),(14,18) (16,16).Then the average of any corresponding pairs will be equal to average of the seq. Given that the series is in AP hence in given problem 4 pair corresponds to 12 pair in a sequence of 15 terms The sum of 4 pair and 12 pair is 8 So, sum= 15(8/2)= 60
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