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If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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02 Apr 2018, 01:56
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If S is the infinite sequence such that \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n1} + 6\),…, what is the sum of all the terms from \(t_{10}\) to \(t_{18}\)? (A) 671 (B) 711 (C) 738 (D) 826 (E) 991
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If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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02 Apr 2018, 02:10
Bunuel wrote: If S is the infinite sequence such that \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n1} + 6\),…, what is the sum of all the terms from \(t_{10}\) to \(t_{18}\)?
(A) 671 (B) 711 (C) 738 (D) 826 (E) 991 As there are specific rules for straightforward calculation of sums of arithmetic sequences, we'll use them. This is a Precise approach. To calculate the sum of an arithmetic sequence, we need to know the first number, the last number and the number of elements in the sequence. Our first number is t_10 which is equal to t_1 + d*9 = 4 + 6*9 = 58. Our last number is t_18 which is equal to t_10 + d*8 = 58 + 6*8 = 106 As there are 9 total elements, our sum is (58+106)*9/2 = 164*9/2 = 82*9 = 82(10  1) = 82082 = 738 (C) is our answer. Note that as all elements of our sequence are even, we could also have eliminated (A), (B), (E) without calculation and then guessed bewteen (C) and (D).
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Re: If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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02 Apr 2018, 02:55
Bunuel wrote: If S is the infinite sequence such that \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n1} + 6\),…, what is the sum of all the terms from \(t_{10}\) to \(t_{18}\)?
(A) 671 (B) 711 (C) 738 (D) 826 (E) 991 Check NEWEST addition to Ultimate GMAT Quantitative Megathread: 'Sequences Made Easy  All in One Topic!'
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Re: If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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02 Apr 2018, 05:56
Bunuel wrote: If S is the infinite sequence such that \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n1} + 6\),…, what is the sum of all the terms from \(t_{10}\) to \(t_{18}\)?
(A) 671 (B) 711 (C) 738 (D) 826 (E) 991 \(t_{10}\) to \(t_{18}\) can be considered an AP with first term as \(t_{10}\), common difference as 6 and with total 9 terms. The first term, \(t_{10} = t_1 + 9*6 = 58\) Sum of 9 terms of AP with first term as 60 is \((n/2) * [2a + (n  1)*d] = (9/2)*[116 + 8*6] = 9*82 = 738\) Answer (C)
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Re: If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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29 Jan 2019, 20:43
Hello VeritasKarishma
A basic question that is making circles in my head...
In order to calculate n (quantity), is it always (Last  First) + 1?
Ex. Number or terms between 10 and 18:
(18  10 ) + 1 = 9?
Kind regards!



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If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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Updated on: 07 Feb 2019, 15:48
Bunuel wrote: If S is the infinite sequence such that \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n1} + 6\),…, what is the sum of all the terms from \(t_{10}\) to \(t_{18}\)?
(A) 671 (B) 711 (C) 738 (D) 826 (E) 991 \(t_{10} = 4 + (101)*6 = 58\) \(t_{18} = 4 + (181)*6 = 106\) \(\frac{9}{2}* (2*58) + (91)*6 =738\)
Originally posted by NinetyFour on 06 Feb 2019, 20:35.
Last edited by NinetyFour on 07 Feb 2019, 15:48, edited 1 time in total.



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Re: If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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07 Feb 2019, 11:24
jfranciscocuencag wrote: Hello VeritasKarishma
A basic question that is making circles in my head...
In order to calculate n (quantity), is it always (Last  First) + 1?
Ex. Number or terms between 10 and 18:
(18  10 ) + 1 = 9?
Kind regards! jfranciscocuencag, Would like to pitch in Your reasoning is correct. from 5 to 10 = 1051 = 4 6,7,8,9 between 5 and 10 = 105+1 = 6 5,6,7,8,9,10 You can always take small numbers to understand the logic.
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Re: If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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10 Feb 2019, 20:37
Bunuel wrote: If S is the infinite sequence such that \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n1} + 6\),…, what is the sum of all the terms from \(t_{10}\) to \(t_{18}\)?
(A) 671 (B) 711 (C) 738 (D) 826 (E) 991 We see we have an evenly spaced set with 6 being the difference between terms. Since t1 = 6 x 1  2 = 4 and t2 = 6 x 2  2 = 10, we see that: t10 = 6 x 10  2 = 58 and t18 = 6 x 18  2 = 106 Thus, the sum of the terms from t10 to t18 is: (58 + 106)/2 x (18  10 + 1) 82 x 9 = 738 Alternate Solution: We see that this is an arithmetic sequence, with common difference d = 6 The formula for the nth term of an arithmetic sequence is t_n = t_1 + (n – 1)d Thus, t_10 = t_1 + (10 1)6 = 4 + (9)(6) = 58 Since each succeeding term is 6 more than the previous, we have: t_11 = 64, t_12 = 70, t_13 = 76, t_14 = 82, t_15 = 88, t_16 = 94, t_17 = 100 and t_18 = 106. The sum of the terms from t_10 to t_18 is: 58 + 64 + 70 + 76 + 82 + 88 + 94 + 100 + 106 = 738 Answer: C
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Re: If S is the infinite sequence such that t(1) = 4, t(2) = 10, …, t(n) =
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10 Feb 2019, 20:37






