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The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
27 Jul 2018, 01:31
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The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A? (A) 243 (B) 14,400 (C) 14,500 (D) 24,300 (E) 24,545
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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
12 Aug 2018, 04:45
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ExplanationThe first term of the sequence is 45, and each subsequent term is determined by adding 2. The problem asks for the sum of the first 100 terms, which cannot be calculated directly in the given time frame; instead, find the pattern. The first few terms of the sequence are 45, 47, 49, 51,… What’s the pattern? To get to the 2nd term, start with 45 and add 2 once. To get to the 3rd term, start with 45 and add 2 twice. To get to the 100th term, then, start with 45 and add 2 ninetynine times: \(45 + (2)(99) = 243.\) Next, find the sum of all odd integers from 45 to 243, inclusive. To sum up any evenly spaced set, multiply the average (arithmetic mean) by the number of elements in the set. To get the average, average the first and last terms. Since \(\frac{45+243}{2}= 144\), the average is 144. To find the total number of elements in the set, subtract 243 – 45 = 198, then divide by 2 (count only the odd numbers, not the even ones): \(\frac{198}{2}= 99\) terms. Now, add 1 (to count both endpoints in a consecutive set, first subtract and then “add 1 before you’re done”). The list has 100 terms. Multiply the average and the number of terms: \(144 \times 100 = 14,400\)
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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
06 Jan 2019, 08:41
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Alternative Short cut approach:
The first item is 45, which is an odd number, each successive item is plus 2, implying that all 100 terms would be odd numbers. As they are evenly spaced and a total of 100, the mean should be the number between the 50th term and the 51st term. It has to be an even number since it would lie between two Odd numbers. Dividing each of the options should reveal their corresponding mean values. Our answer should be the sum that yields an even mean value. There is only one such option.



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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
06 Jan 2019, 08:56
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sandy wrote: The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?
(A) 243 (B) 14,400 (C) 14,500 (D) 24,300 (E) 24,545 Another method.. This is an AP as terms are evenly spaced.. \(A_1=45\), so \(A_{100}=45+2(1001)=45+198=243\) the average of the sequence is \(\frac{1^{st} term+ 2^{nd}}{2}=\frac{45+243}{2}=\frac{288}{2}=144\) SUM = average * number of terms = 144*100 = 14,400 B
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Some useful Theory. 1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressionsarithmeticgeometricandharmonic11574.html#p27048 2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effectsofarithmeticoperationsonfractions11573.html?sid=d570445335a783891cd4d48a17db9825 3. Remainders : https://greprepclub.com/forum/remainderswhatyoushouldknow11524.html 4. Number properties : https://greprepclub.com/forum/numberpropertyallyourequire11518.html 5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolutemodulusabetterunderstanding11281.html



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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
06 Jan 2019, 10:28
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sandy wrote: The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?
(A) 243 (B) 14,400 (C) 14,500 (D) 24,300 (E) 24,545 Solution: Given A_1 = 45 n ≥ 2, A_2 = A_1 + 2 = 45+2 = 47 Series {45, 47, 49.....) First term a = 45 and common difference d = 2 Number of terms = 100 Sum = (n/2)*(2a + (n1)d) = (100/2) * (2 x 45 + 99 x 2) = (100/2) * 288 = 14400



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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
05 Jul 2019, 23:15
chetan2u wrote: sandy wrote: The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?
(A) 243 (B) 14,400 (C) 14,500 (D) 24,300 (E) 24,545 Another method.. This is an AP as terms are evenly spaced.. \(A_1=45\), so \(A_{100}=45+2(1001)=45+198=243\) the average of the sequence is \(\frac{1^{st} term+ 2^{nd}}{2}=\frac{45+243}{2}=\frac{288}{2}=144\) SUM = average * number of terms = 144*100 = 14,400 B the average of the sequence is \(\frac{1^{st} term+ 2^{nd}}{2}=\frac{45+243}{2}=\frac{288}{2}=144\)Here it,s not 2nd term(the 243) it will be last term, edit this, Thanks
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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
05 Jul 2019, 23:18
sbingi wrote: sandy wrote: The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?
(A) 243 (B) 14,400 (C) 14,500 (D) 24,300 (E) 24,545 Solution: Given A_1 = 45 n ≥ 2, A_2 = A_1 + 2 = 45+2 = 47 Series {45, 47, 49.....) First term a = 45 and common difference d = 2 Number of terms = 100 Sum = (n/2)*(2a + (n1)d) = (100/2) * (2 x 45 + 99 x 2) = (100/2) * 288 = 14400 Avoid as much as possible the previous rules that we taught to memorize in School
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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
06 Jul 2019, 07:08
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sandy wrote: The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?
(A) 243 (B) 14,400 (C) 14,500 (D) 24,300 (E) 24,545 Let's examine a few terms to see the pattern: term1 = 45 (we add 0 two's) term2 = 45 + 2 = 47 (we add 1 two) term3 = 45 + 2 + 2 = 49 (we add 2 two's) term4 = 45 + 2 + 2 + 2 = 51 (we add 3 two's) . . . term100 = 45 + 2 + 2 ....... + 2 = 243 (we add 99 two's) So, the sum of the first 100 terms = 45 + 47 + 49 + . . . + 241 + 243 Let's add the values in PAIRS, by pairing up values from each side (left and right) of the sum. That is: 45 + 47 + 49 + . . . + 239 + 241 + 243 = (45 + 243) + (47 + 241) + (49 + 239) + .... = (288) + (288) + (288) + .... Since we have 50 PAIRS that each add to 288, the TOTAL sum = (50)(288) = 14,400 Answer: B Cheers, Brent
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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
06 Jul 2019, 15:02
sandy wrote: Explanation
The first term of the sequence is 45, and each subsequent term is determined by adding 2. The problem asks for the sum of the first 100 terms, which cannot be calculated directly in the given time frame; instead, find the pattern.
The first few terms of the sequence are 45, 47, 49, 51,… What’s the pattern? To get to the 2nd term, start with 45 and add 2 once. To get to the 3rd term, start with 45 and add 2 twice. To get to the 100th term, then, start with 45 and add 2 ninetynine times:
\(45 + (2)(99) = 243.\)
Next, find the sum of all odd integers from 45 to 243, inclusive. To sum up any evenly spaced set, multiply the average (arithmetic mean) by the number of elements in the set. To get the average, average the first and last terms. Since \(\frac{45+243}{2}= 144\), the average is 144.
To find the total number of elements in the set, subtract 243 – 45 = 198, then divide by 2 (count only the odd numbers, not the even ones): \(\frac{198}{2}= 99\) terms.
Now, add 1 (to count both endpoints in a consecutive set, first subtract and then “add 1 before you’re done”). The list has 100 terms. Multiply the average and the number of terms:
\(144 \times 100 = 14,400\) It has been explicitly said that the sum of the first 100 terms. Isn't it enough to move on instead of calculating the total number of terms in the above way?



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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
20 Aug 2019, 12:52
Here is the formula for finding a specific term in an arithmetic sequence. I didn't do the whole problem, because everyone else already has. I just thought this part might have been glossed over a little.
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Re: The sequence A is defined by An = An – 1 + 2 for each intege
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20 Aug 2019, 12:52





