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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
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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 \(A_{100}=45+2(100-1)=45+198=243\)
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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Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
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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 + (n-1)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]
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 \(A_{100}=45+2(100-1)=45+198=243\)
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]
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 + (n-1)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]
2
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]
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 ninety-nine 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]
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 [#permalink]
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