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Math Expert V
Joined: 02 Sep 2009
Posts: 59589
In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 78% (02:25) correct 22% (02:25) wrong based on 138 sessions

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In a certain sequence, the term an is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

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Math Expert V
Joined: 02 Aug 2009
Posts: 8281
Re: In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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Bunuel wrote:
In a certain sequence, the term an is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

It's an AP as each term is 5 more than the previous term..

1) a_1=1 so a_75=1+(75-1)*5=1+370=371..
Sum is average of the first and last number *#=(1+371)/2*75=372/2*75=186*75=13950

2) choices..
AP means the sum would be multiple of # so multiple of 75 that means 3..
Only D and E are multiple of 3..
And D is even and E is odd..
So can we find out if sum is odd or even
Since we are adding odd number one after another alternate will be odd..
74 will have 74/2=37 odd numbers and 75th is also odd..
So total odd numbers =37+1=38 and SUM of 39 odds will be even

D
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In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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Bunuel wrote:
In a certain sequence, the term $$a_n$$ is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

$$a_n$$ is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$,

$$A_1=1 ------>A_1=1$$
$$A_2=A_1+5=6->A_2=1+5$$
$$A_3=A_2+5=11->A_3=1+5+5$$$$A_4=A_3+5=16->A_4 =1+5+5+5$$

The sequence on the right just helps to show that this is an arithmetic series with a common difference of 5 and a start term of 1, that is
$$A_n=1+(n-1)5$$

$$A_2$$ is 1 + one 5
$$A_3$$ is 1 + two 5s
$$A_4$$ is 1 + three 5s
$$A_5$$ will be 1 + four 5s
$$A_{6}$$ = 1 + five 5s

Each term starts with $$A_1=1$$
Each term has a common difference of 5 (increases by 5) -- BUT
Each Term$$_{n}$$ has $$(n-1)$$ 5s.
Each term's number of 5s is one fewer than itself.
$$A_{75}$$ will have $$(n-1) = 74$$ fives

$$A_n=1+(n-1)5$$: sequence rule from above

$$A_{75}=(1+(75-1)*5)=$$
$$A_{75}=(1+(74*5))$$
$$A_{75}=(1+370)=371$$

Sum of the sequence: (average)*(# of terms)

Average in an evenly spaced sequence is $$\frac{(First+ Last)}{2}$$

# of terms, given, is 75

Sum: $$\frac{(First+ Last)}{2}*75$$

Sum = $$(\frac{1+371}{2}*75)=(186*75)$$
Double and halve
Sum = $$(93*150)=$$
Split the 93
Sum of sequence =$$(90*150+3*150)=(13,500+450)= 13,950$$

For an excellent post on sequences that is easy to follow, see this topic thread on GMAT Club, and scroll down to the post by benjiboo

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Re: In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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Bunuel wrote:
In a certain sequence, the term an is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

One of the formulas for Sum of first n members of arithmetic progression is:

S = (2a{1} + d*(n-1))/2 * n

(The othe one is (a{1}+a{n})/2 * n)

a{1} = 1, d = 5 - substitute these numbers and will get option D
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Joined: 19 Jul 2017
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Re: In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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DharLog wrote:
Bunuel wrote:
In a certain sequence, the term an is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

One of the formulas for Sum of first n members of arithmetic progression is:

S = (2a{1} + d*(n-1))/2 * n

(The othe one is (a{1}+a{n})/2 * n)

a{1} = 1, d = 5 - substitute these numbers and will get option D

Would y explain ? plz
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Re: In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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vivapopo wrote:
DharLog wrote:
Bunuel wrote:
In a certain sequence, the term an is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

One of the formulas for Sum of first n members of arithmetic progression is:

S = (2a{1} + d*(n-1))/2 * n

(The othe one is (a{1}+a{n})/2 * n)

a{1} = 1, d = 5 - substitute these numbers and will get option D

Would y explain ? plz

Sure!
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Re: In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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Bunuel wrote:
In a certain sequence, the term an is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

We see that a1 = 1, a2 = 1 + 5 = 6, a3 = 1 + 2 x 5 = 11, …, and a75 = 1 + 74 x 5 = 371.

We will use the formula sum = average x quantity. Recall that the average of an evenly spaced set is (smallest element + largest element)/2; thus, average = (371 + 1)/2 = 186. Thus, the sum is 186 x 75 = 13,950.

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Joined: 28 Jan 2019
Posts: 37
Re: In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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The first thing to note is that this is an arithmetic progression; a fancy way of saying an evenly spaced set of numbers.
When we see an arithmetic progression we should keep a few things in mind.

Average = (first term + last term)/2
Sum = Average * # terms
# terms = last term - first term / increment + 1
nth term = a + d(n-1); where a is the first term, d is the difference. This should intuitively make sense. If you have a series that increases by a constant of 3 (so: 3,6,9,12,15….), then you should be able to calculte a certain term in that series based on the constant difference (d). The 3rd term in this series (9) will = 3 + 3(2).

The above question tests whether you know these properties and how quickly you can use them to solve the question. Given the stimulus, what do we need to calculate?

We know # of terms is 75. We need to know average. In order to know average we need to know last term.

75th term = 1+5(74) = 371
1st term = 1

Average = (371+1)/2 = 186
Sum = average * # terms
Sum = 186 * 75
Sum = (180+6)(70+5) = 12,600 + 900 + 420 + 30
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Re: In a certain sequence, the term an is defined by the formula an = an –  [#permalink]

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Bunuel wrote:
In a certain sequence, the term an is defined by the formula $$a_n = a_{n – 1} + 5$$ for each integer n ≥ 2. If $$a_1 = 1$$, what is the sum of the first 75 terms of this sequence?

(A) 10,150
(B) 11,375
(C) 12,500
(D) 13,950
(E) 15,375

The sequence is arithmetic series with :-
a= 1
n = 75
d = 5

Sum of first 75 terms = n/2 [2a+ (n-1)d] = 75/2 * [2+(75-1)5] = 13950

IMO D

Posted from my mobile device Re: In a certain sequence, the term an is defined by the formula an = an –   [#permalink] 26 Nov 2019, 10:05
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