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Bunuel
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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Bunuel
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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Bunuel
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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Bunuel
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.

Answer: D
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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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Bunuel
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

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