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505-555 (Easy)|   Sequences|      
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Bunuel
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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 17 Sep 2017, 03:20
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If a sequence is given by the expression \(S_n = S_{n-1} + 3\) and \(S_1=1\), what is the sum of the first 30 terms of the series?

A. 1335
B. 885
C. 465
D. 88
E. 58
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A
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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 17 Sep 2017, 07:43
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Let the first term of the sequence be x.
Second term will be x+3, third term is x+3+3

So, the sum will be of form
30x + 3 + 6 + 9 + ....... 42 + 45 + 48 + ...... + 87
= 30x + 90*14 + 45
= 30x + 1305

If x = 1, the sum is 1335(Option A)
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GMAT 1: 640 Q49 V29
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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 17 Sep 2017, 10:06
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Is this question complete???
I mean I am confused ..because we need atleast one term to get the exact answer.. be it the first term or any see term.

Kindly bear with me for my misunderstanding.
Thanks in advance

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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 17 Sep 2017, 10:10
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Is this question complete???
I mean I am confused ..because we need atleast one term to get the exact answer.. be it the first term or any see term.

Kindly bear with me for my misunderstanding.
Thanks in advance

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Edited the question: \(S_1=1\) part was missing. Thank you for noticing it.
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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 17 Sep 2017, 10:21
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S1 = 1,
S2 = S1+3
S3 = S2+3
.
.
.
.
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.
.
.
S30= S29+3

so.. these terms are in A.P with the difference of 3

Lets find Last term S30 = S1 + (n-1)d
S1=1, n=30, d=3... solving we get S30 = 88

Sum of terms in A.P = (first term + Last term) * Number of terms/2
Sum = (1+88)* 30 /2
= 89*15
=1335

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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 19 Sep 2017, 07:38
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Bunuel
If a sequence is given by the expression \(S_n = S_{n-1} + 3\) and \(S_1=1\), what is the sum of the first 30 terms of the series?

A. 1335
B. 885
C. 465
D. 88
E. 58
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\(S_n = S_{n-1} + 3\) and
\(S_1=1\)

\(S_2 = S_{n-1} + 3\)

\(S_2 = S_{1} + 3\) = 4
\(S_3 = S_{2} + 3\) = 7

OR
\(S_1 = 1\)
\(S_2 = 1 + 3\)
\(S_3 = 1 + 3 + 3\)
\(S_4 = 1 + 3 + 3 + 3\)

For \(S_{n}\), then, the first term is \(S_1 = 1\), and the number of 3s is one fewer than \(n\). It's an arithmetic series, first term is 1, common difference of 3, such that

\(S_n = S_1 + 3(n-1)\)

\(S_{30} = 1 + (29)3 = 88\)

Sum of arithmetic series is (average) * (number of terms), where average is (First Term + Last Term)/2

\(\frac{(1+88)}{2}\) * (30) = (89)(15) = 1,335

No need to do the last calculation. Round 89 up to 90: (90)(15). Halve and double:(45)(30) > 1200. Only A is close.

Answer A
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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 19 Sep 2017, 12:14
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Bunuel
If a sequence is given by the expression \(S_n = S_{n-1} + 3\) and \(S_1=1\), what is the sum of the first 30 terms of the series?

A. 1335
B. 885
C. 465
D. 88
E. 58
Show more

\(S_n = S_{n-1} + 3\)

The above can be written as \(S_n - S_{n-1} = 3\)

The conclusion is that this is an AP series, whose first term is 1 and common difference is 3.

Sum of an AP series = \(\frac{n}{2}*[2a + (n-1)*d]\) = \(\frac{30}{2}*[(2*1 + (30-1)*3]\) = 1335


Thus, the correct answer is Option A.
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If a sequence is given by the expression S_n = S_{n-1} + 3 and S_1 = 1 30 Jun 2021, 21:00
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