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kumarparitosh123
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

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app

Edited the question: \(S_1=1\) part was missing. Thank you for noticing it.
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S1 = 1,
S2 = S1+3
S3 = S2+3
.
.
.
.
.
.
.
.
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

If it helps,, Kudos Please :)
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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
\(S_n = S_{n-1} + 3\) 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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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

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

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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