A sequence of numbers a_1, a_2, a_3, \dots is defined by a_1 = 50 and a_{n+1} = \frac{1}{3… — GRE General Test

A sequence of numbers $a_1, a_2, a_3, \dots$ is defined by $a_1 = 50$ and $a_{n+1} = \frac{1}{3} a_n + 15$ for all positive integers $n \ge 1$ . Each term $a_n$ is rounded to the nearest integer to produce a secondary sequence $b_n$ . What is the value of $\sum_{n=1}^{4} b_n$ ?

A
129
B
130
C
131
132Answer
E
133

Answer

132

Evaluating each term yields

a_1 = 50

a_2 = 31.6667...

a_3 = 25.5556...

, and

a_4 = 23.5185...

. Rounding each term individually to the nearest integer gives

b_1 = 50

b_2 = 32

b_3 = 26

, and

b_4 = 24

. Their sum is

50 + 32 + 26 + 24 = 132

Step-by-Step Solution

Calculate the exact value of the first term

a_1

and its rounded value

b_1

a_1 = 50

, which is an integer. Thus,

b_1 = 50

Given initial value.

Calculate the exact value of

a_2

using the recursive formula

a_2 = \frac{1}{3}a_1 + 15

and round to nearest integer.

a_2 = \frac{1}{3}(50) + 15 = \frac{50}{3} + 15 = 31.6667...

. Rounding to the nearest integer gives

b_2 = 32

Apply sequence relation and rounding rule.

Calculate the exact value of

a_3 = \frac{1}{3}a_2 + 15

and round to nearest integer.

a_3 = \frac{1}{3}\left(\frac{95}{3}\right) + 15 = \frac{95}{9} + 15 = \frac{230}{9} = 25.5556...

. Rounding to the nearest integer gives

b_3 = 26

Apply sequence relation and rounding rule.

Calculate the exact value of

a_4 = \frac{1}{3}a_3 + 15

and round to nearest integer.

a_4 = \frac{1}{3}\left(\frac{230}{9}\right) + 15 = \frac{230}{27} + 15 = \frac{635}{27} = 23.5185...

. Rounding to the nearest integer gives

b_4 = 24

Apply sequence relation and rounding rule.

Sum the four rounded terms

b_1 + b_2 + b_3 + b_4

\sum_{n=1}^{4} b_n = 50 + 32 + 26 + 24 = 132

Add the individual rounded terms.

Key Concept

Recursive sequence evaluation and term-by-term rounding vs total sum rounding.

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