Question

Let a1 , a2 , a3 , a4 , a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
If the sum of the numbers in the new sequence is 450, then a5 is

• A. 81
• B. 41
• C. 61
• D. 51

Answer 1

The Correct Option is D

To find the value of \( a_5 \), we begin by identifying the sequences involved:

• Let the sequence of consecutive odd numbers be \( a_1, a_2, a_3, a_4, a_5 \). Represent this as \( a_1 = n, a_2 = n + 2, a_3 = n + 4, a_4 = n + 6, a_5 = n + 8 \), where \( n \) is the first odd number.
• The sequence of consecutive even numbers ending with \( 2a_3 \) is: \( 2a_3 - 8, 2a_3 - 6, 2a_3 - 4, 2a_3 - 2, 2a_3 \).

According to the problem, the sum of these even numbers is 450:

\((2a_3 - 8) + (2a_3 - 6) + (2a_3 - 4) + (2a_3 - 2) + 2a_3 = 450\)

Combine like terms:

\[\begin{align*} 5(2a_3) - (8 + 6 + 4 + 2) &= 450 \\ 10a_3 - 20 &= 450 \\ 10a_3 &= 470 \\ a_3 &= 47 \end{align*}\]

Now, since \( a_3 = n + 4 \):

\( n + 4 = 47 \)

\( n = 43 \)

Thus, \( a_5 = n + 8 = 43 + 8 = 51 \).

Therefore, the value of \( a_5 \) is 51.