Question

Difficulty: HardNumber and Alphabet Series
Consider the following numerical sequence:
5,14,19,54,83,5, 14, 19, 54, 83, \dots
What is the value of the next term in this sequence?

Answer: 202

Answer

The next term in the numerical sequence is 202.
The sequence follows the recurrence pattern an=2an1+(1)nn2a_n = 2 a_{n-1} + (-1)^n n^2 starting with a1=5a_1 = 5. To find the 6th term, double the 5th term (83×2=16683 \times 2 = 166) and add 62=366^2 = 36, resulting in 166+36=202166 + 36 = 202.

Step-by-Step Solution

1
Identify the operation connecting the first two terms (55 and 1414).
a2=2×a1+22=2(5)+4=14a_2 = 2 \times a_1 + 2^2 = 2(5) + 4 = 14
Establishes the initial multiplier and square term offset.
2
Determine the rule for the third, fourth, and fifth terms.
a3=2(14)32=19a_3 = 2(14) - 3^2 = 19, a4=2(19)+42=54a_4 = 2(19) + 4^2 = 54, a5=2(54)52=83a_5 = 2(54) - 5^2 = 83
Shows an alternating addition and subtraction of consecutive squares n2n^2 after doubling the previous term.
3
Apply the rule for n=6n=6 to calculate the missing term.
a6=2(83)+62=166+36=202a_6 = 2(83) + 6^2 = 166 + 36 = 202
Since the 5th term subtracted 525^2, the 6th term must add 626^2 to twice the 5th term.

Key Concept

Alternating Operational and Square Offsets in Recursive Series
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