The problem describes a quadratic number pattern $4, p, 11, q, 22, ...$ with a constant second difference of 1. We need to find the values of $p$ and $q$, determine the general term $T_n$ of the quadratic pattern, find the value of $n$ when $T_n = 232$, and calculate the difference between two consecutive terms whose sum is 1227.

1. Problem Description

The problem describes a quadratic number pattern $4, p, 11, q, 22, ...$ with a constant second difference of

1. We need to find the values of $p$ and $q$, determine the general term $T_n$ of the quadratic pattern, find the value of $n$ when $T_n = 232$, and calculate the difference between two consecutive terms whose sum is

1

2

2

7.

2. Solution Steps

4.1 Showing that $p = 7$ and $q = 16$.

The first differences are $p-4, 11-p, q-11, 22-q, ...$.

The second differences are $(11-p) - (p-4) = 15 - 2p$ and $(q-11) - (11-p) = q - 22 + p$ and $(22-q) - (q-11) = 33 - 2q$.

Since the second difference is constant and equal to 1, we have:

$15 - 2p = 1$

$2p = 14$

$p = 7$

Also, $33 - 2q = 1$

$2q = 32$

$q = 16$

We also have $q - 22 + p = 1$, substituting $p = 7$ and $q = 16$ we get $16 - 22 + 7 = 1$, which confirms our solution.

4.2 Determining the general term $T_n$ of the quadratic pattern.

The general form of a quadratic sequence is $T_n = an^2 + bn + c$.

We know the first few terms: $T_1 = 4, T_2 = 7, T_3 = 11, T_4 = 16, T_5 = 22$.

The first differences are: $3, 4, 5, 6, ...$

The second difference is: $1, 1, 1, ...$

Therefore, $2a = 1$, so $a = \frac{1}{2}$.

$3a + b = 3$, so $3(\frac{1}{2}) + b = 3$, thus $b = 3 - \frac{3}{2} = \frac{3}{2}$.

$a + b + c = 4$, so $\frac{1}{2} + \frac{3}{2} + c = 4$, thus $2 + c = 4$, and $c = 2$.

Therefore, $T_n = \frac{1}{2}n^2 + \frac{3}{2}n + 2 = \frac{n^2 + 3n + 4}{2}$.

4.3 Determining the value of $n$ if $T_n = 232$.

We have $T_n = \frac{n^2 + 3n + 4}{2} = 232$.

$n^2 + 3n + 4 = 464$

$n^2 + 3n - 460 = 0$

We need to find two numbers that multiply to -460 and add to

3. These numbers are 23 and -

2

0. $(n + 23)(n - 20) = 0$

$n = -23$ or $n = 20$.

Since $n$ must be a positive integer, $n = 20$.

4.4 If the sum of two consecutive terms in the pattern is 1227, calculate the difference between these two terms.

Let the two consecutive terms be $T_n$ and $T_{n+1}$.

$T_n + T_{n+1} = 1227$

$\frac{n^2 + 3n + 4}{2} + \frac{(n+1)^2 + 3(n+1) + 4}{2} = 1227$

$n^2 + 3n + 4 + n^2 + 2n + 1 + 3n + 3 + 4 = 2454$

$2n^2 + 8n + 12 = 2454$

$2n^2 + 8n - 2442 = 0$

$n^2 + 4n - 1221 = 0$

We can use the quadratic formula to solve for $n$:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 4(1)(-1221)}}{2} = \frac{-4 \pm \sqrt{16 + 4884}}{2} = \frac{-4 \pm \sqrt{4900}}{2} = \frac{-4 \pm 70}{2}$

$n = \frac{66}{2} = 33$ or $n = \frac{-74}{2} = -37$. Since n must be a positive integer, $n = 33$.

$T_{33} = \frac{33^2 + 3(33) + 4}{2} = \frac{1089 + 99 + 4}{2} = \frac{1192}{2} = 596$

$T_{34} = \frac{34^2 + 3(34) + 4}{2} = \frac{1156 + 102 + 4}{2} = \frac{1262}{2} = 631$

The difference between the two terms is $|T_{34} - T_{33}| = |631 - 596| = 35$.

3. Final Answer

4.1 $p = 7$ and $q = 16$

4.2 $T_n = \frac{n^2 + 3n + 4}{2}$

4.3 $n = 20$

4.4 The difference between the two terms is 35.