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# is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

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1461
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is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

Jul 8, 2015

### Best Answer

#2
+22892
+10

is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

This ist a ARITHMETIC SEQUENCE OF HIGHER ORDER.

The sequence  is arithmetic of order k if the differences of order k are equal.

We have the order k = 2. The second differences are equal = 14.

Let us see:

$$\small{\text{ \begin{array}{lcccccccccc} Number a &a_1={ 14}& & 21& & 42& &77 & &126 & \cdots \\ First difference D^1 & & D_0^1={7}& & 21 & & 35 & & 49 & \cdots \\ Second difference D^2 & & & D_0^2={14}& & 14& &14 & \cdots \\ \end{array} }}$$

If we have a arithmetic sequence of order k, we can find $$a_n$$ by :

$$\boxed{~~a_n = a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2+\cdots+\binom{n-1}{k}\cdot D_0^k ~~}$$

So the nth term is given by:

$$\small{\text{ \begin{array}{rcl} a_n &=& a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2\\\\ a_n &=& {14} + \binom{n-1}{1}\cdot {7} + \binom{n-1}{2}\cdot {14} \qquad | \qquad \binom{n-1}{1} = n-1 \qquad \binom{n-1}{2}=\dfrac{(n-2)(n-1)}{2}\\\\ a_n &=& {14} + (n-1)\cdot {7} + \dfrac{(n-2)(n-1)}{2}\cdot {14} \\\\ a_n &=& {14} + (n-1)\cdot {7} + (n-2)(n-1)\cdot 7 \\\\ a_n &=& {14} + 7(n-1)[1+(n-2)]\\\\ a_n &=& {14} + 7(n-1)(n-1)\\\\ \mathbf{a_n} & \mathbf{=} & \mathbf{14 + 7(n-1)^2} \qquad | \qquad n \ge 1 \\ \end{array} }}$$

Jul 9, 2015

### 2+0 Answers

#1
+102386
+5

This series is not arithmetic, because there is no common difference ......the nth term - for n ≥ 2 - is given by:

14 + 7(n-1)^2

Jul 8, 2015
#2
+22892
+10
Best Answer

is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

This ist a ARITHMETIC SEQUENCE OF HIGHER ORDER.

The sequence  is arithmetic of order k if the differences of order k are equal.

We have the order k = 2. The second differences are equal = 14.

Let us see:

$$\small{\text{ \begin{array}{lcccccccccc} Number a &a_1={ 14}& & 21& & 42& &77 & &126 & \cdots \\ First difference D^1 & & D_0^1={7}& & 21 & & 35 & & 49 & \cdots \\ Second difference D^2 & & & D_0^2={14}& & 14& &14 & \cdots \\ \end{array} }}$$

If we have a arithmetic sequence of order k, we can find $$a_n$$ by :

$$\boxed{~~a_n = a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2+\cdots+\binom{n-1}{k}\cdot D_0^k ~~}$$

So the nth term is given by:

$$\small{\text{ \begin{array}{rcl} a_n &=& a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2\\\\ a_n &=& {14} + \binom{n-1}{1}\cdot {7} + \binom{n-1}{2}\cdot {14} \qquad | \qquad \binom{n-1}{1} = n-1 \qquad \binom{n-1}{2}=\dfrac{(n-2)(n-1)}{2}\\\\ a_n &=& {14} + (n-1)\cdot {7} + \dfrac{(n-2)(n-1)}{2}\cdot {14} \\\\ a_n &=& {14} + (n-1)\cdot {7} + (n-2)(n-1)\cdot 7 \\\\ a_n &=& {14} + 7(n-1)[1+(n-2)]\\\\ a_n &=& {14} + 7(n-1)(n-1)\\\\ \mathbf{a_n} & \mathbf{=} & \mathbf{14 + 7(n-1)^2} \qquad | \qquad n \ge 1 \\ \end{array} }}$$

heureka Jul 9, 2015