Sequences and Series: Arithmetic Progressions

Algebra Level 1

If the $4$ numbers

$14,\quad a,\quad 28,\quad b$

form an arithmetic progression in that order, what is the sum of $a$ and $b?$

7 14 49 56

3 solutions

Mahdi Raza
Aug 1, 2020

We have $d$ as the common difference and $a$ as the second term

$\underbrace{14}_{\red{a-d}}, \underbrace{a}_{\red{a}}, \underbrace{28}_{\red{a+d}}, \underbrace{b}_{\red{a+2d}}$

Subtract first term from the third to get $(a+d) - (a-d) = 2d = 14$ . Now we find $a$ and $b$

$\implies d = 7 \implies a = 21 \implies b = 35$

The sum is

$a+b = \boxed{56}$

Brilliant Mathematics Staff
Aug 1, 2020

Let $d$ be the common difference of the progression. Then since $14, a, 28 , b$ form an arithmetic progression, we can find $d$ as follows: $\begin{aligned} 14 + d &= a \\ a+ d &= 28 \\ 2d &=28 - 14 \\ d &= 7. \end{aligned}$

Now, we can find $a$ and $b$ as follows: $\begin{aligned} a &= 14 + d = 14 + 7 = 21 \\ b &= 28 + d= 28 + 7 = 35. \end{aligned}$

Thus, the sum of $a$ and $b$ is $56.$

Alternative way: Since the three numbers $a , 28 , b$ form an arithmetic progression, the mean of $a$ and $b$ is $28.$ Therefore, sum of $a$ and $b$ is $2 \times 28 = 56.$

I did not think of that alternative way, that's even better! We do not even need to know the common difference in that case

Mahdi Raza - 10 months, 1 week ago

Yes. Can you prove the following?

For any arithmetic progression with odd number of terms, the sum of the first term and last term is twice the middle term.

Brilliant Mathematics Staff - 10 months, 1 week ago

Yes, and as a matter of fact, this is only true for an odd number of terms. Let $n = 2m+1$ number of terms, then:

First term is: $a$
Middle term is: For odd term, $a + \frac{n-1}{2}d \implies a + \frac{(2m+1 - 1)}{2}d = a + md$
Last term is: $a + (2m+1 - 1)d = a + 2md$

$\boxed{(a) + (a + 2md) = 2(a + md)}$

Mahdi Raza - 10 months, 1 week ago

@Mahdi Raza – Well done.

Bonus: Can you work out a similar working for an arithmetic progression with even number of terms?

Bonus: Can you see how to prove the sum of all the terms of an arithmetic progression formula? $S_n= \frac n2 [2a + (n-1)d]$

Brilliant Mathematics Staff - 10 months, 1 week ago

@Brilliant Mathematics – Thanks.

Bonus 1: Yes, but a middle term does not exist for even a number of terms, hence we prove that the sum of first and last term will be equal to the two most middle terms. Here $n=2m$

First term is: $a$
Last term is: $a + (2m-1)d$
Middle term does not exist, hence the two terms are: $a + \bigg(\dfrac{2m-2}{2}\bigg)d$ and $a + \bigg(\dfrac{2m}{2}\bigg)d$

We make the claim that:

$\begin{aligned} u_{\text{First}} + u_{\text{First}} &= u_{\text{Middle 1}} + u_{\text{Middle 2}} \\ (a) + (a + (2m-1)d) &= (a + \dfrac{2m-2}{2}d) + (a + \dfrac{2m}{2}d) \\ 2a + (2m-1)d &= 2a + ((m-1) + m)d \\ 2a + (2m-1)d &= 2a + (2m-1)d \end{aligned}$

Hence the sum of first and last term is $2(a) + (2m-1)d$ , and the sum of two middle terms also $2(a) + d(m + (m-1))$ which is simply $2(a) + (2m-1)d$

Bonus 2: It's a trick that Gauss used to count the sum of 1 to 100. Write $S_{n}$ as a series from the front with the common difference being added on from the first term $(a)$ , and write $S_{n}$ again with the common difference being subtracting from the last term $(l)$ . Add the two and we get

$\begin{aligned} 2S_{n} &= n(a+l) \\ S_{n} &= \dfrac{n}{2} (a + a + (n-1)d) \\ S_{n} &= \dfrac{n}{2} (2a + (n-1)d) \end{aligned}$

Mahdi Raza - 10 months, 1 week ago

@Mahdi Raza – Literally perfect!

Brilliant Mathematics Staff - 10 months, 1 week ago

Yajat Shamji
Aug 1, 2020

$28 - 14 = 14 = 2a$

$a = 7$

$14 + 7 = 21$

$28 + 7 = 35$

$21 + 35 = \fbox{56}$

Sequences and Series: Arithmetic Progressions

3 solutions

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