Question

The sum of the fist 10 terms of an arithmetic series is 155 and the sum of the second 10 terms is 455. Calculate the first three terms.

Answer 1

Answer:

2, 5, and 8

Step-by-step explanation:

To find the first three terms of an arithmetic series, given that the sum of the first 10 terms is 155 and the sum of the second 10 terms is 455, we can use the arithmetic series formula:

$\boxed{\begin{array}{l}\underline{\textsf{Sum of the first n terms of an arithmetic series}}\\\\S_n=\dfrac{n}{2}[2a+(n-1)d]\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ a is the first term.}\\\phantom{ww}\bullet\;\textsf{ d is the common difference.}\\\phantom{ww}\bullet\;\textsf{ n is the position of the term.}\\\end{array}}$

To create an equation for the sum of the first 10 terms, we can substitute Sₙ = 155 and n = 10 into the sum formula:

$\dfrac{10}{2}\left[2a+(10-1)d\right]=155$

Simplify:

$5\left[2a+9d\right]=155$

$2a+9d=31$

If the sum of the first 10 terms is 155, and the sum of the second 10 terms is 455, then the sum of the first 20 terms is the sum of these two sets. Therefore:

$\dfrac{20}{2}\left[2a+(20-1)d\right]=155+455$

Simplify:

$10\left[2a+19d\right]=610$

$2a+19d=61$

Now we have created a system of equations:

$\begin{cases}2a+9d=31\\2a+19d=61\end{cases}$

Subtract the first equation from the first equation to eliminate the term in a, then solve for d:

$2a-2a+19d-9d=61-31$

$10d=30$

$d=3$

Therefore, the common difference is d = 3.

Substitute the found value of d into one of the equations, then solve for a:

$2a+9(3)=31$

$2a+27=31$

$2a=4$

$a=2$

Therefore, the first term is a = 2.

In an arithmetic series, each term is obtained by adding the common difference (d) to the preceding term. Therefore, to find the first three terms, simply add the common difference (d) to the previous term:

$a_1=2$

$a_2=2+3=5$

$a_3=5+3=8$

So, the first three terms of the arithmetic series are 2, 5, and 8.