Question

The sum of the first 10 terms of an arithmetic series is 145 and the sum of its fourth and ninth term is five times the third term. Determine the first term and constant difference.

Answer 1

Answer:

First term = 1

Constant difference = 3

Step-by-step explanation:

To determine the first term and constant difference of an arithmetic series, we can begin by using the arithmetic series formula:

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

Given that the sum of the first 10 terms is 145, we can substitute Sₙ = 145 and n = 10 into the sum formula to create an equation relating a to d:

$\dfrac{10}{2}\left[2a+(10-1)d\right]=145$

Simplify:

$5\left[2a+9d\right]=145$

$2a+9d=29$

Now, we can use the general formula for the nth term of an arithmetic sequence to create expressions for the 3rd, 4th and 9th terms.

$\boxed{\begin{array}{l}\underline{\textsf{General form of the n th term of an arithmetic sequence}}\\\\a_n=a+(n-1)d\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ a_n is the nth term.}\\ \phantom{ww}\bullet\;\textsf{ a is the first term.}\\\phantom{ww}\bullet\;\textsf{ d is the common difference between terms.}\\\phantom{ww}\bullet\;\textsf{ n is the position of the term.}\\\end{array}}$

Substitute n = 3, n = 4 and n = 9 into the formula:

$a_3=a+(3-1)d=a+2d$

$a_4=a+(4-1)d=a+3d$

$a_9=a+(9-1)d=a+8d$

Given that the sum of the 4th and 9th terms is five times the 3rd term, then:

$(a+3d)+(a+8d)=5(a+2d)$

Simplify:

$a+3d+a+8d=5a+10d$

$2a+11d=5a+10d$

$11d=3a+10d$

$d=3a$

Now we have created a system of equations:

$\begin{cases}2a+9d=29\\d=3a\end{cases}$

Substitute the second equation into the first equation, then solve for a:

$2a+9(3a)=29$

$2a+27a=29$

$29a=29$

$a=1$

Therefore, the first term is a = 1.

Substitute the found value of a into the second equation, and solve for d:

$d=3(1)$

$d=3$

Therefore, the common difference is d = 3.