Question

The 7th term of an arithmetic series is 35.The sum to the 10th term of the series is 305.Find a, the first term of the series, and d, the common difference between terms ofthe series.

Answer 1

the common difference d is 3

therefore, the first term is 17

Step-by-step explanation

an arithmetic serie is given by the expression

`\begin{gathered} a_n=a_1+(n-1)d \\ \text{where} \\ a_1\text{ is the first term} \\ \text{and d is the common difference} \end{gathered}`

Also, the sum of the x terms of the serie is given by

`S=n((a_1+a_n)/(2))`

then

Step 1

set the equations

The 7th term of an arithmetic series is 35.

`\begin{gathered} a_7=a_1+(7-1)d \\ so \\ 35=a_1+6d\Rightarrow equation\text{ (1)} \end{gathered}`

and, The sum to the 10th term of the series is 305,

let n= 10

`\begin{gathered} S=n((a_1+a_n)/(2)) \\ 305=10((a_1+a_(10))/(2)) \\ a_(10)=a_1+(10-1)d \\ a_(10)=a_1+9d \\ \text{replace} \\ 305=10((a_1+a_1+9d)/(2)) \\ 305=10((2a_1+9d)/(2)) \\ 305=5(2a_1+9d) \\ 305=10a_1+45d\Rightarrow equation(2) \end{gathered}`

Step 2

solve the equations

`\begin{gathered} 35=a_1+6d\Rightarrow equation\text{ (1)} \\ 305=10a_1+45d\Rightarrow equation(2) \end{gathered}`

a) isolate the a1 value form equation(1) and replace in equation (2)

`\begin{gathered} 35=a_1+6d\Rightarrow equation\text{ (1)} \\ subtract\text{ 6d in both sides} \\ 35-6d=a_1+6d-6d \\ a_1=35-6d \end{gathered}`

replace in eq(2)

`\begin{gathered} 305=10a_1+45d\Rightarrow equation(2) \\ 305=10(35-6d)+45d \\ 305=350-60d+45d \\ 305=350-15d \\ \text{subtract 350 in both sides} \\ 305-350=350-350-15d \\ -45=-15d \\ \text{divide both sides by -15} \\ (-45)/(-15)=(-15d)/(-15) \\ 3=d \end{gathered}`

therefore,

the common difference d is 3

b) now, replace the d value in equation(1) and solve for a1

`\begin{gathered} 35=a_1+6d\Rightarrow equation\text{ (1)} \\ 35=a_1+6(3) \\ 35=a_1+18 \\ \text{subtract 18 in both sides} \\ 35-18=a_1+18-18 \\ 17=a_1 \end{gathered}`

therefore, the first term is 17

I hope this helps you