The first and last terms of an arithmetic series are 5 and 329 respectively. The sum of all terms is 4676. a) What is called mean in arithmetic series? Write it. [1] b) Find the common difference of…

a) Mean in an arithmetic series

The mean (average) of an arithmetic series is the value obtained by dividing the sum of all terms by the number of terms. However, in an arithmetic series, the mean can also be calculated using the first term (a₁) and the last term (aₙ) as follows:

$$\text{Mean}=\frac{\text{First term}+\text{Last term}}{2}=\frac{a_1+a_n}{2}$$

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b) Finding the common difference (d)

Given:

• First term ($a_1$) = 5
• Last term ($a_n$) = 329
• Sum of all terms ($S_n$) = 4676

Step 1: Find the number of terms ($n$)

The sum of an arithmetic series is given by: $$S_n=\frac{n}{2}(a_1+a_n)$$ Substitute the known values: $$4676=\frac{n}{2}(5+329)$$ $$4676=\frac{n}{2}(334)$$ $$4676=167n$$ $$n=\frac{4676}{167}=28$$

Step 2: Find the common difference ($d$)

The $n$th term of an arithmetic series is given by: $$a_n=a_1+(n-1)d$$ Substitute the known values: $$329=5+(28-1)d$$ $$329=5+27d$$ $$324=27d$$ $$d=\frac{324}{27}=12$$

Common difference ($d$) = 12

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c) Number of terms to remove to make the sum 51

We need to find how many terms to remove from the last so that the sum of the remaining terms is 51.

Step 1: Let $k$ be the number of terms removed from the last.

The new sum will be: $$S_{n-k}=51$$ The sum of the first $(n-k)$ terms is: $$S_{n-k}=\frac{(n-k)}{2}[2a_1+(n-k-1)d]$$ Substitute $n=28$, $a_1=5$, $d=12$: $$51=\frac{(28-k)}{2}[2(5)+(28-k-1)(12)]$$ $$51=\frac{(28-k)}{2}[10+(27-k)(12)]$$ $$51=\frac{(28-k)}{2}[10+324-12k]$$ $$51=\frac{(28-k)}{2}[334-12k]$$ Multiply both sides by 2: $$102=(28-k)(334-12k)$$ Expand: $$102=28\times 334-28\times 12k-334k+12k^2$$ $$102=9352-336k-334k+12k^2$$ $$102=9352-670k+12k^2$$ Rearrange: $$12k^2-670k+9352-102=0$$ $$12k^2-670k+9250=0$$ Divide by 2: $$6k^2-335k+4625=0$$

Step 2: Solve the quadratic equation

Use the quadratic formula: $$k=\frac{335\pm \sqrt{335^2-4\times 6\times 4625}}{2\times 6}$$ $$k=\frac{335\pm \sqrt{112225-111000}}{12}$$ $$k=\frac{335\pm \sqrt{1225}}{12}$$ $$k=\frac{335\pm 35}{12}$$

Two possible solutions:

1. $k=\frac{335+35}{12}=\frac{370}{12}=30.83$ (not valid, since $k$ must be ≤ 28)
2. $k=\frac{335-35}{12}=\frac{300}{12}=25$

Number of terms to remove = 25