The first term of an arithmetic series is 12 and the sum of its first five terms is 50. (a) What is the sum of first n terms of an arithmetic series having first term a and common difference d ?…

(a)

The sum of the first $n$ terms of an arithmetic series with first term $a$ and common difference $d$ is given by: $$S_n=\frac{n}{2}\left[2a+(n-1)d\right]$$

(b)

Given:

• First term ($a$) = 12
• Sum of first five terms ($S_5$) = 50

Using the sum formula for $n=5$: $$S_5=\frac{5}{2}\left[2(12)+(5-1)d\right]=50$$ $$\frac{5}{2}\left[24+4d\right]=50$$ Multiply both sides by $\frac{2}{5}$: $$24+4d=20$$ $$4d=20-24=-4$$ $$d=\frac{-4}{4}=-1$$ Common difference ($d$) = $-1$

(c)

Original first three terms of the arithmetic series: $$a_1=12,a_2=12+(-1)=11,a_3=12+2(-1)=10$$

After modification: $$a_1^{\prime }=12,a_2^{\prime }=11-5=6,a_3^{\prime }=10-7=3$$

For a geometric series, the ratio between consecutive terms must be constant: $$\frac{a_2^{\prime }}{a_1^{\prime }}=\frac{6}{12}=\frac{1}{2},\frac{a_3^{\prime }}{a_2^{\prime }}=\frac{3}{6}=\frac{1}{2}$$

Since $\frac{a_2^{\prime }}{a_1^{\prime }}=\frac{a_3^{\prime }}{a_2^{\prime }}$, the modified terms form a geometric series with common ratio $\frac{1}{2}$.

Verification: $$(a_2^{\prime }{)}^2=6^2=36,a_1^{\prime }\cdot a_3^{\prime }=12\cdot 3=36$$ Since $(a_2^{\prime }{)}^2=a_1^{\prime }\cdot a_3^{\prime }$, the terms form a geometric series.