Sum of an Arithmetic Series

From the question we know that; $$a=4$$ , $$d=\frac{1}{2}$$ and $$n=20$$ .

$$S_n=\frac{1}{2}n\left(2a+\left(n-1\right)d\right)$$

$$S_{20}=\frac{1}{2}\times20\left(2\times4+\left(20-1\right)\times\frac{1}{2}\right)$$

$$S_{20}=10\times\frac{35}{2}$$

$$S_{20}=175$$

From the question we know that; $$a=4$$ , $$d=\frac{1}{2}$$ and $$n=100$$ .

$$S_n=\frac{1}{2}n\left(2a+\left(n-1\right)d\right)$$

$$S_{100}=\frac{1}{2}\times100\left(2\times4+\left(100-1\right)\times\frac{1}{2}\right)$$

$$S_{100}=50\times\frac{115}{2}$$

$$S_{20}=2875$$

From the question we know that; $$a=1$$ , $$d=3$$ and $$l=100$$ .
$$l=a+\left(n-1\right)d$$

$$100=1+\left(n-1\right)3$$

$$100-1=3n-3$$

$$99+3=3n$$

$$n=\frac{102}{3}$$

$$n=34$$

Now, we can find the sum of 34 terms in the series by;

$$S_n=\frac{1}{2}n\left(a+l\right)$$

$$S_{34}=\frac{1}{2}\times34\left(1+100\right)$$

$$S_{34}=17\times101$$

$$S_{34}=1717$$

From the question we can write down; $$S_{20}=S_{22}$$ , $$d=-2$$ and we need to know the $$a=\ ...\ ?$$

$$S_n=\frac{1}{2}n\left(2a+\left(n-1\right)d\right)$$

$$S_{20}=\frac{1}{2}\times20\left(2a+\left(20-1\right)\left(-2\right)\right)$$

$$S_{20}=10\left(2a-38\right)$$

$$S_{20}=20a-380$$

$$S_{22}=\frac{1}{2}\times22\left(2a+\left(22-1\right)\left(-2\right)\right)$$

$$S_{22}=11\left(2a-42\right)$$

$$S_{22}=22a-462$$

Now we can use the information by;

$$S_{20}=S_{22}$$

$$20a-380=22a-462$$

$$20a-22a=-462+380$$

$$-2a=-82$$

$$a=41$$

We are given that $$a=3$$ , $$S_8=2S_5$$ ,

$$S_n=\frac{1}{2}n\left(2a+\left(n-1\right)d\right)$$

$$S_8=\frac{1}{2}\times8\left(2\times3+\left(8-1\right)d\right)=24+28d$$

$$S_5=\frac{1}{2}\times5\left(2\times3+\left(5-1\right)d\right)=15+10d$$

$$S_8=2\times S_5$$

$$24+28d=2\left(15+10d\right)$$

$$28d-20d=30-24$$

$$8d=6$$

$$d=\frac{3}{4}$$

We are given that $$a=4$$ , $$d=-7$$ , and $$n=23$$ .

So, we can find that;

$$S_n=\frac{1}{2}n\left(2a+\left(n-1\right)d\right)$$

$$S_{23}=\frac{1}{2}\times23\left(2\times4\left(23-1\right)\left(-7\right)\right)$$

$$S_{23}=\frac{23}{2}\left(-146\right)$$

$$S_{23}=-1679$$