Question

An arithmetic series has the property that the sum of the first ten terms is half the sum of the next ten terms .also its 100th term is 95.find the first term and common difference.

Answer 1

We have:

$$S_{10}=0.5(S_{20}-S_{10})$$

$$3S_{10}=S_{20}$$

We use the formula:

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

$$3\frac{10}{2}(2a+9d)=\frac{20}{2}(2a+19d)$$

$$2a=11d$$

100th term is 

$$95=a+(100-1)d$$

Thus,

$$2(95-99d)=11d$$

The common difference:

$$d=\frac{10}{11}$$

The first term:

$$a=\frac{11}{2} \frac{10}{11}=5$$