Question

The second term of an arithmetic progression is four times the fifth term, and the first term is 10. Find the common difference, and hence the sum of the first 12 terms.

Answer 1

$$a_1=10, a_2=a_1+d=4a_5=4(a_1+4d)$$

$$10+d=4(10+4d)$$

$$15d=-30$$

$$d=-2$$

$$S_{12}=\dfrac{2a_1+11d}{2}(12)=\dfrac{2(10)+11(-2)}{2}(12)$$

$$=-12$$