Question

The 4th and 13th terms of an AP are 5 and -1, find the 8th term of an AP

Answer 1

now, we know the 4th term is 5.... ok... now, what's the common difference? well, we dunno, but notice, from the 4th to the 13th term, you do have to use it 9 times to hop over to the 13th term, let's say is "d", then


`\bf \begin{array}{llll} term&value\\ ------&------\\ a_4&5\\ a_5&5+d\\ a_6&(5+d)+d\\ a_7&(5+d+d)+d\\ a_8&(5+d+d+d)+d\\ a_9&(5+d+d+d+d)+d\\ a_(10)&(5+d+d+d+d+d)+d\\ a_(11)&(5+d+d+d+d+d+d)+d\\ a_(12)&(5+d+d+d+d+d+d+d)+d\\ a_(13)&(5+d+d+d+d+d+d+d+d)+d\\ &5+9d \end{array}`

we also know that the 13th term is -1


`\bf \stackrel{a_(13)}{5+9d}=-1\implies 9d=-6\implies d=\cfrac{-6}{9}\implies \boxed{d=-\cfrac{2}{3}}`

now, recall above what the 8th term is


`\bf a_8=5+4d\implies a_8=5+4\left(-(2)/(3) \right)\implies a_8=5-\cfrac{8}{3}\implies a_8=\cfrac{7}{3}`