Question

1.    The first term of an arithmetic series is 1 . the common difference of the series is 6.Find the tenth term of the series if the sum of the first n terms of the series is 7400. Show that 3n^2-2n-7400=0

Answer 1

$$a_1=1, d=6$$

$$a_n=a_1+d(n-1)=1+6(n-1)$$

$$S_n=\dfrac{a_1+a_n}{2}\cdot n=\dfrac{1+1+6(n-1)}{2}\cdot n$$

$$=n+3n(n-1)=3n^2-2n$$

Given $S_n=7400.$

$$3n^2-2n=7400$$

$$3n^2-2n-7400=0, n\ge 1$$

$$(3n+148)(n-50)=0$$

Since $n\ge 0,$ we take $n=50.$

$$a_{50}=1+6(50-1)=295$$

$$a_{10}=1+6(10-1)=55$$