Question

The sum of the first ten terms of an AP is equal to $155$ and the sum of the first two terms of a GP is $9$. How many values for the first term of an AP is possible if the first term of an AP is equal to the common ratio of a GP and the first term of a GP is equal to to the common difference of a AP

Answer 1

Let the A.P and G.P be respt. a,a+d,a+2d..................and $d,da,d{a}^2..............$

Now $s_{10}ofA.P=\frac{10}{2}[2a+9d]=10a+45d=155\Rightarrow a=\frac{31-9d}{2}$

$S_{2}O)FG.P=d+da=9$

Putting value of a in (2)

WE get $d+d\left(\frac{31-9d}{2}\right)=9\Rightarrow 2d+31d-9d^2=18\Rightarrow 3d^2-11d+6=0$

$\therefore d=\frac{2}{3}or3$

Case one $d=\frac{2}{3}$

$\therefore a=\frac{31-9\frac{2}{3}}{2}=\frac{25}{2}$

$\therefore A.P=\frac{25}{2},\frac{25}{2}+\frac{2}{3},\frac{25}{2}+\frac{4}{3}............i.e\frac{25}{2},\frac{79}{6},\frac{83}{6}.............$

$\therefore G.P=\frac{2}{3},\frac{2}{3}\left(\frac{25}{3}\right),\frac{2}{3}(\frac{25}{2}{)}^2..........\frac{2}{3},\frac{25}{3},\frac{625}{3}......................$