Question

The first term of an arithmetic progression (A. P) is the common ratio of a GP and the first term of the same GP is the common difference of the AP. The sum of the first $10$ terms of the AP is $155$ and the sum of the first two terms of the GP is $9.$ Find the first term of the AP ?

• A. $5or\frac{1}{7}$
• B. $14$
• C. $2or\frac{25}{2}$
  
• D. $22$

Answer 1

The correct option is C $2or\frac{25}{2}$

Let $a,a+d....$be the A.P
and $d,da,d{a}^2....$be the G.P
Sum of n terms in AP is given as,
S${}_n=\frac{n}{2}[2a+(n-1\left)d\right]$
As per the question $\frac{10}{2}(2a+9d)=155$......(1)
Also, sum of first two terms ofGP is 9
So As per question,
$d+da=9$
$\therefore d(1+a)=9$...................(2)
Substituting $d=\frac{9}{1+a}$ from equation 2 in equation 1 we get,
$\frac{10}{2}[2a+9(\frac{9}{1+a}\left)\right]=155$
$\therefore 2a+2a^2+81=31+31a$
$\therefore 2a^2-29a+50=0$
Solving above quadratic equation we get,
$a=\frac{25}{2},a=2$
Hence first term of AP is $a=\frac{25}{2},a=2$