Sequences and Series 1

Let the terms be $1,1+d,1+2d$

Sum of a finite arithmetic series is $S=\frac{n(a_1+a_n)}{2}$

Substitution: $\frac{9(2+8d)}{2}=369$ ............ $d=10$

So the 9th term is $81$

For the geometric series, we can express the terms as $1,1r,1r^2,1r^3,........,1r^8$

Since the 1st and the 9th term are congruent to that of the arithmetic series, we can conclude $1r^8=81$ and $r=\frac{1}{8}$

The 7th term is $1r^6=1(81^\frac{6}{8})=\boxed{27}$

First term of the A.P. is $1$ , common difference is $d$ (let) and sum of first nine terms is $S_{9}=\frac{9}{2}[2*1+(9-1)d]=369$ or, $d=10$ .

So, $9^{\text{th}}$ term of the A.P. is $t_{9}=1+(9-1)*10=81$

If common ratio of the G.P. is $r$ , then $1*r^{9-1}=81$ or, $r^{2}=3$ .

So, seventh term of the G.P. is $1*r^{6}=3^3=\boxed{27}$