The sum of the first ten terms of an A.P. is 160, and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the G.P., and the first term of the G.P. is equal to the common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:
Solutions · 1
Official solution
Let the A.P. have first term a and common difference d. By the given conditions, the G.P. has first term d and common ratio a.
\[S_{10}=\frac{10}{2}[2a+9d]=160 \Rightarrow 2a+9d=32.\quad (1)\]
Sum of the first two terms of the G.P.: d+da=8⇒d(1+a)=8.(2)
From (1): d=932−2a. Substituting into (2):
\[\frac{32-2a}{9}(1+a)=8 \Rightarrow (32-2a)(1+a)=72 \Rightarrow -2a^2+30a-40=0 \Rightarrow a^2-15a+20=0.\]
Let the two roots be a1,a2, with a1+a2=15 (sum of roots). The corresponding G.P. first terms are d1=932−2a1,d2=932−2a2.
\[d_1+d_2=\frac{64-2(a_1+a_2)}{9}=\frac{64-30}{9}=\frac{34}{9}.\]