Sequences & Series
AP and GP intersection — counting pairs (n,m)
MJAT_TS2_P1
Grade 12
Question:
Let $\{a_n\}_{n\geq 1}$ be an arithmetic progression with $a_1=1$ and common difference $d>0$. Suppose $a_2,a_5,a_{14}$ are three consecutive terms of a geometric progression. Let $S_n$ denote the sum of the first $n$ terms. Let $\{b_k\}_{k\geq 1}$ be a GP with $b_1=1$ and common ratio equal to that of the GP formed by $a_2,a_5,a_{14}$. Let $T_m$ denote the sum of the first $m$ terms of $\{b_k\}$. The number of pairs of positive integers $(n,m)$ with $n>1$ such that $S_n=T_m$ is:
Step-by-Step Solution
Key Concept: From $a_5^2=a_2\cdot a_{14}$: $(1+4d)^2=(1+d)(1+13d)$. Solving: $d=2$. Common ratio $r=a_5/a_2=(1+8)/(1+2)=3$. So $S_n=n^2$ and $T_m=(3^m-1)/2$. Need $n^2=(3^m-1)/2$, i.e., $3^m-1=2n^2$, i.e., $3^m=2n^2+1$.
Valid pairs: $(n,m)=(2,2)$ and $(n,m)=(11,5)$. Total: $\mathbf{2}$ pairs.
Correct Answer: 2