For a non-constant arithmetic progression $(a_n)$ there exists a natural $n$ such that $a_{n}+a_{n+1} = a_{1}+…+a_{3n-1}$ . Prove that there are no zero terms in this progression.
$2a+(2n-1)d=(3n-1)a+\frac{(3n-1)(3n)}{2}d$
$(3n-3)a=\frac{9n^2-7n-2}{2}d$
$3a=(9n+2)d$
$d=\frac{3a}{9n+2}$
We need to prove that for every $m$: $a+md=a+\frac{3am}{9n+2}=a \frac{9n+2+3m}{9n+2}$ is non zero, which is true, because $a \neq 0$ and $9n+3m+2 \neq 0$
From the properties of an AP,
a1+(n-1)d + a1 +nd = 3n-1/2 (2a1+3nd-d)
simplifying down yields
a1 = (1-9n)d/6
Since the AP is nonconstant, d is not equal zero.
It follows that there are no zero terms in the AP.
Is this solution correct please?
RagvaloD wrote:
$2a+(2n-1)d=(3n-1)a+\frac{(3n-1)(3n)}{2}d$
$(3n-3)a=\frac{9n^2-7n-2}{2}d$
$3a=(9n+2)d$
$d=\frac{3a}{9n+2}$
We need to prove that for every $m$: $a+md=a+\frac{3am}{9n+2}=a \frac{9n+2+3m}{9n+2}$ is non zero, which is true, because $a \neq 0$ and $9n+3m+2 \neq 0$
The sum of a1+...+a3n-1 seems wrong