Let a_1,a_2,a_3, be an A.P.

Question

Let $a_1,a_2,a_3,\ldots$ be an A.P. If $\dfrac{a_1+a_2+\cdots+a_{10}}{a_1+a_2+\cdots+a_p} = \dfrac{100}{p^2}$, $p
e10$, then $\dfrac{a_{11}}{a_{10}}$ is equal to:

Accepted Answer

$\dfrac{S_{10}}{S_p}=\dfrac{100}{p^2}=\dfrac{10^2}{p^2}$ for all valid $p$ forces $S_n = Kn^2$ (no linear term), so $a_n = S_n-S_{n-1} = K(2n-1)$.

$$\frac{a_{11}}{a_{10}} = \frac{K(21)}{K(19)} = \frac{21}{19}.$$

Solutions · 1