Дана арифметическая прогрессия a 1 , a 2 , … , a n , у которой a 1 = 15 , a n = 4 ....

Ответ: 661

Пошаговое объяснение:

$\dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+\dots+\dfrac{1}{a_{n-1}a_n}=-\dfrac{1}{d}\left(\dfrac{1}{a_1(a_1+d)}+\dfrac{1}{(a_1+d)(a_1+2d)}+\dots +$

$\dfrac{1}{(a_1+(n-2)d)(a_1+(n-1)d)}\bigg)=-\dfrac{1}{d}\left(\dfrac{a_1-(a_1+d)}{a_1(a_1+d)}+\dfrac{a_1+d-(a_2+2d)}{(a_1+d)(a_1+2d)}+$

$+\dots +\dfrac{a_1+(n-2)d-(a_1+(n-1)d)}{(a_1+(n-2)d)(a_1+(n-1)d)}\bigg)=-\dfrac{1}{d}\bigg(\dfrac{1}{a_1+d}-\dfrac{1}{a_1}+\dfrac{1}{a_1+2d}-\\ \\ -\dfrac{1}{a_1+d}+...+\dfrac{1}{a_1+(n-1)d}-\dfrac{1}{a_1+(n-2)d}\bigg)=\dfrac{1}{d}\bigg(\dfrac{1}{a_1}-\dfrac{1}{a_n}\bigg)=11\\ \\ d=\dfrac{1}{11}\times \bigg(\dfrac{1}{15}-\dfrac{1}{4}\bigg)\\ \\ d=-\dfrac{1}{60}\\ \\ a_n=a_1+(n-1)d~~\Leftrightarrow~~ n=661$