Question

Дано аn=43 а1=3,а2=7,а3=11,найти s11-? и n-?

Answer 1

Нахождение суммы 11 членов

$\tt\displaystyle S_{11} = \frac{2a_1 + (n - 1)\cdot d}{2}\cdot n~~~~~~~~~ where~~~n=11\\\\\\d = a_{n + 1} - a_n = a_2 - a_1 = 7 - 3 = 4\\\\\\S_{11} = \frac{2\cdot 3 + (11 - 1)\cdot 4}{2}\cdot 11=\frac{6 + 40}{2}\cdot 11=\frac{46}{2}\cdot 11=23\cdot 11 = 253$

Нахождение N

$\tt\displaystyle a_n = a_1 + (n - 1)\cdot d\\\\\\43 = 3 + (n - 1)\cdot 4\\\\\\40 = (n - 1)\cdot 4\\\\\\n - 1 = 10\\\\\\n = 11$

Проверка

a₁₁ = a₁ + (n - 1) · d = 3 + 40 = 43

ответ

S₁₁ = 253, n = 11