6 uždavinys

5 uždavinys7 uždavinys

Seka a₁, a₂, ... a_n, ... yra aritmetinė progresija, kurios a₅ + a_n = a₂ + a₁₀.

Raskite n.

A 5 B 6 C 7 D 8 E 9

Sprendimas.

a_₅+a__n =
a_₂+a_₁₀

a_₅+a__n = a_₂+a_₁₀ $a_{5}+a_{n}$ = $a_{2}+a_{10}$

a__n = a_₂+a_₁₀-a_₅ $a_{n}$ = $a_{2}+a_{10}-a_{5}$

a__n = a_₁+d+a_₁+ d* 9-(a_₁+ d* (5-1)) $a_{n}$ = $a_{1}+d+a_{1}+d\cdot 9-(a_{1}+d\cdot (5-1))$

a__n = a_₁+d+a_₁+ d* 9-(a_₁+ d* 4) $a_{n}$ = $a_{1}+d+a_{1}+d\cdot 9-(a_{1}+d\cdot 4)$

a__n = a_₁+d+a_₁+ d* 9-a_₁- d* 4 $a_{n}$ = $a_{1}+d+a_{1}+d\cdot 9-a_{1}-d\cdot 4$

a__n = a_₁-a_₁+d+a_₁+ d* 9- d* 4 $a_{n}$ = $a_{1}-a_{1}+d+a_{1}+d\cdot 9-d\cdot 4$

a__n = a_₁-a_₁+a_₁+d+ d* 9- d* 4 $a_{n}$ = $a_{1}-a_{1}+a_{1}+d+d\cdot 9-d\cdot 4$

a__n = a_₁+0+d+ d* 9- d* 4 $a_{n}$ = $a_{1}+0+d+d\cdot 9-d\cdot 4$

a__n = a_₁+d+ d* 9- d* 4 $a_{n}$ = $a_{1}+d+d\cdot 9-d\cdot 4$

a__n = a_₁+ (7-1)* d $a_{n}$ = $a_{1}+(7-1)\cdot d$

$a_{5}+a_{n}$ = $a_{2}+a_{10}$

$a_{n}$ = $a_{2}+a_{10}-a_{5}$

$a_{n}$ = $a_{1}+d+a_{1}+d\cdot (10-1)-(a_{1}+d\cdot (5-1))$

$a_{n}$ = $a_{1}+d+a_{1}+d\cdot 9-(a_{1}+d\cdot (5-1))$

$a_{n}$ = $a_{1}+d+a_{1}+d\cdot 9-(a_{1}+d\cdot 4)$

$a_{n}$ = $a_{1}+d+a_{1}+d\cdot 9-a_{1}-d\cdot 4$

$a_{n}$ = $a_{1}+6\cdot d$

$a_{n}$ = $a_{1}+(7-1)\cdot d$

Pagal aritmetinės progresijos n- o nario formulę $a_{n}$ = $a_{1}+(n-1)\cdot d$ ,

$a_{1}+(7-1)\cdot d$ = $a_{7}$ ,

taigi, n = 7

Atsakymas: C

5 uždavinys7 uždavinys