6 uždavinys
Seka a1, a2, ... an, ... yra aritmetinė progresija, kurios a5 + an = a2 + a10.
Raskite n.
A 5 B 6 C 7 D 8 E 9
Sprendimas.
a_5+a_n =
a_2+a_10
a_2+a_10
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a_5+a_n = a_2+a_10$$a_{5}+a_{n}$$ = $$a_{2}+a_{10}$$
a_n = a_2+a_10-a_5$$a_{n}$$ = $$a_{2}+a_{10}-a_{5}$$
Paaiškinimas: $${\normalsize a_{2} = a_{1}+d}$$.
Pagal progresijos n - o nario formulę $${\normalsize a_{n} = a_{1}+d\cdot (n-1)}$$
Pagal progresijos n - o nario formulę $${\normalsize a_{n} = a_{1}+d\cdot (n-1)}$$
a_n = a_1+d+a_10-a_5$$a_{n}$$ = $$a_{1}+d+a_{10}-a_{5}$$
Paaiškinimas: $${\normalsize a_{10} = a_{1}+d\cdot (10-1)}$$.
Pagal progresijos n - o nario formulę $${\normalsize a_{n} = a_{1}+d\cdot (n-1)}$$
Pagal progresijos n - o nario formulę $${\normalsize a_{n} = a_{1}+d\cdot (n-1)}$$
a_n = a_1+d+a_1+ d* (10-1)-a_5$$a_{n}$$ = $$a_{1}+d+a_{1}+d\cdot (10-1)-a_{5}$$
Paaiškinimas: $${\normalsize a_{5} = a_{1}+d\cdot (5-1)}$$.
Pagal progresijos n - o nario formulę $${\normalsize a_{n} = a_{1}+d\cdot (n-1)}$$
Pagal progresijos n - o nario formulę $${\normalsize a_{n} = a_{1}+d\cdot (n-1)}$$
a_n = a_1+d+a_1+ d* (10-1)-(a_1+ d* (5-1))$$a_{n}$$ = $$a_{1}+d+a_{1}+d\cdot (10-1)-(a_{1}+d\cdot (5-1))$$
10 - 1 = 9
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* (5-1))$$a_{n}$$ = $$a_{1}+d+a_{1}+d\cdot 9-(a_{1}+d\cdot (5-1))$$
5 - 1 = 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+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$$a_{n}$$ = $$a_{1}+d+d\cdot 9-d\cdot 4$$
d + d * 9 = 10 * d
a_n = a_1+ 10* d- d* 4$$a_{n}$$ = $$a_{1}+10\cdot d-d\cdot 4$$
10 * d - d * 4 = 6 * d
a_n = a_1+ 6* d$$a_{n}$$ = $$a_{1}+6\cdot d$$
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