Класс 11 RD Sharma Solutions – Глава 19 Арифметические прогрессии – Упражнение 19.4 | Набор 3
Вопрос 23. Если 5-й и 12-й члены АП равны 30 и 65 соответственно, какова сумма первых 20 членов?
Решение:
According to question we have
a5 = 30
Using the formula
⇒ a + (5 – 1)d = 30
⇒ a + 4d = 30 ……………. (i)
Also, an = 65
Using the formula
⇒ a + (12 – 1)d = 65
⇒ a + 11d = 65 ……………(ii)
On solving eq(i) and (ii), we get
7d = 35
⇒ d = 5
On putting the value of d in (i), we get :
a + 4 x 5 = 30
⇒ a = 10
Now we find the sum of first 20 terms
S20 = 20/2 [2 x 10 + (20 – 1) x 5]
⇒ S20 = 10 [20 + 95]
⇒ S20 = 1150
Вопрос 24. Найдите сумму n членов АП, k-е члены которых равны 5k + 1.
Решение:
According to question we have
ak = 5k + 1
For k = 1, a1 = 5 x 1 + 1 = 6
For k = 2, a2 = 5 x 2 + 1 = 11
For k = n, an = 5n + 1
Now we find the sum of n terms
Sn = n/2 [a + an]
⇒ Sn = n/2[6 + 5n + 1] = n/2 (5n + 7)
Вопрос 25. Найдите сумму всех двузначных чисел, которая при делении на 4 дает в остатке 1.
Решение:
According to question we have
A.P = 13, 17….97
From the A.P we have
a = 13, d = 4, an = 97
Now using the formula
⇒ an = a (n – 1)d
⇒ 97 = 13 + (n – 1)4
⇒ 84 = 4n – 4
⇒ 88 = 4n
⇒ 22 = n …………… (1)
Now we find the sum of all two-digit numbers using the given formula
Sn = n/2[2a + (n – 1)d]
S22 = 22/2 [2 x 13 + (22 – 1) x 4] ( From eq(1))
⇒ S22 = 11[26 + 84]
⇒ S22 = 11[110] = 1210
Вопрос 26. Если сумма некоторого количества членов АП 25, 22, 19, … равна 116. Найдите последний член.
Решение:
According to question
A.P. = 25, 22, 19
From the A.P we have
a = 25, d = 22 – 25 = -3
Sn = 116
Now using the formula
⇒ n/2 [2a + (n – 1)d] = 116
⇒ n [2 x 25 + (n – 1)(-3)] = 232
⇒ 50n -3n2 + 3n = 232
⇒ 3n2 – 53n + 232 = 0
⇒ 3n2 – 29n – 24n + 232 = 0
⇒ n(3n – 29) – 8(3n – 29) = 0
⇒(3n – 29)(n – 8) = 0
⇒ n = 29/ 3 or 8
Since n cannot be a fraction, n = 8.
Now we find the last term using the following formula
an = a + (n – 1)d
⇒ a8 = 25 + (8 – 1)(-3)
⇒ a8 = 4
Вопрос 27. Найдите сумму нечетных целых чисел от 1 до 2001.
Решение:
According to question
A.P = 1, 3, 5 … 2001
From the A.P we have
a = 1 and d = 2
an = 2001
Now using the formula
⇒ 1 + (n – 1)2 = 2001
⇒ 2n – 2 = 2000
⇒ 2n = 2002
⇒ n = 1001
Now we find the sum of odd integers from 1 to 2001
Also, S1001= 1001/2 [2 x 1 + (1001 – 1)2]
⇒ S1001 = 1001/2 [2 + 2000]
⇒ S1001 = 1001 x 1001 = 1002001
Вопрос 28. Сколько членов АП -6, -11/2, -5, … нужно, чтобы получить сумму -25?
Решение:
According to question
A.P = -6, -11/2, -5, …
From the A.P we have
a = – 6 and d =-11/2 – (-6) = 1/2
Sn = -25
Now using the formula
⇒ -25 = n/2 [2 x (-6) + (n -1)(1/2)]
⇒ -25 = n/2 [ -12 + n/2 – 1/2]
⇒ -50 = n [ n/2 – 25/2]
⇒ -100 = n [ n – 25]
⇒ n2 – 25n + 100 = 0
⇒ (n – 20)(n – 5) = 0
⇒ n = 20 or n = 5
Вопрос 29. В AP первое слагаемое равно 2, а сумма первых пяти слагаемых составляет одну четвертую от следующих пяти слагаемых. Покажите, что 20-й член равен -112.
Решение:
According to question we have
a = 2, S5 = 1/4(S10 – S5)
S5 = 5/2 [2 x 2 + (5 – 1)d]
⇒ S5 = 5 [2 + 2d] ………………..(i)
Also, S10 = 10/2 [2 x 2 + (10 – 1)d]
⇒ S10 = 5[4 + 9d] ………………..(ii)
Since S5 = 1/4 (S10 – S5)
So, from eq(i) and (ii), we have:
⇒ 5[2 + 2d] = 1/4 [5(4 + 9d) – 5(2 + 2d)]
⇒ 8 + 8d = 4 + 9d – 2 – 2d
⇒ d = -6
Now we find the 20th term
a20 = a + (20 – 1)d
⇒ a20 = a + 19d
⇒ a20 = 2 + 19(-6) = -112
Hence proved
Вопрос 30. Если S 1 — сумма (2n + 1) членов АП, а S 2 — сумма его нечетных членов, то докажите, что: S 1 : S 2 = (2n + 1) : (n + 1 )
Решение:
Let us assume A. P. be a, a + d, a + 2d…
So, S1 = (2n + 1)/2 [2a + (2n + 1 – 1)d]
Now using the formula, we get
⇒ S1 = (2n + 1)/2 [2a + (2n)d]
⇒ S1 = (2n + 1)(a + nd) …………………(i)
Now using the formula, we get
S2 = (n + 1)/2 [2a + (n + 1 – 1) x 2d]
⇒ S2 = (n +1)/2 [2a + 2nd]
⇒ S2 = (n + 1)[a + nd] …………………..(ii)
From eq(i) and (ii), we get :
S1 / S2 = (2n + 1) / (n + 1)
Hence, proved
Вопрос 31. Найдите АП, в котором сумма любого количества слагаемых всегда в три раза больше квадрата числа этих слагаемых.
Решение:
Given that Sn = 3n2
So, for n = 1, S1 = 3 x 12 = 3
For n = 2, S2 = 3 x 22 = 12
For n = 3, S3= 3 x 32 = 27 and so on
So, S1 = a1 = 3
a2 = S2 – S1 = 12 – 3 = 9
a3 = S3 – S2 = 27 – 12 = 15 and so on
Hence, the A.P. = 3, 9, 15…
Вопрос 32. Если сумма n членов АП равна nP + 1/2 – n (n – 1) Q, где P и Q – константы, найдите общую разность.
Решение:
According to question we have
Sn = nP + 1/2 n(n – 1)Q
For n = 1, S1 = P + 0 = P
For n = 2, S2 = 2P + Q
Also, a1 = S1 = P,
a2 = S2 – S1 = 2P + Q – P = P + Q
Hence, the common difference d = a2 – a1 = P + Q – P = Q
Вопрос 33. Суммы n членов двух арифметических прогрессий относятся как 5n + 4 : 9n + 6. Найдите отношение их 18-х членов.
Решение:
Let us considered we have two A.P’s. So, a1 and a2 are the first terms and d1 and d2 is common difference of the A.P’s
According to question we have
(5n + 4) / (9n + 6) = (Sum of n terms in the first A.P.) / (Sum of n terms in the second A.P.)
⇒ (5n + 4) / (9n + 6) = (2a1+ [(n — 1)d1]) / (2a2 + [(n — 1)d2] ………(1)
Now put n = 2 x 18 – 1 = 35 in eq(1), we get
(5 x 35 + 4) / (9 x 35 + 6) = (2a1 + 34d1) / (2a2 + 34d2)
179 /321 = (a1 + 17d1) / (a2 + 17d2) = (18th term of the first A.P.) / (18th term of the second A.P.)
Hence, the ratio of 18th terms = (a1 + 17d1) / (a2 + 17d2) = 179 /321
Вопрос 34. Суммы n членов двух арифметических прогрессий относятся как 7n + 2 : n + 4. Найдите отношение их 5-х членов.
Решение:
Let us considered we have two A.P’s. So, a1 and a2 are the first terms and S1 and S2 are the sum of the first n terms.
According to question we have
S1 = n/2 [2a1 + (n – 1)d1]
And, S2 = n/2 [2a2 + (n -1)d2]
Given:
S1 / S2 = (n/2 [2a1 + (n – 1)d1])/(n/2 [2a2 + (n – 1)d2]) = (7n + 2) / (n + 4)
Now we find the ratio of their 5th terms
[2a1 + (9 – 1)d1] / [2a2 + (9 – 1)d2] = (7(9) + 2) / (9 + 4)
[2a1 + (8)d1] / [2a2 + (8)d2] = 65/13
[a1 + (4)d1] / [a2 + (4)d2] = 5/1 = 5 : 1
Hence, the ratio of 5th terms = 5 : 1