Класс 11 RD Sharma Solutions — Глава 20 Геометрические прогрессии — Упражнение 20.4
Вопрос 1. Найдите сумму следующего ряда до бесконечности:
(i) 1 – 1/3 + 1/3 2 – 1/3 3 + 1/3 4 + … ∞
Решение:
Given series is an infinite G.P. with first term(a) = 1 and common ratio(r) = –1/3.
We know sum of a G.P. up to infinity is given by, S = a/(1 – r).
= 1/[1 – ( – 1/3)]
= 1/(4/3)
= 3/4
Therefore, sum of the series to infinity is 3/4.
(ii) 8 + 4√2 + 4 + …. ∞
Решение:
Given series is an infinite G.P. with first term(a) = 8 and common ratio(r) = 4√2/8 = 1/√2
We know sum of a G.P. up to infinity is given by, S = a/(1 – r).
= 8/[1 – (1/√2)]
= 8/[(√2 – 1)/√2]
= 8√2/(√2 – 1)
=
=
= 8(2 + √2)
Therefore, sum of the series to infinity is 8(2 + √2).
(iii) 2/5 + 3/5 2 + 2/5 3 + 3/5 4 + …. ∞
Решение:
Given series has sum, S = 2/5 + 3/52 + 2/53 + 3/54 + …. ∞
=> S = (2/5 + 2/53 + … ∞) + (3/52 + 3/54 + … ∞)
We know sum of a G.P. up to infinity is given by, S = a/(1–r).
Let S1 = 2/5 + 2/53 + … ∞
This is an infinite G.P. with first term(a) = 2/5 and common ratio(r) = 1/52 = 1/25.
So, S1 =
=
=
Let S2 = 3/52 + 3/54 + … ∞
This is an infinite G.P. with first term(a) = 3/5 and common ratio(r) = 1/52 = 1/25.
So, S2 =
=
=
Now, required sum, S = S1 + S2
=
= 13/24
Therefore, sum of the series to infinity is 13/24.
(iv) 10 – 9 + 8,1 – 7,29 + …. ∞
Решение:
Given series is an infinite G.P. with first term(a) = 10 and common ratio(r) = – 9/10
We know sum of a G.P. up to infinity is given by, S = a/(1 – r).
= 10/[1 – (–9/10)]
= 10/[1 + 9/10]
= 100/19
= 5.263
Therefore, sum of the series to infinity is 5.263.
(v) 1/3 + 1/5 2 + 1/3 3 + 1/5 4 + + 1/3 5 + 1/5 6 …. ∞
Решение:
Given series has sum, S = 1/3 + 1/52 + 1/33 + 1/54 + + 1/35 + 1/56 …. ∞
=> S = (1/3 + 1/33 + 1/35 … ∞) + (1/52 + 1/54 + 1/56 … ∞)
We know sum of a G.P. up to infinity is given by, S = a/(1–r).
Let S1 = 1/3 + 1/33 + 1/35 … ∞
This is an infinite G.P. with first term(a) = 1/3 and common ratio(r) = 1/32 = 1/9.
So, S1 =
= 3/8
Let S2 = 1/52 + 1/54 + 1/56 … ∞
This is an infinite G.P. with first term(a) = 1/52 and common ratio(r) = 1/52 = 1/25.
So, S2 =
= 1/24
Now, required sum, S = S1 + S2
=
= 10/24
= 5/12
Therefore, sum of the series to infinity is 5/12.
Вопрос 2. Докажите, что:
Решение:
We can write the L.H.S. as,
We have S = 1/3 + 1/9 + 1/27 + . . . .∞
which forms an infinite G.P. with first term(a) = 1/3 and common ratio(r) = 1/3.
Also, sum of a G.P. up to infinity is given by, S = a/(1–r).
=
=
=
So, L.H.S. becomes, 9S = 91/2 = 3 = R.H.S.
Hence proved.
Вопрос 3. Докажите, что:
Решение:
The L.H.S. can be written as,
L.H.S. =
=
We have, S = 1/4 + 2/8 + 3/16 + 4/32 + …∞ . . . . . (1)
Dividing both sides by 2, we get,
S/2 = 1/8 + 2/16 + 3/32 + …∞ . . . . (2)
Subtracting (2) from (1) we get,
Now this is an infinite G.P. with first term(a) = 1/4 and common ratio(r) = 1/2.
Also, sum of a G.P. up to infinity is given by a/(1–r).
=> S/2 = 1/2
=> S = 1
So, L.H.S. becomes, 2S = 2 = R.H.S.
Hence proved.
Вопрос 4. Если S p обозначает сумму ряда 1 + r p + r 2p + … до ∞ и s p сумму ряда 1 – r p + r 2p – … до ∞, докажите, что s p + S p = 2 S 2п .
Решение:
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
So, Sp = 1 + rp + r2p + … to ∞
This is an infinite G.P. with first term(a) = 1 and common ratio(r) = rp.
So, Sp = 1/(1 – rp)
Also, sp = 1 – rp + r2p – … to ∞
This is G.P. with first term(a) = 1 and common ratio(r) = –rp.
So, sp = 1/[1 – (-rp)] = 1/(1 + rp)
Thus, L.H.S. = sp + Sp
=
=
=
Now, R.H.S. = 2 S2p
=
= L.H.S.
Hence proved.
Вопрос 5. Найдите сумму членов бесконечно убывающей ЗП, в которой все члены положительны, первый член равен 4, а разница между третьим и пятым членами равна 32/81.
Решение:
Given G.P. has first term(a) = 4 and,
=> ar4 – ar2 = 32/81
=> 4(r4 – r2) = 32/81
=> r2(r2 – 1) = 8/81
Let’s suppose r2 = x, so the equation becomes,
=> x(x – 1) = 8/81
=> 81x2 – 81x – 8 = 0
Solving for x, we get,
=>
=>
=> x = 1/9 or x = 8/9
So, r2 = 1/9 or r2 = 8/9
=> r = 1/3 or r = 2√2/3
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
When a = 4 and r = 1/3,
S = 4/(1 – (1/3))
= 4/(2/3)
= 6
When a = 4 and r = 2√2/3,
S = 4/[1 – (2√2/3)]
= 12/(3 – 2√2)
Вопрос 6. Выразите повторяющееся десятичное число 0,125125125… в виде рационального числа.
Решение:
We are given, 0.125125125 = 0.125 + 0.000125 + 0.000000125 + …
= 125/103 + 125/106 + 125/109 + …
This is an infinite G.P. with first term(a) = 125/103 and common ratio(r) = 1/103 = 1/1000.
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
S =
=
=
Therefore, the recurring decimal 0.125125125 can be expressed in rational number as 125/999.
Вопрос 7. Найдите рациональное число, десятичное разложение которого равно 
Решение:
We are given,
= 0.4 + 0.0232323 . . . . . .
= 0.4 + 0.023 + 0.00023 + 0.0000023 . . . .
= 0.4 + 23/103 + 23/105 + 23/107 + . . . .
We have, S = 23/103 + 23/105 + 23/107 + . . . .
This is an infinite G.P. with first term(a) = 23/103 and common ratio(r) = 1/102 = 1/100.
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
S =
=
=
Thus, the rational number becomes,
=
=
Therefore, 419/990 is the rational number for this decimal expansion.
Вопрос 8. Найдите рациональные числа, имеющие следующие десятичные разложения:
(я) 
Решение:
We have,
= 0.3333…
= 0.3 + 0.33 + 0.333 + . . . .
= 3/10 + 3/102 + 3/103 + . . . .
This is an infinite G.P. with first term(a) = 3/10 and common ratio(r) = 1/10.
We know, sum of a G.P. up to infinity is given by, S = a/(1–r).
=
=
=
Therefore, 1/3 is the rational number for this decimal expansion.
(ii) 
Решение:
We have,
= 0.231231231
= 0.231 + 0.000231 + 0.000000231 + . . . .
= 231/103 + 231/106 + 231/109 + . . . .
This is an infinite G.P. with first term(a) = 231/103 and common ratio(r) = 1/103 = 1/1000.
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
=
=
=
Therefore, 231/999 is the rational number for this decimal expansion.
(iii) 
Решение:
We have,
= 3.522222
= 3.5 + 0.02 + 0.002 + 0.0002 + . . . .
= 3.5 + (2/102 + 2/103 + 2/104 + . . . )
We have, S = 2/102 + 2/103 + 2/104 + . . .
This is an infinite G.P. with first term(a) = 2/102 and common ratio(r) = 1/10.
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
So, S =
=
=
Thus, the number becomes,
=
=
Therefore, 317/90 is the rational number for this decimal expansion.
(4) 
Решение:
We have,
= 0.68888
= 0.6 + 0.08 + 0.888 + 0.8888 + . . . .
= 0.6 + (8/102 + 8/103 + 8/104 + . . . .)
We have, S = 8/102 + 8/103 + 8/104 + . . . .
This is an infinite G.P. with first term(a) = 8/102 and common ratio(r) = 1/10.
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
S =
=
=
Thus, the number becomes,
=
=
Therefore, 31/45 is the rational number for this decimal expansion.
Вопрос 9. Одна сторона равностороннего треугольника равна 18 см. Середины его сторон соединяются, образуя другой треугольник, середины которого, в свою очередь, соединяются, образуя еще один треугольник. Процесс продолжается бесконечно. Найдите сумму:
(i) периметры всех треугольников.
Решение:
The sides of all these triangles form an infinite G.P., 18, 9, 9/2, . . . .
Its first term is first term(a) = 18 and common ratio(r) = 9/18 = 1/2
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
Therefore, S = 18/[1 – 1/2]
= 18/(1/2)
= 36
Now, sum of perimeters of all the triangles, Sp = 3a1 + 3a2 + 3a3 + . . . .,
where a1,a2,a3, . . . are sides of these triangles.
= 3 (a1 + a2 + a3 + . . . .)
= 3S
= 3(36)
= 108
Therefore, the sum of perimeters of all the triangles is 108 cm.
(ii) площади всех треугольников.
Решение:
Sum of areas of all the triangles,
=
Here the series 324 + 81 + (81/4)+ . . .
forms an infinite G.P. with first term(a) = 324 and common ratio(r) = 81/324 = 1/4.
We know, sum of a G.P. up to infinity is given by, S = a/(1–r).
So, Sa =
=
= 108√3
Therefore, the sum of areas of all the triangles is 108√3 cm2.
Вопрос 10. Найдите бесконечную ЗП, у которой первый член равен 1 и каждый член является суммой всех следующих за ним членов.
Решение:
Given first term of G.P.(a) = 1 and,
an = an+1 + an+2 + an+3 + . . . .
We know nth term of a G.P. is given by an = arn-1. So, we get,
=> arn–1 = arn + arn+1 + arn+2 + . . . .
=> arn-1 = arn(1 + r + r2 + . . . .)
=> (1 + r + r2 + . . . .) = 1/r
Here the series on L.H.S forms an infinite G.P. with first term(a) = 1 and common ratio(r) = r.
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
=> 1/(1 – r) = 1/r
=> 2r = 1
=> r = 1/2
As first term(a) is 1 and common ratio is 1/2, the required infinite G.P. is,
1, 1/2, 1/4, 1/8, . . .
Вопрос 11. Сумма первых двух членов бесконечной ЗП равна 5, и каждый член в три раза больше суммы последующих членов. Найдите терапевта
Решение:
Given a + ar = 5 => a(1 + r) = 5 . . . . (1)
Also, an = 3(an+1 + an+2 + an+3 + . . . .)
We know nth term of a G.P. is given by an = arn-1. So, we get,
=> arn-1 = 3(arn + arn+1 + arn+2 + . . . .)
=> arn-1 = 3arn(1 + r + r2 + . . . .)
=> (1 + r + r2 + . . . .) = 1/3r
Here the series on L.H.S forms an infinite G.P. with first term(a) = 1 and common ratio(r) = r.
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
=> 1/(1 – r) = 1/3r
=> 4r = 1
=> r = 1/4
From (1), we get, a(1 + 1/4) = 5
=> a(5/4) = 5
=> a = 4
As first term(a) is 4 and common ratio is 1/4, the required infinite G.P. is,
4, 1, 1/4, 1/16, . . . .
Вопрос 12. Покажите, что в бесконечной ЗП со знаменателем r (|r| < 1) каждый член имеет постоянное отношение к сумме всех следующих за ним членов.
Решение:
According to the question, we have,
=
=
Here the series in the denominator forms an infinite G.P. with first term(a) = 1 and common ratio(r) = r.
We know, sum of a G.P. up to infinity is given by, S = a/(1–r). So our expression becomes,
=
=
Since common ratio(r) of the G.P. remains constant, so value of the ratio
Therefore, in an infinite G.P., each term bears a constant ratio to the sum of all terms that follow it.
Вопрос 13. Если S обозначает сумму бесконечной ВП, S 1 обозначает сумму квадратов ее членов, то докажи, что первый член и знаменатель равны соответственно 
а также 
Решение:
We know, sum of a G.P. up to infinity is given by, S = a/(1 – r).
Here, S = a + ar + ar2 + ar3 + . . . . = a/(1 – r) . . . . (1)
Also given that S1 = a2 + a2r2 + a2r4 + a2r6 + . . . . = a2/(1 – r2) . . . . (2)
Squaring both sides of (1), we get,
=>
From eq(2), we get,
=>
=>
=> S2 − S2r = S1 + S1r
=> (S1 + S2)r = S2 − S1
=> r =
On putting value of r in eq(1), we get
=> a = S(1 − r)
=> a =
=> a =
=> a =
Hence proved.
