Класс 11 RD Sharma Solutions – Глава 20 Геометрические прогрессии – Упражнение 20.3 | Набор 2

Вопрос 12. Найдите сумму:

Решение:

Given summation can be written as,

S =  

= (1 + 1/2 + 1/2² + . . . . 10 terms) + (1/5² + 1/5³ + 1/5⁴ + . . . . 10 terms)

= 

= 

Therefore, sum of the series is .

Вопрос 13. Пятый член ГП равен 81, тогда как второй член равен 24. Найдите ряд и сумму первых восьми членов.

Решение:

We know nth term of G.P. is given by, a_n = ar^n-1.

According to the question, we have,

=> ar⁴ = 81 . . . . (1)

=> ar = 24 . . . . (2)

Dividing (2) by (1),

=> r³ = 81/24

=> r³ = 27/8

=> r = 3/2

Putting r = 3/2 in (2), we get,

=> a = (24)(2)/3

=> a = 16

We know sum of n terms of G.P. is given by, S_n = a(rⁿ−1)/(r−1).

So, S₈ = 16[(3/2)⁸−1]/[(3/2)−1]

= 16[3⁸ − 2⁸]/2⁷

= 6305/8

As a = 16 and r = 3/2, series would be 16, 24, 36, 54, . . . . 

Also, the sum of first 8 terms of G.P. is 6305/8.

Вопрос 14. Если S ₁ , S ₂ , S ₃ — сумма n, 2n, 3n членов ГП соответственно, то докажите, что S ² ₁ + S ² ₂ = S ₁ (S ₂ + S ₃ ).

Решение:

We are given,

S₁ = Sum of n terms = a[1−rⁿ]/(1−r)

S₂ = Sum of 2n terms = a[1−r²ⁿ]/(1−r)

S₃ = Sum of 3n terms = a[1−r³ⁿ]/(1−r)

We have, 

L.H.S. = S²₁+S²₂

= 

= 

= 

= 

And R.H.S. = S₁(S₂+S₃)

= 

= 

= 

= 

= L.H.S.

Hence, proved.

Вопрос 15. Покажите, что отношение суммы n членов ГП к сумме членов от (n+1) ^-го до (2n) ^-го равно 1/r ⁿ .

Решение:

We know sum of n terms of a G.P. is given by, S₁ = a[1−rⁿ]/(1−r).

And sum of terms from (n+1)^th to (2n)^th term will be,

S₂ = arⁿ + arⁿ⁺¹ + arⁿ⁺² + . . . . ar^2n-1

= arⁿ[1−rⁿ]/(1−r)

L.H.S. = 

= 1/rⁿ

= R.H.S.

Hence, proved.

Вопрос 16. Если a и b – корни x ² – 3x + p = 0, а c, d – корни x ² – 12x + q = 0, где a, b, c, d образуют ГП. Докажите, что ( д + р): (д - р) = 17:15.

Решение:

We are given that a, b, c, d are in G.P. Let’s suppose the common ratio is r.

So, b=ar, c=ar² and d=ar³

Now a and b are the roots of x² – 3x + p = 0.

Sum of roots = a + b = 3  

=> a + ar = 3  

=> a(1+r) = 3 ….. (1)

Product of roots = ab = p

=> a(ar) = p

=> a²r = p  ….. (2)

And c, d are the roots of x² − 12x + q = 0

Sum of roots = c + d = 12

=> ar² + ar³ = 12

=> ar²(1+r) = 12  ….. (3)

Product of roots = cd = q

=> ar²(ar³) = q

=> a²r⁵ = q  ….. (4)

Dividing equation (3) by (1), we get,

=>  = 

=> r² = 4

=> r = ±2

When r=2, from (1), we get,

=> a(1+2) = 3

=> a = 1

Putting a=1 and r=2 in (2),

=> p = (1)²(2) = 2

From (4) we get,

q = (1)²(2)⁵ = 32

Now L.H.S. =  =  =  = 

= R.H.S.  

When r=−2, from (1), we get,

=> a(1−2) = 3

=> a = −3

Putting a=−3 and r=−2 in (2),

p = (−3)²(−2) = −18

From (4) we get,

q = (−3)²(−2)⁵ = −288

Now L.H.S. =  =  =  = 

= R.H.S.

Hence, proved.

Вопрос 17. Сколько членов ГП 3, 3/2, 3/4, … нужно, чтобы получить сумму 3069/512?

Решение:

Given G.P. has first term(a) = 3, common ratio(r) = (3/2)/3 = 1/2 and sum of terms(S_n) = 3069/512.

We know sum of n terms of a G.P. is given by S_n = a(rⁿ–1)/(r–1).

=> 3069/512 = 3[1–(1/2)ⁿ] / [1–(1/2)]

=> 2(2ⁿ–1)/(2ⁿ) = 1023/512

=> 1023(2)ⁿ = 1024(2)ⁿ – 1024

=> 2ⁿ = 1024

=> n = 10

Therefore, 10 terms of the G.P. should be taken together to make 3069/512.

Вопрос 18. У человека 2 родителя, 4 бабушки и дедушки, 8 прадедов и так далее. Найдите число его предков в течение десяти поколений, предшествующих его собственному.

Решение:

We have the sequence, 2,4,8, . . . . which forms a G.P. with first term(a) = 2 and common ratio(r) = 4/2 = 2.

 We know sum of n terms of a G.P. is given by S_n = a(rⁿ–1)/(r–1).

Number of ancestors during the ten generations = Sum of first 10 terms of the G.P. 

S₁₀ = 2(2¹⁰–1)/(2–1)

= 2(1023)

= 2046

Therefore, the number his ancestors during the ten generations preceding his own is 2046.

Вопрос 19. S ₁ , S ₂ , S ₃ , . . . ., S _n являются суммами n членов GP, первый член которых равен 1 в каждом, а обыкновенные отношения равны 1, 2, 3, . . . ., n соответственно, то докажите, что S ₁ + S ₂ + 2S ₃ + 3S ₄ + . . . . + (n–1)S _n = 1 ⁿ + 2 ⁿ + 3 ⁿ + . . . . + ^н .

Решение:

S₁, S₂, S₃, . . . ., S_n are the sums of n terms of G.P.’s whose first term is 1 in each and common ratios are 1,2,3, . . . ., n respectively. 

Here for S₁, series will be 1,1,1,1, . . . up to n terms as first term(a) and common ratio(r) both are equal to 1.

So, S₁ = 1+1+1+1+ . . . . up to n terms = n

So, L.H.S. = S₁ + S₂ + 2S₃ + 3S₄ + . . . . + (n–1)S_n

= 

= n + (2ⁿ – 1) + (3ⁿ – 1) + (4ⁿ – 1) + . . . . (nⁿ – 1)

= n + (2ⁿ + 3ⁿ + 4ⁿ +. . . . + nⁿ) – (1 + 1 +1 + 1 + . . . . (n–1) terms)

= n + (2ⁿ + 3ⁿ + 4ⁿ +. . . . + nⁿ) – n + 1

= 1 + 2ⁿ + 3ⁿ + 4ⁿ +. . . . + nⁿ

= 1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ +. . . . + nⁿ

= R.H.S.

Hence, proved.

Вопрос 20. ВП состоит из четного числа терминов. Если сумма всех слагаемых в 5 раз больше суммы слагаемых, стоящих на нечетных местах. Найдите обыкновенное отношение GP

Решение:

As number of terms is even, let the number of terms of the G.P. be 2n.

According to the question,

Sum of all terms = 5 (Sum of the terms occupying the odd places)

=> a₁ + a₂ + a₃ + . . . . + a_n = 5 (a₁ + a₃ + a₅ + . . . . a_2n–1)

=> a + ar + ar² + . . . . + ar^n–1 = 5 (a + ar² + ar⁴ + . . . . + ar^2n–2)

=> a(1–r²ⁿ)/(1–r) = 5a(1–r²ⁿ)/(1–r²)

=> a/(1–r) = 5a/(1–r²)

=> a/(1–r) = 5a/[(1–r)(1+r)]

=> 5/(1+r) = 1

=> 1+r = 5

=> r = 4

Therefore, the common ratio of the G.P. is 4.

Вопрос 21. Пусть _n — n-й член ГП положительных чисел. Позволять а также , такое что α ≠ β . Докажите, что обыкновенное отношение GP равно α/β.

Решение:

We have, 

=> a₂ + a₄ + a₆ + . . . . + a₂₀₀ = α 

=> ar + ar³ + ar⁵ + . . . . + ar¹⁹⁹ = α

=> ar(r²⁽¹⁰⁰⁾–1)/(r–1) = α

=> ar(r²⁰⁰–1)/(r–1) = α . . . . (1)

Also we have, 

=> a₁ + a₃ + a₅ + . . . . + a₁₉₉ = β

=> a + ar² + ar⁴ + . . . . + ar¹⁹⁸ = β

=> a(r²⁽¹⁰⁰⁾–1)/(r–1) = β

=> a(r²⁰⁰–1)/(r–1) = β . . . . (2)

Dividing (2) by (1), we get,

=>  = 

=> r = 

Hence, proved.

Вопрос 22. Найдите сумму 2n членов ряда, каждый четный член которого равен a, умноженному на предшествующий член, а каждый нечетный член равен c, умноженному на предшествующий член, причем первый член равен единице.

Решение:

Suppose we have the series, a₁,a₂,a₃, . . . . a_n.

According to the question, we have, a₁ = 1, a₂ = a, a₃ = ca, a₄ = a²c, a₅ = a²c² and so on.

Now, sum of 2n terms of the series = a₁ + a₂ + a₃ + . . . . + a_2n

S_2n = 1 + a + ca + a²c + a²c² + . . . . 2n terms

= (1+a) + ca(1+a) + a²c²(1+a) + . . . . n terms

Now this is a G.P. with first term(a) = (1+a) and common ratio(r) = ca. So, we get

S_2n = 

Therefore, sum of 2n terms of the series is .