Решения NCERT для класса 11 - Глава 9 Последовательности и серии - Разное упражнение по главе 9 | Набор 1

Вопрос 1. Покажите, что сумма (m + n) ^-го и (m – n) ^-го членов АП равна удвоенному m ^-му члену.

Решение:

Let the first term and common difference of the A.P. be a and d respectively.

(m+n)^th term of the A.P. = a+(m+n−1)d

(m−n)^th term of the A.P. = a+(m−n−1)d 

Thus, L.H.S = a+(m+n−1)d + a+(m−n−1)d 

= 2a+(m+n−1+m−n−1)d

= 2a+(2m−2)d

= 2[a+(m−1)d] = 2a_m  

= R.H.S.

Hence, proved.

Вопрос 2. Если сумма трех чисел в АП равна 24, а их произведение равно 440, найдите эти числа.

Решение:

Let three numbers be a − d, a, a + d.

Sum of these numbers = 24

=> a−d+a+a+d = 24

=> 3a = 24

=> a = 8

Product of these numbers = 440

=> (a−d) a (a+d) = 440

=> (8−d) 8 (8+d) = 440

=> 64−d² = 55

=> d² = 9

=> d = ±3

When a=8, d=3, the numbers are 

8−3, 8, 8+3, i.e., 5, 8, 11

When a=8, d=−3, the numbers are

8+3, 8, 8-3, i.e., 11, 8, 5

Вопрос 3. Пусть сумма n, 2n, 3n членов АП равна S ₁ , S ₂ и S ₃ соответственно, покажем, что S ₃ = 3(S ₂ – S ₁ ).

Решение:

According to the question,

S₁ = Sum of n terms =n[2a+(n−1)d]/2

S₂ = Sum of 2n terms =  2n[2a+(2n−1)d]/2

S₃ = Sum of 3n terms = 3n[2a+(3n−1)d]/2 

Therefore, R.H.S.= 3(S₂−S₁) 

= 

= [4a+(4n−2)d−2a−(n−1)d]

= [4a−2a+(4n−2−n+1)d]

= [2a+(3n−1)d]

= S₃ = L.H.S.

Hence, proved.

Вопрос 4. Найдите сумму всех чисел от 200 до 400, которые делятся на 7.

Решение:

The numbers between 200 and 400 divisible by 7 form an A.P. with common difference(d)=7 and first term (a)=203.

Last term of this series will be, a_n = 399

We also know, a_n = a+(n−1)d

=> 399 = 203+(n−1)7

=> (n–1)7 = 196

=> n−1 = 28

=> n = 29

Sum of the series = (a + a_n)

= (203+399) = 8729

Therefore, required sum is 8729.

Вопрос 5. Найдите сумму целых чисел от 1 до 100, которые делятся на 2 или 5.

Решение:

Let S₁ be the sum of integers from 1 to 100 divisible by 2.

For S_1, Common difference(d)=2 and first term(a)=2

Last term will be, a_n = 100. Therefore,

=> 100 = 2+(n–1)2

=> n = 50

Hence, S₁ = (a + a_n)

 = [2+100] = 2550 

Let S₂ be the sum of integers from 1 to 100 divisible by 5.

For S_2, Common difference(d)=5 and first term(a)=5

Last term will be, a_n = 100

=> 100 = 5+(n–1)5

=> n = 20

Hence, S₂ = [5+100] = 1050

Let S₃ be the sum of integers from 1 to 100 divisible by both 2 and 5.

For S₃, Common difference(d)=10 and first term(a)=10

Last term will be, a_n = 100

=> 100 = 10+(n–1)10

=> n = 10

Hence, S₃ = [10+100] = 550

Required sum = S₁ + S₂ – S₃

= 2550 + 1050 – 550 

= 3050

Вопрос 6. Найдите сумму всех двузначных чисел, которая при делении на 4 дает в остатке 1.

Решение:

Two-digit numbers, which when divided by 4, yield 1 as remainder are:

13, 17, 21 … 97.

This series forms an A.P. with first term(a) = 13 and common difference(d) = 4.

Let n be the number of terms of the A.P.

We know the nth term of an A.P. is, a_n= a+(n–1)d

=> 97 = 13 + (n–1)4

=> 4(n–1) = 84

=> n = 22

Now, the sum of n terms of an A.P. is given by,

S_n = [a + a_n]

= [13+97] = 1210

Therefore, required sum is 1210.

Вопрос 7. Если функция f удовлетворяет условию f(x + y) = f(x) f(y) для всех x, y ∈ N таких, что f(1) = 3 и , найдите значение n.

Решение:

We are given that,

f(x + y) = f(x) × f(y) for all x, y ∈ N … (1)

f(1) = 3

Putting x = y = 1 in (1), we have

f(1 + 1) = f(2) = f(1) f(1) = 3 × 3 = 9

Similarly,

f(1+1+1) = f(3) = f(1+2) = f(1) f(2) = 3 × 9 = 27

And, f(4) = f(1+3) = f(1) f(3) = 3 × 27 = 81

We have, f(1), f(2), f(3).., i.e., 3, 9, 27, …, 

This series forms a G.P. with the first term(a)=3 and common ratio(r)=3.

We know that sum of terms in G.P is given by,

S_n = 

And also it’s given that the sum of terms of the function is 120.

=> 120 = 

=> 3ⁿ–1 = 80 

=> n=4

Therefore, number of terms is 4.

Вопрос 8. Сумма некоторых членов ГП равна 315, у которых первый член и знаменатель равны 5 и 2 соответственно. Найдите последний член и количество терминов.

Решение:

Let the total number of terms be n.

We know that, sum of terms of a G.P. is,

S_n = 

Also, we know that the first term(a) is 5 and common ratio(r) is 2.

=> 315 = 

=> 2ⁿ–1 = 63

=> n=6

The last term of the G.P = 6th term = ar^6-1 = (5)(2)⁵ = 160

Therefore, last term of the G.P. is 160 and the number of terms is 6.  

Вопрос 9. Первый член ВП равен 1. Сумма третьего члена и пятого члена равна 90. Найдите знаменатель ВП

Решение:

Let the first term and common ratio of the G.P. be a and r respectively.

We have, a = 1. So,

a₃ = ar² = r²  …. (1)

a₅ = ar⁴ = r^{4 _..} …. (2)

According to the question, we have

a₃ + a₅ = 90

From (1) and (2),

r² + r⁴ = 90

r⁴ + r² – 90 = 0

Solving for r², we get,

r² = –10 or r² = 9

Taking only real roots, we get,

r = ±3

Therefore, the common ratio of the G.P. is ±3.

Вопрос 10. Сумма трех чисел в GP равна 56. Если из этих чисел вычесть 1, 7, 21 в указанном порядке, то получится арифметическая прогрессия. Найдите числа.

Решение:

Suppose the three numbers in G.P. are a, ar, and ar².

According to question, we have

=> a + ar + ar² = 56

=> a (1 + r + r²) = 56

=> a =  …. (1)

Also, we are given that,

a – 1, ar – 7, ar² – 21 forms an A.P. 

which implies,

=> (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)

=> ar – a – 6 = ar² – ar – 14

=> ar² – 2ar + a = 8

=> a (r – 1)² = 8 ….. (2)

Putting value of a from (1) in (2), we get,

=> (r – 1)² = 8

=> 7(r² – 2r + 1) = 1 + r + r²

=> 6r² – 15r + 6 = 0

=> (6r – 3) (r – 2) = 0

=> r = 2, 1/2

Now if r = 2, then a = 8 and the three numbers in G.P. are 8, 16, and 32.

If r = 1/2, then a = 32 and the three numbers in G.P. are 32, 16, and 8.

Therefore, in both the cases, the three numbers are 8, 16, and 32.  

Вопрос 11. ВП состоит из четного числа терминов. Если сумма всех слагаемых в 5 раз больше суммы слагаемых, стоящих на нечетных местах, то найти ее знаменатель.

Решение:

Suppose the terms in the G.P. are a_1, a₂, a₃, a₄, … a_2n.

Number of terms = 2n

According to the question, we have

=> a₁ + a₂ + a₃ + …+ a_2n = 5 [a₁ + a₃ + … + a_2n–1]

=> a₁ + a₂ + a₃ + … + a_2n – 5 [a₁ + a₃ + … + a_2n–1] = 0

=> a₂ + a₄ + … + a_2n = 4 [a₁ + a₃ + … + a_{2n – 1}]      …. (1)

Now, let the terms of our G.P. be a, ar, ar², ar³, …..

Using sum of terms of a G.P., equation (1) becomes,

After solving we get,

ar = 4a

r = 4

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

Вопрос 12. Сумма первых четырех членов АП равна 56. Сумма последних четырех членов равна 112. Если его первый член равен 11, то найдите количество членов.

Решение:

Suppose the terms in A.P. are a, a + d, a + 2d, a + 3d, … a + (n – 2) d, a + (n – 1)d.

According to the question, we have,

a = 11,

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d,

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]

= 4a + (4n – 10) d

Then according to the given condition,

=> 4a + 6d = 56

=> 4(11) + 6d = 56

=> d = 2

Therefore, 4a + (4n –10)d = 112

=> 4(11) + (4n – 10)2 = 112

=> (4n – 10)2 = 68

=> 4n = 44

=> n = 11

Therefore, the number of terms of the A.P. is 11.

Вопрос 13. Если, , затем покажите, что a, b, c и d принадлежат GP

Решение:

We are given,

On cross multiplication, we have,

=> (a+bx)(b-cx) = (b+cx)(a-bx)

=> ab–acx+b²x–bcx² = ab–b²x+acx–bcx²

=> 2b²x = 2acx

=> b² = ac

=> b/a = c/b   ….. (1)

We are also given,

On cross multiplication, we have,

=> (b+cx)(c−dx) = (c+dx)(b−cx)

=> bc−bdx+c²x-cdx² = bc+bdx−c²x−cdx²

=> 2c²x = 2bdx

=> c² = bd

=> c/d = d/c   ….. (2)

From (1) and (2), we get

b/a = c/b = d/c

Therefore, a, b, c and d are in G.P.

Вопрос 14. Пусть S — сумма, P — произведение , а R — сумма обратных величин n членов ГП. Докажите, что P ² R ⁿ = S ⁿ .

Решение:

Suppose the terms in the G.P. are a, ar, ar², ar³, … ar^{n – 1}…

According to the question, we have,

S = a(rⁿ−1)/(r−1)  ….. (1)

P = aⁿ × r^1+2+….+n-1

= aⁿ r ^n(n-1)/2

R = 1/a + 1/ar + 1/ar² + …. + 1/ar^n-1

= [Tex]left(frac{1}{ar^{n-1}} ight)[/Tex]

So, P²Rⁿ = a²ⁿ r ^n(n-1) = 

= 

= 

From (1), we get,

P²Rⁿ = Sⁿ

Hence, proved

Вопрос 15. P ^-й , q ^-й и r ^-й термины AP равны a, b, c соответственно. Покажите, что (q – r) a + (r – p) b + (p – q) c = 0.

Решение:

Let us take t and d to be the first term and the common difference of the A.P. respectively.

The n^th term of the A.P. is given by, a_n = t + (n – 1) d

Therefore,

p^th term will be, a_p = t+ (p – 1) d = a  ….. (1)

q^th term will be, a_q = t + (q – 1) d = b  ….. (2)

r^th term will be, a_r = t + (r – 1) d = c  ….. (3)

On subtracting equation (2) from (1), we get

=> (p – 1 – q + 1) d = a – b

=> (p – q) d = a – b

=> d =  ….. (4)

On subtracting equation (3) from (2), we get

=> (q – 1 – r + 1) d = b – c

=> (q – r) d = b – c

=>  d =  .…. (5)

Equating both the values of d obtained in (4) and (5), we get

=>  = 

=> (a–b)(q–r) = (b–c)(p–q)

=> bp–cp+cq–aq+ar–br = 0

Rearranging terms, we get,

=> (–aq+ar)+(bp–br)+(–cp+cq) = 0  

=> a(q–r)+b(r–p)+c(p–q) = 0

Hence, proved

Вопрос 16. Если a(1/b+1/c), b(1/c+1/a), c(1/a+1/b) лежат в AP, докажите, что a, b, c лежат в AP

Решение:

We are given, a(1/b+1/c), b(1/c+1/a), c(1/a+1/b) are in A.P.

=> b(1/c+1/a)–a(1/b+1/c) = c(1/a+1/b)–b(1/c+1/a)

=> 

Multiplying both sides by abc and rearranging, we get,

=> b²a–a²b+b²c–a²c = c²a–b²a+c²b–b²c 

=> ab(b–a)+c(b²–a²) = a(c²–b²)+bc(c–b)

=> (b–a)(ab+bc+ca) = (c–b)(ab+bc+ca)

=> b–a = c–b

Therefore, a, b and c are in A.P.