11类NCERT解决方案-第9章序列和序列–练习9.3 |套装1

问题1.找到GP 5 / 2、5 / 4、5 / 8，…的^第20和^第n个项

解决方案：

According to the question

G.P: 5/2, 5/4, 5/8, …

So, first term(a) = 5/2

So, the common ratio(r) = $\frac{a_1}{a}=\frac{\frac54}{\frac52}= \frac12$    

Find: 20^th and n^th terms of the given G.P

So, the n^th term of G.P can be expressed using formula:

a_n = ar^{n – 1}  

Where a is 1^st term and r is the common ratio.

Now we find the 20^th terms of the given G.P:

a₂₀ = (5/2)(1/2)^20-1 = (5/2)(1/2)¹⁹= $\frac{5}{2^{20}}$

Find the n^th terms of the given G.P:

a_n = $\frac52\frac1{2^{n-1}}=\frac5{2^{n}}$

问题2。找到^{GP的第12个}项，其第8个项是192，共同比率是2。

解决方案：

According to the question

Common ration(r) = 2

and 8^th term is 192

So let us considered a be the first term

So, 

a₈ = ar⁷

ar⁷ = 192

a(2)⁷ = 192

a = 3/2

Find: 12^th term of a G.P.

As we know that the n^th term of G.P can be expressed using formula:

a_n = ar^{n – 1}  

Where a is 1^st term and r is the common ratio

So, we find 12^th term of a G.P. 

 a₁₂ = ar^{12 – 1}

 a₁₂ = ar¹¹

 a₁₂ = (3/2)(2)¹¹

a₁₂ = 3072

问题3，^第5，^第8和^第11一个GP的术语是分别为p，q和s。证明q ² = ps。

解决方案：

According to the question

The 5^th, 8^th and 11^th terms of a G.P. are p, q and s

Prove: q² = ps

Proof:

 Let a G.P. with first term a and common ratio r,

So, a₅ = ar⁴ = p  ….(1)

a₈ = ar⁷= q  ….(2)

a₁₁ = ar¹⁰ = s  ….(3)

Now divide eq(2) by (1), we get

ar⁷/ar⁴ = q/p

r³ = q/p  ….(4)

Now divide eq(3) by (2), we get

ar¹⁰/ar⁷= s/q

r³ = s/q  ….(5)

From eq(4) and (5), we get

q/p = s/q

q² = ps

Hence Proved

问题4.一个GP的^第4项是正方形其第二项的和的第一项是- (3)确定的^第7^次项。

解决方案：

According to the question

First term(a) = – 3

and the 4^th term of a G.P. is square of its second term

Find: 7^th term

Let us considered r be the common ratio 

As we know that the n^th term of G.P can be expressed using formula:

a_n = ar^{n – 1}  

Where a is 1^st term and r is the common ratio

So, a₄ = ar³

It is given that the 4^th term of a G.P. is square of its second term

a₄ = (a₂)²

ar³ = (a₂)²

ar³ = (ar)²

ar³ = a²r²

r = a

Now put the value of a, we get

r = -3

Now, we find the 7^th term

a₇ = ar^{7 – 1}  

a₇ = ar⁶  

= −3 x (−3)⁶ =−2187

问题5.以下序列中的哪个术语：

(a)2、2√2、4，…是128吗?

解决方案：

According to the question

G.P.: 2, 2√2, 4, …

So, first term(a) = 2

So, the common ratio(r) = √2

As we know that, the n^th term of G.P can be expressed using formula:

a_n = ar^{n – 1}  

Where a is 1^st term and r is the common ratio.

128 = 2(√2)^{n – 1}

(2)⁷ = 2(√2)^{n – 1} 

(2)⁷ = 2((2)^1/2)^{n – 1} 

(2)⁷ = 2(2)^{(n – 1)/2}

(2)⁶ = (2)^{(n – 1)/2}

6 = (n – 1)/2

12 = n – 1

12 + 1 = n

n = 13

Hence, the 13th term of the G.P. is 128

(b)√3，3，3√3，…。是729吗?

解决方案：

According to the question

G.P.: √3, 3, 3√3, …. 

So, first term(a) = √3

So, the common ratio(r) = √3

As we know that, the n^th term of G.P can be expressed using formula:

a_n = ar^{n – 1}  

Where a is 1^st term and r is the common ratio.

729 = √3(√3)^n-1

(3)⁶ = √3(√3)^n-1

(3)⁶ = (3)^1/2((3)^1/2)^n-1

(3)⁶ = (3)^1/2(3)^n-1/2

(3)⁶ = (3)^1/2+(n-1)/2

6 = 1/2 + (n – 1)/2

6 – 1/2 = (n – 1)/2

(12 – 1)/2 = (n – 1)/2

11 = n – 1

n = 12

Hence, the 12th term of the G.P. is 729

(C) $\frac13,\frac19,\frac1{27},...$是$\frac1{19683}$ ?

解决方案：

According to the question

G.P.: $\frac13,\frac19,\frac1{27},...$

So, first term(a) = 1/3

So, the common ratio(r) = 1/3

As we know that, the n^th term of G.P can be expressed using formula:

a_n = ar^{n – 1}  

Where a is 1^st term and r is the common ratio.

$\frac1{19683}=(\frac13)(\frac13)^{n-1}$

(1/3)⁹ = (1/3)ⁿ

n = 9

Hence, the 9th term of the G.P. is 1/19683

问题6.对于x的什么值，数字-2 / 7，x。 -7 / 2…在GP中吗?

解决方案：

According to the question

Numbers are -2/7, x. -7/2…

The common ration is (r) = $\frac{x}{\frac{-2}{7}}$= -7x/2

Again the common ration(r) = = $\frac{\frac{-7}{2}}{x}$= -7/2x

So, 

-7x/2 = -7/2x

7x/2 = 2/7x

14x² = 14

x² = 14/14

x = ±1

在练习7到10中的每个几何级数中找到表示的项数之和：

问题7. 0.15、0.015、0.0015，…20个字词。

解决方案：

According to the question

G.P.: 0.15, 0.015, 0.0015, … 

So, first term(a) = 0.15

So, the common ratio(r) = 0.1

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(1 - r^n)}{1-r}$

So, S₂₀ = (0.15)[(1 – (0.1)²⁰)/(1 – 0.1)]

= (0.15/0.9)(1 – (0.1)²⁰)

= 1/6(1 – (0.1)²⁰)

问题8：7√，√21，3√7，… n项。

解决方案：

According to the question

G.P.: √7​, √21​, 3√7​,…

So, first term(a) = √7​

So, the common ratio(r) = √3​

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(1 - r^n)}{1-r}$

So, S_n = (√7​)[(1 – (√3)ⁿ)/(1 – √3)]

= (√7​)[(1 – (√3)ⁿ)/(1 – √3)] x [(1 + √3)/(1 + √3)]

= -(√7​)(1 + √3)/2(1 – (3)^n/2)

= (√7​)(1 + √3)/2((3)^n/2 – 1)

问题9. 1，-a，a ² ，-a ³ ，… n项(如果≠– 1)。

解决方案：

According to the question

G.P.: 1, -a, a², -a³,… 

So, first term(a) = 1

So, the common ratio(r) = -a​

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(1 - r^n)}{1-r}$

So, $S_n=\frac{(1)(1-(-a)^n)}{1 -(-a)}=\frac{1-(-a)^n}{1+a}$

问题10. x ³ ，x ⁵ ，x ⁷ ，…n个项(如果x≠±1)。

解决方案：

According to the question

G.P.: x³, x⁵, x⁷, …

So, first term(a) = x³

So, the common ratio(r) = x²​

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(1 - r^n)}{1-r}$

So,

$S_n=\frac{(x^3)(1-(x^2)^n)}{1-x^2}$

$S_n=\frac{(x^3)(1-x^{2n})}{1-x^2}$

问题11：评估$\sum_{k=1}^{11} (2+3^k)$

解决方案：

$\sum_{k=1}^{11} (2+3^k)$= (2 + 3¹) + (2 + 3²) + (2 + 3³) + …. + (2 + 3¹¹)

= (2 + 2 + …11 terms) + (3 + 3² + 3³+…11 terms)

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(1 - r^n)}{1-r}$

So, S_n = 2 x 11 + 3(3¹¹ – 1)/3 – 1 

S_n = 22 + 3/2((3¹¹ – 1))

问题12. GP的前三个项的总和为39/10，其乘积为1。找到共同比率和项。

解决方案：

Let us considered the three terms a/r, a, ar of the G.P. 

According to the question 

$\frac{a}r\times a\times ar$ = 1 

a³ = 1

a = 1

Also. 

$\frac{a}{r}+a+ar=\frac{39}{10}$ 

Now put the value of a, we get

$\frac1r+1+r=\frac{39}{10}$ 

 $\frac{r^2+r+1}r=\frac{39}{10}$

10r² + 10r + 10 = 39r

10r² – 29r + 10 = 0

On solving the equation, we get 

r = 2/5, 5/2

So, the G.P. is 5/2, 1, 2/5.

问题13 GP 3的多少而言，3 ^{2，3 3，…}都需要给总和120?

解决方案：

According to the question

G.P.: 3, 3², 3³, …

So, first term(a) = 3

So, the common ratio(r) = 3

S_n = ​120

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(1 - r^n)}{1-r}$

S_n = (3)(1 – (3)ⁿ)(1-3)

120 = (3)(1 – (3)ⁿ)

-240 = (3)(1 – (3)ⁿ)

-80 = 1 – (3)ⁿ

-80 – 1 = – (3)ⁿ

-81 = – (3)ⁿ

(3)⁴ = (3)ⁿ

n = 4   

Hence, 4 terms of G.P. are needed to give the sum 120

问题14. GP的前三个项的总和是16，接下来的三个项的总和是128。确定GP的第一个项，公共比率和n个项的总和

解决方案：

Let us considered the first three terms of the G.P. are a, ar, ar² and

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

Find: the first term, the common ratio, and the sum to n terms of the G.P. 

According to the question 

 a + ar + ar² = 16   

a(1 + r + r²) = 16 …….(1)

 ar³ + ar⁴ + ar⁵ = 128 

ar³(1 + r + r²) = 128  …….(2)

Now divide eq(2) by (1), we get

ar³(1 + r + r²)/ a(1 + r + r²) = 128/16

r³ = 8

r = 2

Now put the value of r in eq(1), we get

a(1 + (2) + (2)²) = 16

a(1 + 2 + 4) = 16

a(7) = 16

a = 16/7

Now we find the sum to n terms of the G.P. 

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(r^n-1)}{r-1}$

$S_n=\frac{16(2^n-1)}{7(2-1)}=\frac{16(2^n-1)}7$

问题15.给定一个GP = 729和^第7项64的GP，确定S ₇ 。

解决方案：

According to the question 

The first term(a) = 729

and 7^th term 64

Find: S₇

Let the common ratio be r.

a₇ = ar⁶ = 64

r⁶ = 64/729

r = ±2/3

Now we find the sum to 7th terms of the G.P. 

As we know that, the sum of n terms of G.P. with 1^st term a & common ratio r is given by  

$S_n=\frac{a(r^n-1)}{r-1}$

On taking r = +2/3   

$S_7=\frac{729((\frac23)^7-1)}{\frac23-1}$ = 2059

On taking r = -2/3  

$S_7=\frac{729((-\frac23)^7-1)}{(-\frac23-1)}$ = 463

问题16。找到一个GP，其前两项之和为– 4，而第五项为第三项的4倍。

解决方案：

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

According to the question 

a + ar = -4   

a₅ = 4(a₃)

So, ar⁴ = 4(ar²)  

r² = 4 

r = ±2 

When r = +2 then 

a + ar = a + 2a = 3a = −4

 a =  -4/3

Then the G.P. is $\frac{-4}3,\frac{-8}3,\frac{16}3,...$

When r = -2 then 

a + ar = a − 2a = −a = −4 

a = 4

Then the G.P. is 4, -8, 16, -32,…