11类NCERT解决方案-第9章序列和序列–练习9.2

问题1.找到1到2001之间的奇数整数之和。

解决方案：

The odd integers forms an A.P. with 

Common difference(d)=2  and  first term (a)=1

Here, a+(n-1)d=2001

⇒1+(n-1)2=2001

⇒2n-2=2000

⇒n=1001

S_n = n[2a + (n – 1)d]/2

S₁₀₀₁=1001[2(1)+(1001-1)2]/2

=1001[2+2000]/2

=1001[2002]/2

=1001[1001]

=1002001

Therefore, the sum of odd integers from 1 to 2001 is 1002001.

问题2.在AP中，第一项为2，而前5个项的总和为接下来5个项的四分之一。证明第20^个学期是-112。

解决方案：

The first term(a)=2.

Let the common difference be d.

Sum of first 5 terms=10+10d.

Sum of next 5 terms=10+35d.

According to given condition ,

10+10d=¼(10+35d)

⇒40+40d=10+35d

⇒5d=-30

⇒d=-6

a₂₀=2+(20-1)(-6)=2-(19)6=2-114=-112

Hence, proved

问题3.如果AP的前n个项之和为(pn + qn ² )，则p和q为常数。找到共同的区别。

解决方案：

We know that, 

S_n = n[2a + (n – 1)d]/2

According to the given condition,

n[2a+(n-1)d]/2 = pn+qn²

⇒n[2a+nd-d]/2 = pn+qn²

⇒na+n²(d/2)-n(d/2) = pn+qn²

Comparing the coefficients of n² on both sides we get,

d/2=q ⇒ d=2q

Therefore, the common difference is 2q.

问题4.得出总和-25需要多少个AP -6，-11 / 7，-5，…项?

解决方案：

Let the number of terms to give the sum -25 be n.

We know S_{​​​​​​n} = n(2a+(n-1)d)/2 where n is the number of terms, a is the first term and d is the common difference.

Here, a=-6 and d=(-11/2)+6=(-11+12)/2=1/2.

Substituting these values in the formula of sum we get,

-25=n[2(-6)+(n-1)(1/2)]/2

⇒-50=n[-12+(n/2)-(1/2)]

⇒-50=n[(-25/2)+(n/2)]

⇒-100=n(-25+n)

⇒n²-25n+100=0

⇒n²-5n-20n+100=0

⇒n(n-5)-20(n-5)=0

⇒n=20 or 5

Therefore, the number of terms required is 5 or 20.

问题5.在AP中，第p^个项是1 / q，第q^个项是1 / p。证明第一个pq项的总和为(pq + 1)/ 2，其中p≠q。

解决方案：

It is known that the general term of an A.P. is a_n =a+(n-1)d.

Given,

p^th term=a_p =a+(p-1)d=1/q               —(1)

q^th term = a_q= a+(q-1)d = 1/p           —(2)

Subtracting (2) from (1):

(p-1)d-(q-1)d=(1/q)-(1/p)

⇒(p-1-q+1)d=(p-q)/pq

⇒(p-q)d=(p-q)/pq

⇒d=1/pq

Putting value of d in (1):

a+(p-1)/pq=1/q

⇒a=(1/q)-(1/q)+(1/pq)=1/pq

Therefore,

S_pq = pq[2a+(pq-1)d]/2 

= pq[(2/pq)+((pq-1)/pq)]/2 

=1+(pq-1)/2

=(pq+1)/2

The sum of the first pq terms of the A.P. is (pq+1)/2.

问题6.如果AP 25、22、19，…的某些术语的总和是116。找到最后一个术语。

解决方案：

Let the number of terms required to give the sum of 116 be n

We know, S_n=n[2a+(n-1)d]/2

Given, a=25 and d= 22-25=-3

Substituting the values we get,

116=n[2(25)+(n-1)(-3)]/2

⇒232=n[50-3n+3]

⇒232=n(53-3n)=53n-3n²

⇒3n²-53n+232=0

⇒3n²-24n-29n+232=0

⇒3n(n-8)-29(n-8)=0

⇒(n-8)(3n-29)=0

⇒n=8 or 29/3

But, n cannot be 29/3.Therefore, n=8.

a₈ is the last term=a+(n-1)d=25+(8-1)(-3)=25+(7)(-3)=25-21=4

Therefore, the last term is 4.

^{问题7.如果第k个}项的给定为5k + 1，则求出AP的n个项之和。

解决方案：

Given, k^th term is 5k+1

k^th term = a_k = a+(k-1)d

a+(k-1)d=5k+1

⇒ a + kd – d=5k+1

Comparing the coefficients on both sides, we get d=5

a-d=1⇒a-5=1⇒a=6

S_n=n[2a+(n-1)d]/2 

= n[2(6)-(n-1)5]/2 

=n[12+5n-5]/2 

= n(5n+7)/2.

问题8。找到100到1000之间的所有自然数的和，这是5的倍数。

解决方案：

The natural numbers between 100 and 1000 which are multiples of 5 are, 105, 110, . . . , 995. It can be seen as an A.P. starting with a= 105 and common difference d=5.

a+(n-1)d=995

⇒105+(n-1)5=995

⇒(n-1)5=995-105=890

⇒n-1=178

⇒n=179

S_n=179[2(105)+(179-1)5]/2

=179[2(105)+(178)5]/2

=179[105+(89)5]

=179[105+445]

=179[550]

=98450

The required sum is 98450.

问题9.两个AP的n个项之和为5n + 4：9n + 6。找出他们^第18个学期的比率。

解决方案：

Let a₁, a₂ and d₁, d₂ be the first term and the common difference of the first and the second A.P. respectively.

Given,

(Sum of n terms of first A.P)/(Sum of n terms of second A.P.)=(5n+4)/(9n+6)

⇒{n[2a₁-(n-1)d₁]/2}/{n[2a₂+(n-1)d₂]/2}=(5n+4)/(9n+6)

⇒[2a₁+(n-1)d₁]/[2a₂+(n-1)d₂]=(5n+4)/(9n+6)                                                         —(1)

Substituting n=35 in (1), 

(2a₁+34d₁)/(2a₂+34d₂)=[5(35)+4]/[9(35)+6]

⇒[a₁+17d₁]/[a₂+17d₂]=179/321                                                                             —(2)

(18th term of the first A.P)/(18th term of the second A.P.)=(a₁+17d₁)/(a₂+17d₂)    —(3)

R.H.S. of (3) =L.H.S. of (2). So, we get,

(18th term of the first A.P)/(18th term of the second A.P.)=179/321

Therefore, the ratio of the 18th term of both the A.P. is 179:321.

问题10.如果AP的前p个项的总和等于前q个项的总和，则求出前(p + q)个项的总和。

解决方案：

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

Given,

p[2a+(p-1)d]/2=q[2a+(q-1)d]/2

⇒p[2a+(p-1)d]=q[2a+(q-1)d]

⇒2ap+pd(p-1)=2aq+qd(q-1)

⇒2a(p-q)+d[p(p-1)-q(q-1)]=0

⇒2a(p-q)+d[p²-p-q²+q]=0

⇒2a(p-q)+d[(p-q)(p+q)-(p-q)]=0

⇒2a(p-q)+d[(p-q)(p+q-1)]=0

⇒2a+d(p+q-1)=0

⇒d=-2a/(p+q-1)                                         —(1)

So, S_p+q =(p+q)[2a+(p+q-1)d]/2

Putting value of d ,

⇒S_p+q=(p+q)[2a-2a]/2=0

Therefore, the sum of first (p+q) terms of the A.P. is 0.

问题11. AP的第一个p，q和r项的总和分别为a，b和c。证明{a(qr)/ p} + {b(rp)/ q} + {c(pq)/ r = 0。

解决方案：

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

Given,

S_p=p[2a1+(p-1)d]/2=a

⇒2a1=(p-1)d=2a/p                                                      —(1)

Sq=q[2a1+(q-1)d]/2=b

⇒2a1+(q-1)d=2b/q                                                      —(2)

Sr=r[2a1+(r-1)d]/2=c

⇒2a1+(r-1)d=2c/r                                                         —(3)

Subtracting (2) from (1), we get

(p-1)d-(q-1)d=(2a/p)-(2b/q)

⇒d(p-1-q+1)=(2aq-2bq)/pq

⇒d=2(aq-bp)/pq(p-q)                                                    —(4)

Subtracting (3) from (2),we get

(q-1)d-(r-1)d=(2b/q)-(2c/r)

⇒d(q-1-r+1)=(2b/q)-(2c/r)

⇒d(q-r)=(2br-2qc)/qr

⇒d={2(br-qc)}/{qr(q-r)}                                                  —(5)

Equating both values of d in (4) and (5),we obtain

(aq-bp)/(pq(p-q))=(br-qc)/(qr(q-r))

⇒qr(q-r)(aq-bq)=pq(p-q)(br-qc)

⇒r(aq-bp)(q-r)=p(br-qc)(p-q)

⇒(aqr-bpr)(q-r)=(bpr-pqc)(p-q)

Dividing both sides by pqr,

{(a/p)-(b/q)}(q-r)={(b/q)-(c/r)}(p-q)

⇒(a/p)(q-r)-(b/q)(q-r+p-q)+(c/r)(p-q)=0

⇒(a(q-r)/p)+(b(r-p)/q)+(c(p-q)/r)=0. 

Hence, proved

问题12. AP的m和n项之和的比率为m ² ：n ² 。表明，^第m和比^第n^个项是(2M-1):( 2N-1)。

解决方案：

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

Given,

(Sum of m terms)/(Sum of n terms)=m²/n²

⇒[m(2a+(m-1)d)/2]/[n(2a+(n-1)d)/2]=m²/n²

⇒[2a+(m-1)d]/[2a+(n-1)d]=m/n                                              —(1)

Putting m=2m-1 and n=2n-1 in (1), we get

[2a+(2m-2)d]/[2a+(2n-2)d]=(2m-1)/(2n-1)

⇒[a+(m-1)d]/[a+(n-1)d]=[2m-1]/[2n-1]                                   —(2)

[m^th term of A.P.]/[n^th term of A.P.]=[a+(m-1)d]/[a+(n-1)d]      —(3)

R.H.S. of (3) = L.H.S. of (2). So, we get,

[m^th term of A.P.]/[n^th term of A.P.]=(2m-1)/(2n-1) 

Hence, proved

问题13.如果AP的n个项的总和为3n ² + 5n，而第m^个项为164，则找到m的值。

解决方案：

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

a_m =a+(m-1)d=164                                                                   —(1)

S_n =n[2a+(n-1)d]/2

Given,

n[2a+nd-d]/2=3n² +5n

⇒na+(n²d/2)=3n² +5n

Comparing coefficients on both sides,

d/2=3⇒d=6    and a-(d/2)=5⇒a-3=5 ⇒ a=8

From (1),

8+(m-1)6=164

⇒ (m-1)6=164-8=156

⇒ m-1=26

⇒ m=27

Therefore, the value of m is 27.

问题14：在8到26之间插入5个数字，以使所得序列位于AP中

解决方案：

Let A₁, A₂, A₃, A₄ and A₅ be 5 numbers between 8 and 26 such that it forms an A.P.

a=8, b=26, n=7

We can see, 26=8+(7-1)d

⇒6d=26-8=18

⇒d=3

A₁=a+d=8+3=11

A₂=a+2d=8+2(3)=8+6=14

A₃=a+3d=8+3(3)=8+9=17

A₄=a+4d=8+4(3)=8+12=20

A₅=a+5d=8+5(3)=8+15=23

Therefore, the required 5 numbers between 8 and 26 to form an A.P. are 11, 14, 17, 20 and 23.

问题15.如果(a ⁿ + b ⁿ )/(a ^n-1 + b ^n-1 )是a和b之间的AM。找到n的值。

解决方案：

A.M. of a and b=(a+b)/2

According to given,

(a+b)/2=(aⁿ +bⁿ )/(a^n-1 +b^n-1)

⇒(a+b)(a^n-1+b^n-1)=2((aⁿ+bⁿ)

⇒aⁿ+ab^n-1+ba^n-1+bⁿ=2aⁿ+2bⁿ

⇒ab^n-1+a^n-1b=aⁿ+bⁿ

⇒ab^n-1-bⁿ=aⁿ-a^n-1b

⇒b^n-1(a-b)=a^n-1(a-b)

⇒b^n-1=a^n-1

⇒(a/b)^n-1=1=(a/b)⁰

⇒n-1=0

⇒n=1

问题16：在1到31之间插入了m个数字，以使所得序列为AP，第7^个项与第(m-1)^个项的比率为5：9。找出m的值。

解决方案：

Let A₁, A₂, . . ., A_m be m numbers such that 1, A₁, A₂, . . ., A_m, 31 is an A.P.

a=1, b=31, n=m+2

31=1+(m+2-1)d

⇒30=(m+1)d

⇒d=30/(m+1)                                                                —(1)

A₁=a+d

A₂=a+2d

A₃=a+3d

A₄=a+4d . . .

A₇=a+7d

A_m=a+(m-1)d

According to given,

(a-7d)/[a+(m-1)d]=5/9

⇒[1+7(30/(m+1))]/[1+(m-1)(30/m+1)]=5/9             [From (1)]

⇒[m+1+7(30)]/[m+1+30(m-1)]=5/9

⇒[m+1+210]/[m+1+30m-30]=5/9

⇒[m+211]/[31m-29]=5/9

⇒9m+1899=155m-145

⇒155m-9m=1899+145

⇒146m=2044

⇒m=14

Therefore, the value of m is 14.

问题17.一个人开始偿还第一期卢比的贷款。 100.如果他将分期付款增加Rs。每月5日，他将在30^日分期付款金额是多少?

解决方案：

Given, he first installment of the loan is Rs 100. The second installment of the loan in Rs 105 and so on. We can see that the amount that the man pays every month forms an A.P.

First term, a=100. Common difference, d=5

a₃₀ =a+(30-1)d

=100+(29)5

=100+145

=245

Therefore, the installment to be paid in the 30^th installment is Rs 245.

问题18：多边形的角度将形成AP，其共同差d为5°，第一项a为120°。找到多边形的边数。

解决方案：

It is known that the sum of all the angles of a polygon with n sides is 180°(n-2).

That is, S_n =180°(n-2)

⇒n[2a+(n-1)d]/2=180°(n-2)

⇒n[240+(n-1)5]=360(n-2)

⇒240n+5n² -5n=360n-720

⇒5n²+235n-360n+720=0

⇒5n²-125n+144=0

⇒n²-16n-9n+144=0

⇒n(n-16)-9(n-16)=0

⇒(n-9)(n-16)=0

⇒n = 9 or 16