问题1.查找：
(i)AP 1、4、7、10…的第十个帐篷。
(ii)AP的第18届√2、3√2、5√2，……。
(iii)AP 13、8、3，-2，……的第n个任期。
(iv)AP -40，-15、10、35，…………的第十个任期。
(v)AP 11、104、91、78，………的第八届任期
(vi)AP 10.0、10.5、11.0、11.2，…………的第11个男高音。
(vii)AP 3 / 4、5 / 4、7 / 4 + 9/4，………。的第9个学期
解决方案：
(i) Given that,
A.P. is 1, 4, 7, 10, ……….
First term(a) = 1
Common difference (d) = Second term – First term
= 4 – 1 = 3.
As we know that, to find nth term in an A.P = a + (n – 1)d
Therefore, 10th term in the A.P is 1 + (10 – 1)3
= 1 + 9×3 = 1 + 27 = 28
Hence 10th term of A.P is 28
(ii) Given that,
A.P. is √2, 3√2, 5√2, …….
First term (a) = √2,
Common difference = Second term – First term
d = 3√2 – √2 = 2√2
As we know that, to find nth term in an A.P = a + (n – 1)d
Therefore, 18th term of A. P. = √2 + (18 – 1)2√2
= √2 + 17.2√2 = √2 (1+34) = 35√2
Hence 18th term of A.P is 35√2
(iii) Given that,
A. P. is 13, 8, 3, – 2, …………
First term (a) = 13,
Common difference (d) = Second term first term
d = 8 – 13 = – 5
As we know that, to find nth term in an A.P = a + (n – 1)d
= 13 + (n – 1) – 5 = 13 – 5n + 5
Hence nth term of the A.P is an = 18 – 5n
(iv) Given that,
A. P. is – 40, -15, 10, 35, ……….
First term (a) = -40,
Common difference (d) = Second term – fast term
d = -15 – (- 40) = 40 – 15 = 25
As we know that, to find nth term in an A.P = a + (n – 1)d
Therefore, 10th term of A. P. = -40 + (10 – 1)25
= – 40 + 9.25 = – 40 + 225 = 185
Hence 10th term of the A.P is 185
(v) Given that,
Sequence is 117, 104, 91, 78, ………….
First term (a) = 117,
Common difference (d) = Second term – first term
d = 104 – 117 = –13
As we know that, to find nth term in an A.P = a + (n – 1)d
Therefore, 8th term = a + (8 – 1)d
= 117 + 7(-13) = 117 – 91 = 26
Hence 8th term of the A.P is 26
(vi) Given that,
A. P is 10.0, 10.5, 11.0, 11.5,
First term(a) = 10.0,
Common difference (d) = Second term – first term
d = 10.5 – 10.0 = 0.5
As we know that, to find nth term in an A.P = a + (n – 1)d
Therefore, 11th term a11 = 10.0 + (11 – 1)0.5
= 10.0 + 10 x 0.5 = 10.0 + 5 =15.0
Hence 11th term of the A. P. is 15.0
(vii) Given that,
A. P is 3/4, 5/4, 7/4, 9/4, …………
First term (a) = 3/4,
Common difference (d) = Second term – first term
d = 5/4 – 3/4 = 2/4
As we know that, to find nth term in an A.P = a + (n – 1)d
Therefore, 9th term a9 = a + (9 – 1)d
= 3/4 + 16/4 = 19/4
Hence 9th term of the A.P is 19/4.
问题2：查找：
(i)AP 3、8、13的哪个任期。是248吗?
(ii)AP 84、80、76 …的哪个项为0?
(iii)AP的第4、9、14任期。是254吗?
(iv)AP 21、42、63、84，…中的哪一个是420?
(v)AP 121、117。113，…中的哪一个是其第一个否定术语?
解决方案：
(i) Given that,
A.P. is 3, 8, 13, ………..
First term (a) = 3,
nth term is 248
Common difference (d) = Second term – first term
d = 8 – 3 = 5
As we know that, to find nth term in an A.P = a + (n – 1)d
248 = 3+(n – 1)5
248 = -2 + 5n
5n = 250
n =250/5 = 50
Hence 50th term in the A.P is 248.
(ii) Given that,
A. P is 84, 80, 76, …………
First term (a) = 84
nth term is 0
Common difference (d) = a2 – a
d = 80 – 84 = – 4
As we know that, to find nth term in an A.P = a + (n – 1)d
0 = 84 + (n – 1) – 4
84 = +4(n – 1)
n – 1 = 84/4 = 21
n = 21 + 1 = 22
Hence 22nd term in the A.P is 0.
(iii) Given A. P 4, 9, 14, …………
First term (a) = 4,
nth term is 254
Common difference (d) = a2 – a1
d = 9 – 4 = 5
As we know that, to find nth term in an A.P = a + (n – 1)d
4 + (n – 1)5 = 254
(n – 1)∙5 = 250
n – 1 = 250/5 = 50
n = 51
Hence 51th term in the A.P is 254.
(iv) Given that,
A. P 21, 42, 63, 84, ………
a = 21,
nth term = 420,
d = a2 – a1
= 42 – 21 = 21
As we know that, to find nth term in an A.P = a + (n – 1)d
21 + (n – 1)21 = 420
(n – 1)21 = 399
n – 1 = 399/21 = 19
n = 20
Hence 20th term is 420.
(v) Given that,
A.P is 121, 117, 113, ………..
First term (a) = 121,
nth term is negative i.e. an < 0,
Common difference (d) = 117 – 121 = – 4
As we know that, to find nth term in an A.P = a + (n – 1)d
121 + (n – 1) – 4 < 0
121 + 4 – 4n < 0
125 – 4n < 0
4n > 125
n > 125/4
n > 31.25
The integer which comes after 31.25 is 32.
Hence 32nd term in the A.P will be the first negative term.
问题3。
(i)68是AP 7、10、13，…的任期吗?
(ii)302是AP 3、8、13，…的任期。 ?
(iii)-150是AP 11、8、5、2…的任期吗?
解决方案：
(i) Given that,
A.P. 7, 10, 13,…
from given series,
a = 7 and d = a2 – a1 = 10 – 7 = 3
As we know that, to find nth term in an A.P = a + (n – 1)d
we have to find at which position 68 is present in given series,
a + (n – 1)d = 68
7 + (n – 1)3 = 68
7 + 3n – 3 = 68
3n + 4 = 68
3n = 64
n = 64/3, which is not a whole number.
Hence, 68 is not a term in the A.P.
(ii) Given, A.P. 3, 8, 13,…
from given series, a = 3 and d = a2 – a1 = 8 – 3 = 5
As we know that, to find nth term in an A.P = a + (n – 1)d
we have to find at which position 302 is present in given series,
a + (n – 1)d = 302
3 + (n – 1)5 = 302
3 + 5n – 5 = 302
5n – 2 = 302
5n = 304
n = 304/5, which is not a whole number.
Hence, 302 is not a term in the A.P.
(iii) Given, A.P. 11, 8, 5, 2, …
from given series,
a = 11 and d = a2 – a1 = 8 – 11 = -3
As we know that, to find nth term in an A.P = a + (n – 1)d
we have to find at which position -150 is present in given series,
a + (n – 1)d = -150
11 + (n – 1)(-3) = -150
11 – 3n + 3 = -150
3n = 150 + 14
3n = 164
n = 164/3, which is not a whole number.
Hence, -150 is not a term in the A.P.
问题4. AP中有多少个术语?
(i)7，10，13，…..，43
(ii)-1，-5 / 6，-2 / 3，-1 / 2，…，10/3
(iii)7，13，19，…，205
(iv)18、15½，13，….，-47
解决方案：
(i) Given that,
A.P. 7, 10, 13, ….., 43
where, a = 7 and d = a2 – a1 = 10 – 7 = 3
As we know that, to find nth term in an A.P = a + (n – 1)d
a + (n – 1)d = 43
7 + (n – 1)(3) = 43
7 + 3n – 3 = 43
3n = 43 – 4
3n = 39
n = 13
Hence, there are 13 terms in the given A.P.
(ii) Given that,
A.P. -1, -5/6, -2/3, -1/2, … , 10/3
where, a = -1 and d = a2 – a1 = -5/6 – (-1) = 1/6
As we know that, to find nth term in an A.P = a + (n – 1)d
a + (n – 1)d = 10/3
-1 + (n – 1)(1/6) = 10/3
-1 + n/6 – 1/6 = 10/3
n/6 = 10/3 + 1 + 1/6
n/6 = (20 + 6 + 1)/6
n = (20 + 6 + 1)
n = 27
Hence, there are 27 terms in the given A.P.
(iii) Given that,
A.P. 7, 13, 19, …, 205
where, a = 7 and d = a2 – a1 = 13 – 7 = 6,
nth term is 205
As we know that, to find nth term in an A.P = a + (n – 1)d
a + (n – 1)d = 205
7 + (n – 1)(6) = 205
7 + 6n – 6 = 205
6n = 205 – 1
n = 204/6
n = 34
Hence, there are 34 terms in the given A.P.
(iv) Given that,
A.P. 18, 15½, 13, …., -47
where, a = 7 and d = 15½ – 18 = 5/2,
As we know that, to find nth term in an A.P = a + (n – 1)d
a + (n – 1)d = 43
18 + (n – 1)(-5/2) = -47
18 – 5n/2 + 5/2 = -47
36 – 5n + 5 = -94
5n = 94 + 36 + 5
5n = 135
n = 27
Hence, there are 27 terms in the given A.P.
问题5. AP的第一项是5，共同的差是3，最后一项是80；查找条款的数量。
解决方案：
Given that,
a = 5 and d = 3,
last term = 80
As we know that, to find nth term in an A.P = a + (n – 1)d
therefore, for the given A.P. an = 5 + (n – 1)3 = 3n + 2
=3n + 2 = 80
3n = 78
n = 78/3 = 26
Hence, there are 26 terms in the A.P.
问题6. AP的第六和第17项分别是19和41，找到第40项。
解决方案：
Given that,
a6 = 19 and a17 = 41
As we know that, to find nth term in an A.P = a + (n – 1)d
therefore,
a6 = a + (6-1)d
= a + 5d = 19 ——-(i)
Similarly,
a17 = a + (17 – 1)d
= a + 16d = 41 ———-(ii)
Solving (i) and (ii),
(ii) – (i)
a + 16d – (a + 5d) = 41 – 19
11d = 22
d = 2
Using d in eqn(i), we get
a + 5(2) = 19
a = 19 – 10 = 9
Now, the 40th term is given by a40 = 9 + (40 – 1)2 = 9 + 78 = 87
Hence the 40th term is 87.
问题7.如果AP的第9项为零，则证明其第29项是第19项的两倍。
解决方案：
Given that,
a9 = 0
As we know that, to find nth term in an A.P = a + (n – 1)d
therefore, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ———-(i)
Now,
29th term is given by a29 = a + (29 – 1)d
=a29 = a + 28d
And, a29 = (a + 8d) + 20d (using (i))
= a29 = 20d ———-(ii)
Similarly, 19th term is given by a19 = a + (19 – 1)d
=a19 = a + 18d
And, a19 = (a + 8d) + 10d (using (i))
=a19 = 10d ———(iii)
On comparing (ii) and (iii), we observe that
a29 = 2(a19)
Hence, 29th term is double the 19th term.
问题8.如果AP的第10项的10倍等于第15项的15倍，则表明AP的第25项为零。
解决方案：
Given that,
10 times the 10th term of an A.P. is equal to 15 times the 15th term.
As we know that, to find nth term in an A.P = a + (n – 1)d
10(a10) = 15(a15)
10(a + (10 – 1)d) = 15(a + (15 – 1)d)
10(a + 9d) = 15(a + 14d)
10a + 90d = 15a + 210d
5a + 120d = 0
5(a + 24d) = 0
a + 24d = 0
a + (25 – 1)d = 0
a25 = 0
Hence, the 25th term of the A.P. is zero.
问题9. AP的第10和第18项分别是41和73。查找第26个学期。
解决方案：
Given that,
A10 = 41 and a18 = 73
As we know that, to find nth term in an A.P = a + (n – 1)d
therefore,
a10 = a + (10 – 1)d
= a + 9d = 41 ———(i)
Similarly,
a18 = a + (18 – 1)d
= a + 17d = 73 ——-(ii)
Solving (i) and (ii),
(ii) – (i)
a + 17d – (a + 9d) = 73 – 41
8d = 32
d = 4
Using d in (i), we get
a + 9(4) = 41
a = 41 – 36 = 5
Now, the 26th term is given by a26 = 5 + (26 – 1)4 = 5 + 100 = 105
Hence the 26th term is 105.
问题10.在某个AP中，第24个学期是第10个学期的两倍。证明第72个学期是第34个学期的两倍。
解决方案：
Given that,
24th term is twice the 10th term.
As we know that, to find nth term in an A.P = a + (n – 1)d
a24 = 2(a10)
a + (24 – 1)d = 2(a + (10 – 1)d)
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
a = 5d …. (1)
Now, the 72nd term can be expressed as:
a72 = a + (72 – 1)d
= a + 71d
= a + 5d + 66d
= a + a + 66d [using (1)]
= 2(a + 33d)
= 2(a + (34 – 1)d)
= 2(a34)
⇒ a72 = 2(a34)
Hence, the 72nd term is twice the 34th term of the given A.P.
问题11. AP的第26、11和最后一项分别为0、3和-1/5。找到共同的区别和术语的数量。
解决方案：
Given that,
a26 = 0, a11 = 3 and an (last term) = -1/5 of an A.P.
As we know that, to find nth term in an A.P = a + (n – 1)d
Therefore,
a26 = a + (26 – 1)d
a + 25d = 0 ——–(1)
a11 = a + (11 – 1)d
a + 10d = 3 ———(2)
Solving (1) and (2),
(1) – (2)
a + 25d – (a + 10d) = 0 – 3
15d = -3
d = -1/5
Using d in (1), we get
a + 25(-1/5) = 0
a = 5
Now, given that the last term is -1/5
5 + (n – 1)(-1/5) = -1/5
5 + -n/5 + 1/5 = -1/5
25 – n + 1 = -1
n = 27
Hence, the A.P has 27 terms and its common difference is -1/5.
问题12.如果AP第n个任期为9、7、5，…。与AP 15、12、9的第n个项相同……找到n。
解决方案：
Given that,
A.P1 = 9, 7, 5, …. and A.P2 = 15, 12, 9, …
nth term of the A.P1 = nth term of the A.P2,
As we know that, to find nth term in an A.P = a + (n – 1)d
For A.P1,
a = 9, d = Second term – first term = 9 – 7 = -2
And, its nth term an = 9 + (n – 1)(-2) = 9 – 2n + 2
an = 11 – 2n ———-(i)
Similarly, for A.P2
a = 15, d = Second term – first term = 12 – 15 = -3
And, its nth term an = 15 + (n – 1)(-3) = 15 – 3n + 3
an = 18 – 3n ——–(ii)
11 – 2n = 18 – 3n
n = 7
Hence, the 7th term of the both the A.Ps are equal.
问题13：从以下算术级数的结尾查找第12项：
(i)3、5、7、9…。 201
(ii)3,8,13，…，253
(iii)1，4，7，10，…，88
解决方案：
In order the find the 12th term from the end of an A.P which has n terms, it is done by simply finding the ((n -12) + 1)th of the A.P.
As we know that, to find nth term in an A.P = a + (n – 1)d
(i) Given that,
A.P = 3, 5, 7, 9, …. 201
last term is 201
where, a = 3 and d = (5 – 3) = 2
an = 3 + (n – 1)2 = 201
3 + 2n – 2 = 201
2n = 200
n = 100
Hence the A.P has 100 terms.
therefore, the 12th term from the end is same as (100 – 12 + 1)th of the A.P which is the 89th term.
a89 = 3 + (89 – 1)2
= 3 + 88(2)
= 3 + 176 = 179
Hence, the 12th term from the end of the A.P is 179.
(ii) Given that,
A.P = 3,8,13, … ,253
last term is 253
where, a = 3 and d = (8 – 3) = 5
an = 3 + (n – 1)5 = 253
3 + 5n – 5 = 253
5n = 253 + 2 = 255
n = 255/5
n = 51
Hence, the A.P has 51 terms.
therefore, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term.
a40 = 3 + (40 – 1)5
= 3 + 39(5)
= 3 + 195 = 198
Hence, the 12th term from the end of the A.P is 198.
(iii) Given that,
A.P = 1, 4, 7, 10, … ,88
where, a = 1 and d = (4 – 1) = 3
last term is 88
an = 1 + (n – 1)3 = 88
1 + 3n – 3 = 8
3n = 90
n = 30
Hence, the A.P has 30 terms.
therefore, the 12th term from the end is same as (30 – 12 + 1)th of the A.P which is the 19th term.
= a89 = 1 + (19 – 1)3
= 1 + 18(3) = 1 + 54 = 55
Hence the 12th term from the end of the A.P is 55.