第 11 课 RD Sharma 解决方案 - 第 19 章算术级数 - 练习 19.2 |设置 1
General Formula
Tn = a + (n – 1)*d
where,
Tn is the nth term
a is the first term
d = common difference
问题 1. 查找
i ) AP 1、4、7、10 的第 10 期......
解决方案:
So we have a1 = a = 1
d= 4-1 = 3
n= 10
So we can calculate 10th term of AP by using general formula-
T10= a + (10-1)d
= 1 + 9*3
= 28
Hence, the 10th term is 28.
ii) 第 18届AP √2、3√2、5√2…..
解决方案:
So we have a1 = a = √2
d= 3√2 – √2 = 2√2
n= 18 (given in Question )
So we can calculate 18th term of AP by using general formula-
T18= a + (18-1)d
= √2 + 17*(2√2 )
= 35√2
Hence the 18th term is 35√2.
iii) AP 13、8、3、-2…的第 n 项。
解决方案:
So we have a1 = a = 13
d= a2 – a1 = 8 – 13 = -5
So nth term is
Tn= 13 + (n-1)( – 5 )
= 13 -5n + 5
= 18 – 5n
Hence the nth term is 18-5n.
问题 2. 在 AP 中显示 a m+n + a mn = 2a m 。
解决方案:
We can prove this with the help of general formula. Let’s solve LHS.
am+n= a + (m+n-1)d
am-n = a + (m-n-1)d
and am = a + (m-1)d
So now,
LHS = 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)
= 2am= RHS
问题 3. i) AP 的哪一项是 3, 8, 13,…… 是 248。
解决方案:
So we are given a = 3
d= 8 – 3 = 5
Tn = 248
a + (n -1)d = 248
3 + (n-1)5 = 248
(n-1)5 = 245
n-1 = 49
n =50
Hence 50th term of this AP is 248.
ii) AP 的哪一项是 84、80、76……是 0。
解决方案:
So we are given a = 84
d= 80 – 84 = -4
Tn = 0
a + (n -1)d = 0
84 + (n-1)(-4) =0
(n-1)(-4) = -84
n-1 = 21
n =22
Hence 22th term of this AP is 248.
iii) AP 4、9、14….. 中的哪一项是 254?
解决方案:
So we are given a = 4
d= 9 – 4 = 5
Tn = 254
a + (n -1)d = 254
4 + (n-1)5 = 254
(n-1)5 = 250
n-1 = 50
n =51
Hence 51th term of this AP is 248.
问题 4. i) 68 是 AP 7、10、13、……的一个术语吗?
解决方案:
We are given a = 7
d= 10 – 7 = 3
We can find whether 68 is a term of this AP by finding the valid n for 68 by using general formula. If there is no valid integer n value for 68 then it will not be a term of this AP.
Tn = 68
a+ (n-1)d = 68
7 + (n-1)3 = 68
(n-1)3= 61
n-1 = 61/3
n= 64/3
Hence we can see there is no valid integer value of n for 68 so 68 is not the term of the AP.
ii) 302 是 AP 3, 8, 13, ....的一个术语吗?
解决方案:
We are given a = 3
d= 8 – 3 = 5
We can find whether 302 is a term of this AP by finding the valid n for 302 by using general formula. If there is no valid integer n value for 302 then it will not be a term of this AP.
Tn = 302
a+ (n-1)d = 302
3 + (n-1)5 = 302
(n-1)5= 299
n-1 = 299/5
n= 304/5
Hence we can see there is no valid integer value of n for 302 so 302 is not the term of the AP.
问题 5. i) 序列 24, 23¼, 22½, 22¾, ...... 中的第一个负数是什么?
解决方案:
We are given a = 24
d= 23¼ – 24 = 93/4- 24 = -3/4
So to find first negative number we can find n value from Tn < 0
a + (n- 1)d < 0
24 + (n-1)(-3/4) < 0
24 -3n/4 +3/4 < 0
99/4 – 3n/4 <0
3n/4> 99/4
By solving this inequality we found
n>33
So we can say 34th term of AP will be first negative number.
ii) 序列 12 + 8i , 11+ 6i, 10 +4i, …… 的哪一项是 (a) 纯实数 (b) 纯虚数
解决方案:
We are given a = 12 + 8i
d= (11 + 6i) -(12 + 8i) = 11 +6i – 12 – 8i =-1 – 2i
So Tn for this sequence will be
Tn = a + (n-1)d
= 12 + 8i + (n-1)(-1-2i)
= 12 + 8i -n +1 -2in + 2i
= 13 – n + 10i -2in
= 13 – n + (10 – 2n)i
(a) For Tn to be purely real, imaginary part should be equal to 0.
So we know that 10 – 2n must be 0
10 – 2n =0
so n= 5
Hence 5th term of A.P. is purely real.
(b) For Tn to be purely imaginary, real part should be equal to 0.
So we know that 13 – n must be 0
13 – n =0
so n= 13
Hence 13th term of A.P. is purely imaginary.
问题 6. i) AP 7, 10, 13, ....., 43 中有多少项?
解决方案:
We are given a= 7
d= 10 – 7 = 3
Tn = 43
Last term of AP is 43. So if we calculate the position of 43 then we will get the terms in this AP.
a + (n-1)d =43
7 + (n-1)3 = 43
(n-1)3 = 36
n-1 = 12
n = 13
So there are total 13 terms in this AP.
ii) AP -1、-5/6、-2/3、-1/2……、10/3 中有多少项?
解决方案:
We are given a= -1
d= -5/6 – (-1) = 1-5/6 = 1/6
Tn = 10/3
Last term of AP is 10/3. So if we calculate the position of 10/3 then we will get the terms in this AP.
a + (n-1)d =10/3
-1 + (n-1)(1/6) = 10/3
(n-1)(1/6) = 13/3
n-1 = 13*6/3
n-1= 26
n = 27
So there are total 27 terms in this AP.
问题 7. AP 的第一项是 5。共同差是 3,最后一项是 80。求词数。
解决方案:
We are given
first term a1= a= 5
common difference d= 3
Tn = 80
Last term of AP is 80. So if we calculate the position of 80 then we will get the terms in this AP.
a + (n-1)d =80
5 + (n-1)3 = 80
(n-1)3 = 75
n-1 = 75/3
n = 25+1
n = 26
So there are total 26 terms in this AP.
问题 8. AP 的第6和第 17项分别是 19 和 41。找到第 40项。
解决方案:
We are given T6=19 and T17=41.
a + 5d = 19 —– 1
a+ 16d = 41 —– 2
On solving equation 1 and 2
a + 5d – a – 16d = 19 – 41
-11d = – 22
d = 2
and a =19 – 10
a= 9
So T40 = a + 39d
= 9 + 39*2
= 9 + 78
= 87
So 40th term is 87 in this AP.
问题 9. 如果 AP 的第9 项为零,证明其第29项是第 19项的两倍。
解决方案:
We are given 9th term of AP is 0.
a9=a+ 8d = 0 ——1
We have to prove, a29 = 2a19
We know a29= a+ 28d ——-2
a19= a+ 18d ——-3
From equation 1 we get
a+ 8d = 0
a= -8d
So putting a = -8d in eq 2 & 3 will give us the value of a29 and a19
a29 = -8d +28d = 20d
a19 = -8d +18d = 10d
So its proved 2a19 = a29.
问题 10. 如果 AP 的第10 项的 10 倍等于第 15项的 15 倍,证明 AP 的第25 项为零。
解决方案:
We are given 10 times the 10th term of an AP is equal to 15 times the 15th term.
10a10 = 15a15
10(a + 9d) = 15( a +14d)
10a + 90d = 15a + 210d
5a= -120d
5a + 120d= 0
a + 24d = 0
a25 = 0
So its proved a25 is equal 0.
问题 11. AP 的第 10 和第18项分别是 41 和 73,求第26 项。
解决方案:
We are given a10 = a+ 9d = 41 ——-1
a18 = a+ 17d = 73 ——-2
On solving 1 & 2
a + 9d – a -17d = 41 – 73
-8d = -32
d= 4
By substituting value of d in equation 1 we get
a= 41-9*4
a= 5
So value of 26th term can be calculated by,
a26 = a + 25d
= 5 + 25*4
= 105
So the 26th term of AP is 105.
问题 12. 在某些 AP 中,第 24项是第 10项的两倍。证明第72项是第34项的两倍。
解决方案:
We are given 24th term of an AP is twice the 10th term.
a24 = 2a10
So a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
a = 5d
And we know, a34= a +33d
= 5d + 33d
= 38d
Similarly, a72= a +71d
= 5d + 71d
= 76d
Now we can see that a72 = 2a34 . Hence proved 72nd term is twice the 34th term.