Class 11 RD Sharma解决方案–第19章算术级数-练习19.4 |套装3

问题23.如果AP的第5和第12个学期分别是30和65，那么前20个学期的总和是多少?

解决方案：

According to question we have

a₅ = 30 

Using the formula

⇒ a + (5 – 1)d = 30 

⇒ a + 4d = 30        ……………. (i) 

Also, a_n = 65 

Using the formula

⇒ a + (12 – 1)d = 65 

⇒ a + 11d = 65      ……………(ii) 

On solving eq(i) and (ii), we get 

7d = 35 

⇒ d = 5 

On putting the value of d in (i), we get : 

a + 4 x 5 = 30 

⇒ a = 10 

Now we find the sum of first 20 terms

S₂₀ = 20/2 [2 x 10 + (20 – 1) x 5] 

⇒ S₂₀ = 10 [20 + 95]

⇒ S₂₀ = 1150

问题24.找到第k个项为5k + 1的AP的n个项之和。

解决方案：

According to question we have

a_k = 5k + 1 

For k = 1, a₁ = 5 x 1 + 1 = 6 

For k = 2, a₂ = 5 x 2 + 1 = 11 

For k = n, a_n = 5n + 1 

Now we find the sum of n terms

Sn = n/2 [a + an] 

⇒ Sn = n/2[6 + 5n + 1] = n/2 (5n + 7)  

问题25.找到所有两位数字的总和，当它们除以4时，得出1的余数。

解决方案：

According to question we have

A.P = 13, 17….97

From the A.P we have

a = 13, d = 4, a_n = 97 

Now using the formula

⇒ a_n = a (n – 1)d 

⇒ 97 = 13 + (n – 1)4 

⇒ 84 = 4n – 4 

⇒ 88 = 4n 

⇒ 22 = n          …………… (1) 

Now we find the sum of all two-digit numbers using the given formula

S_n = n/2[2a + (n – 1)d] 

S₂₂ = 22/2 [2 x 13 + (22 – 1) x 4]      ( From eq(1)) 

⇒ S₂₂ = 11[26 + 84]

⇒ S₂₂ = 11[110] = 1210

问题26.如果AP 25、22、19，…的一定数量的术语的总和是116。找到最后一个术语。

解决方案：

According to question 

A.P. = 25, 22, 19 

From the A.P we have

a = 25, d = 22 – 25 = -3 

S_n = 116 

Now using the formula

⇒ n/2 [2a + (n – 1)d] = 116 

⇒ n [2 x 25 + (n – 1)(-3)] = 232 

⇒ 50n -3n² + 3n = 232 

⇒ 3n² – 53n + 232 = 0 

⇒ 3n² – 29n – 24n + 232 = 0 

⇒ n(3n – 29) – 8(3n – 29) = 0 

⇒(3n – 29)(n – 8) = 0 

⇒ n = 29/ 3 or 8 

Since n cannot be a fraction, n = 8. 

Now we find the last term using the following formula

a_n = a + (n – 1)d

⇒ a₈ = 25 + (8 – 1)(-3)

⇒ a₈ = 4

问题27.找出1到2001之间的奇数整数之和。

解决方案：

According to question 

A.P = 1, 3, 5 … 2001

From the A.P we have

a = 1 and d = 2 

a_n = 2001 

Now using the formula

⇒ 1 + (n – 1)2 = 2001 

⇒ 2n – 2 = 2000 

⇒ 2n = 2002 

⇒ n = 1001  

Now we find the sum of odd integers from 1 to 2001

Also, S₁₀₀₁= 1001/2 [2 x 1 + (1001 – 1)2]

⇒ S₁₀₀₁ = 1001/2 [2 + 2000]

⇒ S₁₀₀₁ = 1001 x 1001 = 1002001 

问题28.要加上-25，需要多少个AP -6，-11 / 2，-5，…?

解决方案：

According to question 

A.P = -6, -11/2, -5, … 

From the A.P we have

a = – 6 and d =-11/2 – (-6) = 1/2

S_n = -25  

Now using the formula

 ⇒ -25 = n/2 [2 x (-6) + (n -1)(1/2)]

⇒ -25 = n/2 [ -12 + n/2 – 1/2]

⇒ -50 = n [ n/2 – 25/2]

⇒ -100 = n [ n – 25]

⇒ n² – 25n + 100 = 0

⇒ (n – 20)(n – 5) = 0

⇒ n = 20 or n = 5

问题29.在AP中，第一项为2，而前五项之和为后五项的四分之一。表明第20个学期是-112。

解决方案：

According to question we have

a = 2, S₅ = 1/4(S₁₀ – S₅)  

S₅ =  5/2 [2 x 2 + (5 – 1)d]  

⇒ S₅ = 5 [2 + 2d]    ………………..(i) 

Also, S₁₀ = 10/2 [2 x 2 + (10 – 1)d] 

⇒ S₁₀ = 5[4 + 9d]   ………………..(ii) 

Since S₅ = 1/4 (S₁₀ – S₅) 

So, from eq(i) and (ii), we have: 

⇒ 5[2 + 2d] = 1/4 [5(4 + 9d) – 5(2 + 2d)] 

⇒ 8 + 8d = 4 + 9d – 2 – 2d  

⇒ d = -6

Now we find the 20th term

a₂₀ = a + (20 – 1)d 

⇒ a₂₀ = a + 19d

⇒ a₂₀ = 2 + 19(-6) = -112

Hence proved

问题30.如果S ₁是AP的(2n +1)项之和，而S ₂是其奇数项的和，则证明：S ₁ ：S ₂ =(2n +1)：(n +1) )

解决方案：

Let us assume A. P. be a, a + d, a + 2d… 

So, S₁ = (2n + 1)/2 [2a + (2n + 1 – 1)d] 

Now using the formula, we get

⇒ S₁ = (2n + 1)/2 [2a + (2n)d] 

⇒ S₁ = (2n + 1)(a + nd)          …………………(i) 

Now using the formula, we get

S₂ = (n + 1)/2 [2a + (n + 1 – 1) x 2d] 

⇒ S₂ = (n +1)/2 [2a + 2nd] 

⇒ S₂ = (n + 1)[a + nd]           …………………..(ii)

 From eq(i) and (ii), we get : 

S₁ / S₂ = (2n + 1) / (n + 1) 

Hence, proved 

问题31.找到一个AP，其中任意数量的项之和始终是这些项的平方数的三倍。

解决方案：

Given that S_n = 3n²

So, for n = 1, S₁ = 3 x 1² = 3 

For n = 2, S₂ = 3 x 2² = 12 

For n = 3, S₃= 3 x 3² = 27 and so on  

So, S₁ = a₁ = 3 

a₂ = S₂ – S₁ = 12 – 3 = 9 

a3 = S₃ – S₂ = 27 – 12 = 15 and so on  

Hence, the A.P. = 3, 9, 15… 

问题32.如果一个AP的n个项之和为nP + 1/2 – n(n – 1)Q，其中P和Q为常数，则求出共同的差。

解决方案：

According to question we have

S_n = nP + 1/2 n(n – 1)Q 

For n = 1, S₁ = P + 0 = P 

For n = 2, S₂ = 2P + Q 

Also, a₁ = S₁ = P,

a₂ = S₂  –  S₁ = 2P + Q – P = P + Q 

Hence, the common difference d = a₂ – a₁ = P + Q – P = Q  

问题33.两个算术级数的n个项之和为5n + 4：9n +6。找到它们的第18个项之比。

解决方案：

Let us considered we have two A.P’s. So, a₁ and a₂ are the first terms and d₁ and d₂ is common difference of the A.P’s

According to question we have

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

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

Now put n = 2 x 18 – 1 = 35 in eq(1), we get 

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

179 /321 = (a₁ + 17d₁) / (a₂ + 17d₂) = (18th term of the first A.P.) / (18th term of the second A.P.)

Hence, the ratio of 18th terms = (a₁ + 17d₁) / (a₂ + 17d₂) = 179 /321

问题34.两个算术级数的n个项之和为7n + 2：n +4。找到它们的第五项之比。

解决方案：

Let us considered we have two A.P’s. So, a₁ and a₂ are the first terms and S₁ and S₂ are the sum of the first n terms. 

According to question we have

S₁ = n/2 [2a₁ + (n – 1)d₁] 

And, S₂ = n/2 [2a₂ + (n -1)d₂] 

Given:  

S₁ / S₂ = (n/2 [2a₁ + (n – 1)d₁])/(n/2 [2a₂ + (n – 1)d₂]) = (7n + 2) / (n + 4)

Now we find the ratio of their 5th terms

[2a₁ + (9 – 1)d₁] / [2a₂ + (9 – 1)d₂] = (7(9) + 2) / (9 + 4)

[2a₁ + (8)d₁] / [2a₂ + (8)d₂] = 65/13

[a₁ + (4)d₁] / [a₂ + (4)d₂] = 5/1 = 5 : 1

Hence, the ratio of 5th terms = 5 : 1