10类NCERT解决方案-第5章算术级数–练习5.3 |套装1

问题1.找到以下AP的总和。

(i)2、7、12，……至10个学期。

(ii)− 37，− 33，− 29，…，至12个词

(iii)0.6、1.7、2.8，……..至100个条款

(iv)1 / 15、1 / 12、1 / 10，……至11个学期

解决方案：

(i) Given, 2, 7, 12,…, to 10 terms

For this A.P., we have,

first term, a = 2

common difference, d = a2 − a1 = 7−2 = 5

no. of terms, n = 10

Sum of nth term in AP series is,

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

Substituting the values,

S₁₀ = 10/2 [2(2)+(10 -1)×5]

= 5[4+(9)×(5)]

= 5 × 49 = 245

(ii) Given, −37, −33, −29,…, to 12 terms

For this A.P.,we have,

first term, a = −37

common difference, d = a2− a1

= (−33)−(−37)

= − 33 + 37 = 4

no. of terms, n = 12

Sum of nth term in AP series is,

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

Substituting the values,

S₁₂ = 12/2 [2(-37)+(12-1)×4]

= 6[-74+11×4]

= 6[-74+44]

= 6(-30) = -180

(iii) Given, 0.6, 1.7, 2.8,…, to 100 terms

For this A.P.,

first term, a = 0.6

common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1

no. of terms, n = 100

Sum of nth term in AP series is,

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

S12 = 50/2 [1.2+(99)×1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

(iv) Given, 1/15, 1/12, 1/10, ……, to 11 terms

For this A.P.,

first term, a = 1/5

common difference, d = a2 –a1 = (1/12)-(1/5) = 1/60

number of terms, n = 11

Sum of nth term in AP series is,

Sn = n/2 [2a + (n – 1) d]

Substituting the values, we have,

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

= 33/20

问题2。找到以下给出的总和：

(i)7+ 10 1/2 + 14 +………+ 84

(ii)34 + 32 + 30 +……….. + 10

(iii)− 5 +(− 8)+(− 11)+…………+(− 230)

解决方案：

(i) Given,

First term, a = 7

nth term, a_n = 84

Common difference, d = 10 1/2 – 7 = 21/2 – 7 = 7/2

Let 84 be the nth term of this A.P.

Then,

a_n = a(n-1)d

Substituting these values,

84 = 7+(n – 1)×7/2

77 = (n-1)×7/2

22 = n−1

n = 23

We know that, sum of n term is;

S_n = n/2 (a + l), l = 84

S_n = 23/2 (7+84)

= (23×91/2) = 2093/2

= 1046 1/2

(ii) Given,

first term, a = 34

common difference, d = a₂−a₁ = 32−34 = −2

nth term, an= 10

Let 10 be the nth term of this A.P.,

Now,

a_n = a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

12 = n −1

n = 13

Sum of n terms is;

S_n = n/2 (a +l), l = 10

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

= 286

(iii) Given:

First term, a = −5

nth term, a_n= −230

Common difference, d = a₂−a₁ = (−8)−(−5)

⇒d = − 8+5 = −3

Let us assume −230 be the nth term of this A.P.

Since,

a_n= a+(n−1)d

−230 = − 5+(n−1)(−3)

−225 = (n−1)(−3)

(n−1) = 75

n = 76

Sum of n terms, is equivalent to,

S_n = n/2 (a + l)

= 76/2 [(-5) + (-230)]

= 38(-235)

= -8930

问题3.在AP中

(i)给定a = 5，d = 3，an = 50，求n和Sn。

(ii)给定a = 7，a13 = 35，找到d和S13。

(iii)给定a12 = 37，d = 3，找到a和S12。

(iv)给定a3 = 15，S10 = 125，找到d和a10。

(v)给定d = 5，S9 = 75，找到a和a9。

(vi)给定a = 2，d = 8，Sn = 90，求n和an。

(vii)给定a = 8，an = 62，Sn = 210，求n和d。

(viii)给定an = 4，d = 2，Sn = − 14，求n和a。

(ix)给定a = 3，n = 8，S = 192，求d。

(x)给定l = 28，S = 144，共有9个条件。找到一个。

解决方案：

(i) Given values, we have,

a = 5, d = 3, a_n = 50

The nth term in an AP,

a_n = a +(n −1)d,

Substituting the given values, we have,

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

Obtaining the value of n, we get,

⇒ n = 16

Now, sum of n terms is equivalent to,

S_n = n/2 (a +an)

S_n = 16/2 (5 + 50) = 440

(ii) Given values, we have,

a = 7, a₁₃ = 35

The nth term in an AP,

a_n = a+(n−1)d,

Substituting the given values, we have,

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now, S_n = n/2 (a+an)

Obtaining the final value, we get,

S₁₃ = 13/2 (7+35) = 273

(iii) Given values, we have,

a₁₂ = 37, d = 3

The nth term in an AP,

a_n = a+(n −1)d,

Substituting the given values, we have,

⇒ a₁₂ = a+(12−1)3

⇒ 37 = a+33

Obtaining the value of a, we get,

⇒ a = 4

Now, sum of nth term,

S_n = n/2 (a+an)

= 12/2 (4+37)

Obtaining the final value,

= 246

(iv) Given that,

a3 = 15, S10 = 125

The formula of the nth term in an AP,

a_n = a +(n−1)d,

Substituting the given values, we have,

a₃ = a+(3−1)d

15 = a+2d ………….. (i)

Also,

Sum of the nth term,

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

S₁₀ = 10/2 [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9d …………….. (ii)

Solving eq (i) by (ii),

30 = 2a+4d ………. (iii)

And, by subtracting equation (iii) from (ii), we get,

−5 = 5d

that is,

d = −1

Substituting in equation (i),

15 = a+2(−1)

15 = a−2

a = 17 =

And,

a₁₀ = a+(10−1)d

a₁₀ = 17+(9)(−1)

a₁₀ = 17−9 = 8

(v) Given:

d = 5, S₉ = 75

Sum of n terms in AP is,

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

Substituting values, we get,

S₉ = 9/2 [2a +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3a = 25−60

a = -35/3

Also,

a_n = a+(n−1)d

Substituting values, we get,

a₉ = a+(9−1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

(vi) Given:

a = 2, d = 8, S_n = 90

Sum of n terms in an AP is,

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

Substituting values, we get,

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

⇒ 180 = n(4+8n -8) = n(8n-4) = 8n²-4n

Solving the eq, we get,

⇒ 8n²-4n –180 = 0

⇒ 2n²–n-45 = 0

⇒ 2n²-10n+9n-45 = 0

⇒ 2n(n -5)+9(n -5) = 0

⇒ (n-5)(2n+9) = 0

Since, n can only be a positive integer,

Therefore,

n = 5

Now,

∴ a₅ = 8+5×4 = 34

(vii) Given:

a = 8, an = 62, S_n = 210

Since, sum of n terms in an AP is equivalent to,

S_n = n/2 (a + an)

210 = n/2 (8 +62)

Solving,

⇒ 35n = 210

⇒ n = 210/35 = 6

Now, 62 = 8+5d

⇒ 5d = 62-8 = 54

⇒ d = 54/5 = 10.8

(viii) Given :

nth term, an = 4, common difference, d = 2, sum of n terms, S_n = −14.

Formula of the nth term in an AP,

a_n = a+(n −1)d,

Substituting the values, we get,

4 = a+(n −1)2

4 = a+2n−2

a+2n = 6

a = 6 − 2n …………………. (i)

Sum of n terms is;

S_n = n/2 (a+an)

-14 = n/2 (a+4)

−28 = n (a+4)

From equation (i), we get,

−28 = n (6 −2n +4)

−28 = n (− 2n +10)

−28 = − 2n²+10n

2n² −10n − 28 = 0

n² −5n −14 = 0

n² −7n+2n −14 = 0

n (n−7)+2(n −7) = 0

Solving for n,

(n −7)(n +2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

Since, we know, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we get

a = 6−2n

a = 6−2(7)

= 6−14

= −8

(ix) Given values are,

first term, a = 3,

number of terms, n = 8

sum of n terms, S = 192

We know,

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

Substituting values,

192 = 8/2 [2×3+(8 -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

Solving for d, we get,

d = 6

(x) Given values are,

l = 28,S = 144 and there are total of 9 terms.

Sum of n terms,

S_n = n/2 (a + l)

Substituting values, we get,

144 = 9/2(a+28)

(16)×(2) = a+28

32 = a+28

Calculating, we get,

a = 4

问题4. AP的多少个条款。必须取9、17、25…的总和为636?

解决方案：

Let us assume that there are n terms of the AP. 9, 17, 25 …

For this A.P.,

We know,

First term, a = 9

Common difference, d = a2−a1 = 17−9 = 8

Sum of n terms, is;

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

Substituting the values,

636 = n/2 [2×a+(8-1)×8]

636 = n/2 [18+(n-1)×8]

636 = n [9 +4n −4]

636 = n (4n +5)

4n² +5n −636 = 0

4n² +53n −48n −636 = 0

Solving, we get,

n (4n + 53)−12 (4n + 53) = 0

(4n +53)(n −12) = 0

that is,

4n+53 = 0 or n−12 = 0

On solving,

n = (-53/4) or n = 12

We know,

n cannot be negative or fraction, therefore, n = 12 is the only plausible value.

问题5. AP的第一项为5，最后一项为45，总和为400。找到项数和共同差异。

解决方案：

Given:

first term, a = 5

last term, l = 45

Also,

Sum of the AP, Sn = 400

Sum of AP is equivalent to

S_n = n/2 (a+l)

Substituting the values,

400 = n/2(5+45)

400 = n/2(50)

Number of terms, n =16

Since, the last term of AP series is equivalent to

l = a+(n −1)d

45 = 5 +(16 −1)d

40 = 15d

Solving for d, we get,

Common difference, d = 40/15 = 8/3

问题6. AP的第一项和最后一项分别为17和350。如果共同的差异是9，那么有多少个术语，它们的总和是多少?

解决方案：

Given:

First term, a = 17

Last term, l = 350

Common difference, d = 9

The last term of the AP can be written as;

l = a+(n −1)d

Substituting the values, we get,

350 = 17+(n −1)9

333 = (n−1)9

Solving for n,

(n−1) = 37

n = 38

S_n = n/2 (a+l)

S₃₈ = 13/2 (17+350)

= 19×367

= 6973

问题7。找到一个AP的前22个项之和，其中d = 7，第22个项是149。

解决方案：

Given:

Common difference, d = 7

Also,

22nd term, a₂₂ = 149

By the formula of nth term of an AP,

a_n = a+(n−1)d

Substituting values, we get,

a₂₂ = a+(22−1)d

149 = a+21×7

149 = a+147

a = 2 = First term

Sum of n terms,

S_n = n/2(a+an)

S₂₂ = 22/2 (2+149)

= 11×151

= 1661

问题8.找出第二项和第三项分别为14和18的AP的前51个项之和。

解决方案：

Given:

Second term, a₂ = 14

Third term, a₃ = 18

Also,

Common difference, d = a₃−a₂ = 18−14 = 4

a₂ = a+d

14 = a+4

Therefore,

a = 10 = First term

And,

Sum of n terms;

S_n = n/2 [2a + (n – 1)d]

Substituting values,

S₅₁ = 51/2 [2×10 (51-1) 4]

= 51/2 [2+(20)×4]

= 51 × 220/2

= 51 × 110

= 5610

问题9.如果AP的前7个项的总和为49，而17个项的总和是289，则求出前n个项的总和。

解决方案：

Given:

S₇ = 49

S₁₇ = 289

Since, we know

S_n = n/2 [2a + (n – 1)d]

Substituting values, we get,

S₇= 7/2 [2a +(n -1)d]

S₇ = 7/2 [2a + (7 -1)d]

49 = 7/2 [2a + 6d]

7 = (a+3d)

a + 3d = 7 ………………. (i)

Similarly,

S₁₇ = 17/2 [2a+(17-1)d]

Substituting values, we get,

289 = 17/2 (2a +16d)

17 = (a+8d)

a +8d = 17 ………………. (ii)

Solving (i) and (ii),

5d = 10

Solving for d, we get,

d = 2

Now, obtaining value for a, we get,

a+3(2) = 7

a+ 6 = 7

a = 1

Therefore,

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

= n/2[2(1)+(n – 1)×2]

= n/2(2+2n-2)

= n/2(2n)

= n²

问题10显示的是A _{1，A 2} …，A _N，…形成的地方被定义为下面的AP

(i) _n = 3 + 4n

(ii) _n = 9-5n

此外，在每种情况下，请找到前15个项的总和。

解决方案：

(i) a_n = 3+4n

Calculating,

a₁ = 3+4(1) = 7

a₂ = 3+4(2) = 3+8 = 11

a₃ = 3+4(3) = 3+12 = 15

a₄ = 3+4(4) = 3+16 = 19

Now d =

a2 − a1 = 11−7 = 4

a3 − a2 = 15−11 = 4

a4 − a3 = 19−15 = 4

Hence, a_{k + 1} − a_k holds the same value between all pairs of successive terms. Therefore, this is an AP with common difference as 4 and first term as 7.

Sum of nth term is;

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

Substituting the value, we get,

S₁₅ = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) a_n = 9−5n

Calculating, we get,

a₁ = 9−5×1 = 9−5 = 4

a₂ = 9−5×2 = 9−10 = −1

a₃ = 9−5×3 = 9−15 = −6

a₄ = 9−5×4 = 9−20 = −11

Common difference, d

a2 − a1 = −1−4 = −5

a3 − a2 = −6−(−1) = −5

a4 − a3 = −11−(−6) = −5

Hence, a_{k + 1} − a_k holds the same value between all pairs of successive terms. Therefore, this is an AP with common difference as −5 and first term as 4.

Sum of nth term is;

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

S₁₅ = 15/2[2(4) +(15 -1)(-5)]

Substituting values,

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465