Question

NCERT Solutions Class 11 Maths Chapter 9 Sequence and Series is one of the important chapters of class 11. Our NCERT Solutions are made in very concise form.

One must refer to our NCERT Solutions Class 11 for complete understanding of all the concepts. We have covered all the important concepts such as

• Sequence – sequence is orderly arrangement of grouped numbers which is dictated by some specific rule. Example of sequence 1, 3, 5, 7, 9, 11.  
• Series – things or the events of similar class which comes one after one another in spatial or temporal succession is known as series. It is also the sum of the elements in the sequence.

• Arithmetic Progression (A.P.) – an arithmetic progression is a sequence in which the difference between every successive term is same. In arithmetic progression every next term can be obtained by adding some fixed number to the previous term. Example: 1, 3, 5, 7, 9, 11, 13

• Arithmetic Mean (A.M.) – arithmetic mean is the average sum of set of given numbers. In Arithmetic Progression the arithmetic mean is the middle term of the arithmetic progression. In a given arithmetic progression when there are three terms a, b, and c then the arithmetic mean is the middle term, i.e. ‘b’, hence the b is the arithmetic mean of the given arithmetic progression.

• Geometric Progression (G.P.) – in Geometric Progression each new term is obtained by multiplying a fixed number to the last term. The number which is being multiplied is also called the common ratio. Example: 1,2,4,8,16,32,64.

• Sum of Geometric series of infinite terms – in infinite geometric series there is no particular last term. There are two such cases of infinite geometric series. In one case the ratio of geometric series is greater than 1 and other where the ratio of geometric series is less than 1.
• Geometric mean (G.M.) – the geometric mean is a peculiar type of average where we take the reciprocal of number of number of term and put it over the product of numbers. Example: geometric mean of 4 and 16 is 8. \(4\times 16\) = 64, \(\sqrt[]{64} \) = 8.

This NCERT Solutions Class 11 Maths provides a holistic approach to provide solution of all queries of students.

Answer 1

NCERT Solutions Class 11 Maths Chapter 9 Sequence and Series

1. Write the first five terms of the sequences whose n^th term is a_n= n(n + 2).

Answer:

a_n= n(n + 2) 

Substituting n = 1, 2, 3, 4, and 5, we obtain

Therefore, the required terms are 3, 8, 15, 24, and 35.

2. Write the first five terms of the sequences whose n^th term is \(a_n = \frac n{n+1}\) 

Answer:

\(a_n = \frac n{n+1}\) 

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are \(\frac12,\frac23,\frac34,\frac45\) and \(\frac56\) 

3. Write the first five terms of the sequences whose n^th term is a_n = 2ⁿ

Answer:

a_n = 2ⁿ 

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are 2, 4, 8, 16, and 32.

4. Write the first five terms of the sequences whose n^th term is \(a_n=\frac{2n-3}{6}\)

Answer:

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are.

 

5. Write the first five terms of the sequences whose n^th term is \(a_n = (-1)^{n-1}5^{n+1}\)

Answer:

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are 25, –125, 625, –3125, and 15625.

6. Write the first five terms of the sequences whose n^th term is \(a_n=n\frac{n^2+5}{4}\)

Answer:

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are

7. Find the 17^th term in the following sequence whose n^th term is a_n = 4n -3 ; a₁₇,a₂₄

Answer:

Substituting n = 17, we obtain

a₁₇ = 4(17) - 3 = 68 - 3 = 65

Substituting n = 24, we obtain

a₂₄ = 4(24) - 3 = 96 - 3 = 93

8. Find the 7^th term in the following sequence whose n^th term is \(a_n = \frac{n^2}{2n}; a_7\)

Answer:

Substituting n = 7, we obtain

\(a_7 = \frac{7^2}{9^2}=\frac{49}{128}\)

9. Find the 9^th term in the following sequence whose n^th term is

\(a_n=(-1)^{n-1}n^3;a_9\)

Answer:

Substituting n = 9, we obtain

\(a_9 = (-1)^{9-1}(9)^3=(9)^3= 729\)

10. Find the 20^th term in the following sequence whose n^th term is

\(a_n=\frac{n(n-2)}{n+3};a_{20}\)

Answer:

Substituting n = 20, we obtain

11. Write the first five terms of the following sequence and obtain the corresponding series:

a₁ = 3, a_n = 3a_n-1 +2 for all n > 1

Answer:

a₁ = 3, a_n = 3a_n-1 +2 for all n > 1

 

Hence, the first five terms of the sequence are 3, 11, 35, 107, and 323. 

The corresponding series is 3 + 11 + 35 + 107 + 323 + …

12. Write the first five terms of the following sequence and obtain the corresponding series:

\(a_1=-1\), \(a_n = \frac{a_{n-1}}{n}\) , \(n \ge2\)

Answer:

 \(a_1=-1\), \(a_n = \frac{a_{n-1}}{n}\) , \(n \ge2\) 

Hence, the first five terms of the sequence are 

The corresponding series is

13. Write the first five terms of the following sequence and obtain the corresponding series:

a₁ = a₂ = 2, a_n = a_n-1 -1, n > 2

Answer:

a₁ = a₂ = 2, a_n = a_n-1 -1, n > 2

Hence, the first five terms of the sequence are 2, 2, 1, 0, and –1. 

The corresponding series is 2 + 2 + 1 + 0 + (–1) + …

14. The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_n-1+ a_n-2, n > 2

FInd \(\frac{a_{n+1}}{a_n}\) , for n = 1, 2, 3, 4, 5

Answer:

15. Find the sum of odd integers from 1 to 2001.

Answer:

The odd integers from 1 to 2001 are 1, 3, 5 …1999, 2001. 

This sequence forms an A.P. 

Here, first term, a = 1 

Common difference, d = 2

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

Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Answer 2

16. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Answer:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

17. In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20^th term is –112.

Answer:

First term = 2 

Let d be the common difference of the A.P.

Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d … 

Sum of first five terms = 10 + 10d 

Sum of next five terms = 10 + 35d 

According to the given condition,

Thus, the 20^th term of the A.P. is –112.

18. How many terms of the A.P. \(-6,-\frac{11}2,-5....\) are needed to give the sum –25?

Answer:

Let the sum of n terms of the given A.P. be –25. 

It is known that,

Where n = number of terms, a = first term, and d = common difference 

Here, a = –6

Therefore, we obtain

19. In an A.P., if p^th term is 1/q and q^th term is 1/p, prove that the sum of first pq terms is 1/2 (pq +1), where p ≠ q.

Answer:

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

\(\therefore\) According to the given information,

Subtracting (2) from (1), we obtain

Putting the value of d in (1), we obtain

Thus, the sum of first pq terms of the A.P. is \(\frac12(pq+1)\).

20. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term

Answer:

If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term

Here, a = 25 and d = 22 – 25 = – 3

However, 

n cannot be equal to \(\frac{29}3\). Therefore, n = 8

\(\therefore\) a₈ = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3) 

= 25 + (7) (– 3) = 25 – 21 

= 4 

Thus, the last term of the A.P. is 4.

21. Find the sum to n terms of the A.P., whose k^th term is 5k + 1.

Answer:

It is given that the k^th term of the A.P. is 5k + 1. 

k^th term = ak = a + (k – 1)d

\(\therefore\) a + (k – 1)d = 5k + 1 a + kd – d = 5k + 1 

\(\therefore\) Comparing the coefficient of k, we obtain d = 5 a – d = 1 

⇒ a – 5 = 1 

⇒ a = 6

22. If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference.

Answer:

It is known that: 

According to the given condition,

Comparing the coefficients of n² on both sides, we obtain

\(\frac d2 = q\)

∴ d = 2q

Thus, the common difference of the A.P. is 2q.

23. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18^th terms.

Answer:

Let a₁, a_2, and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression respectively. 

According to the given condition,

Substituting n = 35 in (1), we obtain

From (2) and (3), we obtain

Thus, the ratio of 18^th term of both the A.P.s is 179: 321.

24. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

Answer:

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

Here,

According to the given condition,

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

Answer 3

25. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that 

\(\frac ap(q-r)+\frac bq(r-p)+\frac cr (p-q) = 0\)

Answer:

Let a₁ and d be the first term and the common difference of the A.P. respectively. 

According to the given information,

Subtracting (2) from (1), we obtain

Subtracting (3) from (2), we obtain

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

Dividing both sides by pqr, we obtain

Thus, the given result is proved.

26. The ratio of the sums of m and n terms of an A.P. is m²: n² . Show that the ratio of m^th and n^th term is (2m – 1): (2n – 1).

Answer:

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

According to the given condition,

Putting m = 2m – 1 and n = 2n – 1 in (1), we obtain

From (2) and (3), we obtain

Thus, the given result is proved.

27. If the sum of n terms of an A.P. is 2n² + 5n and its m^th term is 164, find the value of m.

Answer:

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

a_m = a + (m – 1)d = 164 … (1) 

Sum of n terms:

Here,

Comparing the coefficient of n² on both sides, we obtain

\(\frac d2 = 3\)

⇒ d = 6

Comparing the coefficient of n on both sides, we obtain

\(a - \frac d2 = 5\)

⇒ a - 3 = 5

⇒ a = 8

Therefore, from (1), we obtain

8 + (m – 1) 6 = 164 

⇒ (m – 1) 6 = 164 – 8 = 156 

⇒ m – 1 = 26 

⇒ m = 27 

Thus, the value of m is 27.

28. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer:

Let A₁, A₂, A₃, A₄, and A₅ be five numbers between 8 and 26 

such that 8, A₁, A₂, A₃, A₄, A₅, 26 is an A.P.

Here, a = 8, b = 26, n = 7 

Therefore, 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 

Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

29. If \(\frac{a^n+b^n}{a^{n-1}+b^{n-1}}\) is the A.M. between a and b, then find the value of n.

Answer:

A.M. of a and b = \(\frac{a=b}2\)

According to the given condition,

30. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^th and (m – 1)^th numbers is 5:9. Find the value of m.

Answer:

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

Here, a = 1, b = 31, n = m + 2 

∴ 31 = 1 + (m + 2 – 1) (d) 

⇒ 30 = (m + 1) d

⇒ \(d = \frac{30}{m+1}\)       ......(1)

A₁ = a + d 

A₂ = a + 2d 

A₃ = a + 3d … 

∴ A₇ = a + 7d

A_m–1 = a + (m – 1) d 

According to the given condition,

Thus, the value of m is 14.

31. A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs 5 every month, what amount he will pay in the 30th installment?

Answer:

The first installment of the loan is Rs 100. 

The second installment of the loan is Rs 105 and so on. 

The amount that the man repays every month forms an A.P. 

The A.P. is 100, 105, 110 …

First term, a = 100 

Common difference, d = 5 

A₃₀ = a + (30 – 1)d 

= 100 + (29) (5) 

= 100 + 145 

= 245 

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

32. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

Answer:

The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°. 

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

33. Find the 20^th and n^th terms of the G.P. \(\frac52,\frac54,\frac58,...\)

Answer:

The given G.P. is \(\frac52,\frac54,\frac58,...\) 

Here, a = First term = \(\frac52\)

r = Common ratio = \(\cfrac{\frac54}{\frac52}=\frac12\)

34. Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

Answer:

Common ratio, r = 2 

Let a be the first term of the G.P. 

∴ a₈ = ar^8–1 = ar⁷ ⇒ ar⁷ = 192 a(2)⁷ = 192 a(2)⁷ = (2)⁶ (3)

Answer 4

35. The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

Answer:

Let a be the first term and r be the common ratio of the G.P. 

According to the given condition,

a₅ = a r^5–1 = a r⁴ = p ………..… (1)

a₈ = a r^8–1 = a r⁷ = q ………….. (2)

a₁₁ = a r^11–1 = a r¹⁰ = s ………. (3)

Dividing equation (2) by (1), we obtain

Dividing equation (3) by (2), we obtain

Equating the values of r³ obtained in (4) and (5), we obtain

Thus, the given result is proved.

36. The 4^th term of a G.P. is square of its second term, and the first term is –3. Determine its 7^th term.

Answer:

Let a be the first term and r be the common ratio of the G.P. 

∴ a = –3 

It is known that, an = ar^n–1 

∴ a₄ = ar³ = (–3) r³ 

a₂ = a r¹ = (–3) r

According to the given condition, 

(–3) r³ = [(–3) r]² 

⇒ –3r³ = 9 r² 

⇒ r = –3 a₇ = a r^7–1 = a 

r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187

Thus, the seventh term of the G.P. is –2187.

37. Which term of the following sequences:

(a) \(2,2\sqrt2,4,... is \,128? \)

(b) \(\sqrt3,3,3\sqrt3,...is \,729? \)

(c) \(\frac13,\frac19,\frac1{27},...is\frac{1}{19683}?\)

Answer:

(a) The given sequence is 2, 2√2, 4, … 

Here, a = 2 and r = (2√2)/2 = √2

Let the n^th term of the given sequence be 128.

Thus, the 13^th term of the given sequence is 128.

(b) The given sequence is \(\sqrt3,3,3\sqrt3,...\)   

\(a = \sqrt3\) and r = \(\frac3{\sqrt3}\) =  \(\sqrt3\)

Let the n^th term of the given sequence be 729.

Thus, the 12^th term of the given sequence is 729.

(c) The given sequence is \(\frac13,\frac19,\frac1{27}...\)

Here, 

Let the n^th term of the given sequence be \(\frac1{19683}\) .

Thus, the 9^th term of the given sequence is \(\frac1{19683}\) .

38. For what values of x, the numbers \(\frac27,x,-\frac72\) are in G.P?

Answer:

The given numbers are  \(\frac{-2}7,x,\frac{-7}2\) 

Common ratio = \(\frac{x}{\frac{-2}{7}}= \frac{-7x}{2}\) 

Also, common ratio = \(\frac{\frac{-7}2}{x}=\frac{-7}{2x}\)

Thus, for x = ± 1, the given numbers will be in G.P.

39. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

Answer:

The given G.P. is 0.15, 0.015, 0.00015 … 

Here, a = 0.15 and \(r = \frac{0.015}{0.15}= 0.1 \)

40. Find the sum to n terms in the geometric progression √7 , √21, 3√7, …

Answer:

The given G.P. is √7 , √21, 3√7, … 

Here, a = √7 and = \(\frac{\sqrt{21}}{7}\) = √3

 41. Find the sum to n terms in the geometric progression

1, -a, a², -a³, ... (if a ≠ -1)

Answer:

The given G.P. is 1, -a, a², -a³, ...

Here, first term = a₁ = 1 

Common ratio = r = – a

42. Find the sum to n terms in the geometric progression 

x³, x⁵, x⁷ ....(if x ≠ ± 1)

Answer:

The given G.P. is x³, x⁵, x⁷, ....

Here, a = x³ and r = x²

43. Evaluate \(\displaystyle\sum^{11}_{k=1}(2+3^k)\) 

Answer:

The terms of this sequence 3, 3² , 3³ … forms a G.P.

Substituting this value in equation (1), we obtain

Answer 5

44. The sum of first three terms of a G.P. is \(\frac{39}{10}\) and their product is 1. Find the common ratio and the terms.

Answer:

Let \(\frac ax, a, ar\) be the first three terms of the G.P.

From (2), we obtain 

a³ = 1 

⇒ a = 1 (Considering real roots only) 

Substituting a = 1 in equation (1), we obtain

Thus, the three terms of G.P. are \(\frac52,1\) and \(\frac25\) 

45. How many terms of G.P. 3, 3² , 3³ … are needed to give the sum 120?

Answer:

The given G.P. is 3, 3², 3³ … 

Let n terms of this G.P. be required to obtain the sum as 120.

\(S_n = \frac{a(1-r^n)}{1-r}\)

Here, a = 3 and r = 3

∴ n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.

46. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. 

Determine the first term, the common ratio and the sum to n terms of the G.P.

Answer:

Let the G.P. be a, ar, ar², ar³, … 

According to the given condition, 

a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128

⇒ a (1 + r + r²) = 16 ……………. (1) 

ar³(1 + r + r²) = 128 …………….. (2)

Dividing equation (2) by (1), we obtain

Substituting r = 2 in (1), we obtain 

a (1 + 2 + 4) = 16 

⇒ a (7) = 16

47. Given a G.P. with a = 729 and 7^th term 64, determine S₇.

Answer:

a = 729 a₇ = 64 

Let r be the common ratio of the G.P. It is known that, 

an = a r^n–1 

a₇ = ar^7–1 = (729)r⁶ 

⇒ 64 = 729 r⁶

Also, it is known that,

\(S_n = \frac{a(1-r^n)}{1-r}\) 

 48. Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.

Answer:

Let a be the first term and r be the common ratio of the G.P. 

According to the given conditions,

\(S_2=-4 = \frac{a(1-r^2)}{1-r}\)         ....(1)

a₅ = 4 × a₃ 

⇒ ar⁴ = 4ar² 

⇒ r² = 4 

∴ r = ± 2 

From (1), we obtain

Thus, the required G.P. is \(\frac{-4}{3},\frac{-8}{3},\frac{-16}{3},...\) or 4, –8, 16, –32 …

49. If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Answer:

Let a be the first term and r be the common ratio of the G.P. 

According to the given condition,

a₄ = a r³ = x ……………… (1)

a₁₀ = a r⁹ = y ……………… (2)

a₁₆ = a r¹⁵ = z ……………… (3)

Dividing (2) by (1), we obtain

Dividing (3) by (2), we obtain

Thus, x, y, z are in G. P.

50. Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Answer:

The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. 

However, it can be changed to G.P. by writing the terms as Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

51. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

Answer:

Required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32  x  \(\frac12\)

= \(64\left[4 + 2+1+\frac12+\frac1{2^2}\right]\)

Here, 4, 2, 1, \(\frac12,\frac1{2^2}\) is a G.P.

First term, a = 4

Common ratio, r = \(\frac12\)

It is known that,

\(S_n = \frac{a(1-r^n)}{1-r}\) 

∴Required sum = \(64\left(\frac{31}4\right)= (16)(31)= 496\)

52. Show that the products of the corresponding terms of the sequences form a, ar, ar² ..... ar^n-1 and A , AR, AR², ...AR^n-1 a G.P, and find the common ratio.

Answer:

It has to be proved that the sequence: aA, arAR, ar²AR² , …ar^n–1AR^n–1 , forms a G.P.

Thus, the above sequence forms a G.P. and the common ratio is rR.

53. Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.

Answer:

Let a be the first term and r be the common ratio of the G.P. 

a₁ = a, a₂ = ar, a₃ = ar₂ , a₄ = ar₃

By the given condition, 

a₃ = a₁ + 9 ⇒ ar² = a + 9 ……………. (1) 

a₂ = a₄ + 18 ⇒ ar = ar³ + 18 ………… (2)

From (1) and (2), we obtain 

a(r² – 1) = 9 …………………….. (3) 

ar (1– r²) = 18 …………………. (4) 

Dividing (4) by (3), we obtain

Substituting the value of r in (1), we obtain

 4a = a + 9 

⇒ 3a = 9 

∴ a = 3

Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)² , and 3(–2)³

i.e., 3¸–6, 12, and –24.

Answer 6

54. If p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that a^{q – r} .b^{r – p} .c^{p – q} = 1.

Answer:

Let A be the first term and R be the common ratio of the G.P. 

According to the given information, 

AR^p–1 = a 

AR^q–1 = b 

AR^r–1 = c

a^{q – r} .b^{r – p} .c^{p – q} 

= A^q–r × R^{(p–1)(q–r)} × A^r–p × R^(q–1)(r-p) × A^p–q × R^{(r –1)(p–q)} 

= A^{q – r + r – p + p – q} × R^{(pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)}

= A⁰ × R⁰ 

= 1 

Thus, the given result is proved.

55. If the first and the n^th term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.

Answer:

The first term of the G.P is a and the last term is b. 

Therefore, the G.P. is a, ar, ar² , ar³ … ar^n–1, where r is the common ratio. 

b = ar^n–1 ………………. (1) 

P = Product of n terms

= (a) (ar) (ar²) … (ar^n–1) 

= (a × a ×…a) (r × r² × …r^n–1) 

= an r 1 + 2 +…(n–1) … (2) 

Here, 1, 2, …(n – 1) is an A.P. 

\(\therefore\) 1 + 2 + ……….+ (n – 1)

Thus, the given result is proved.

56. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)^th to (2n)^th term is 1/rⁿ . 

Answer:

Let a be the first term and r be the common ratio of the G.P.

Sum of first n terms = \(\frac{a(1-r^n)}{(1-r)}\)

Since there are n terms from (n +1)^th to (2n)^th term, 

Sum of terms from(n + 1)^th to (2n)^th term

\(S_n=\frac{a_{n+1}(1-r^n)}{1-r}\)

Thus, required ratio = 

Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)^th to (2n)^th term is 1/rⁿ.

56. If a, b, c and d are in G.P. show that: (a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

Answer:

a, b, c, d are in G.P. 

Therefore,

bc = ad ……………… (1) 

b² = ac ……………… (2) 

c² = bd ……………… (3) 

It has to be proved that, 

(a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

R.H.S. =

 (ab + bc + cd)² 

= (ab + ad + cd)²         [Using (1)] 

= [ab + d (a + c)]² 

= a²b² + 2abd (a + c) + d² (a + c)² 

= a²b² +2a²bd + 2acbd + d²(a² + 2ac + c²)

= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c²        [Using (1) and (2)] 

= a²b² + a²c² + a²c² + b²c² + b²c² + d²a² + d²b² + d²b² + d²c² 

= a²b² + a²c² + a²d² + b² × b² + b²c² + b²d² + c²b² + c² × c² + c²d²

[Using (2) and (3) and rearranging terms]

= a²(b² + c² + d²) + b² (b² + c² + d²) + c² (b²+ c² + d²) 

= (a² + b² + c²) (b² + c² + d²) = L.H.S. 

∴ L.H.S. = R.H.S. 

∴ (a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

57. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Answer:

Let G₁ and G₂ be two numbers between 3 and 81 such that the series, 3, G₁, G₂, 81, forms a G.P. 

Let a be the first term and r be the common ratio of the G.P. 

∴ 81 = (3) (r)³ 

⇒ r³ = 27 

∴ r = 3 (Taking real roots only) 

For r = 3, 

G₁ = ar = (3) (3) = 9 

G₂ = ar² = (3) (3)² = 27 

Thus, the required two numbers are 9 and 27.

58. Find the value of n so that \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\) may be the geometric mean between a and b.

Answer:

M. of a and b is \(\sqrt{ab}\)

By the given condition: 

\(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}= \sqrt{ab}\) 

Squaring both sides, we obtain

59. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 - 2√2)

Answer:

Let the two numbers be a and b. 

G.M. = \(\sqrt{ab}\)

According to the given condition,

a + b = 6\(\sqrt{ab}\)           ...........(1)

⇒ (a + b)² = 36(ab)

Also,

Adding (1) and (2), we obtain

Substituting the value of a in (1), we obtain

Thus, the required ratio is (3 + 2√2) : (3 - 2√2)

Answer 7

60. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are \(A \pm\sqrt{(A+G)(A-G)}\) 

Answer:

It is given that A and G are A.M. and G.M. between two positive numbers. L

et these two positive numbers be a and b.

∴ AM = A = \(\frac{a+b}2\)     .......(1)

GM = G = \(\sqrt{ab}\)       .......(2)

From (1) and (2), we obtain 

a + b = 2A …………… (3) 

ab = G2 ……………… (4) 

Substituting the value of a and b from (3) and (4) in the identity

(a – b)² = (a + b)² – 4ab, 

we obtain 

(a – b)² = 4A² – 4G² = 4 (A²–G²) 

(a – b)² = 4 (A + G) (A – G)

(a – b) = \(2\sqrt{(A+G)(A-G)}\)     .........(5)

From (3) and (5), we obtain

2a = 2A + \(2\sqrt{(A+G)(A-G)}\)

⇒ a = A + \( \sqrt{(A+G)(A-G)}\)

Substituting the value of a in (3), we obtain

b = 2A - A - \( \sqrt{(A+G)(A-G)}\) = A - \( \sqrt{(A+G)(A-G)}\)

Thus, the two numbers are \(A \pm\sqrt{(A+G)(A-G)}\) 

61. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour, 4^th hour and n^th hour?

Answer:

It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P. 

Here, a = 30 and r = 2 a₃ = ar² = (30) (2)² = 120 

Therefore, the number of bacteria at the end of 2^nd hour will be 120. 

a₅ = ar⁴ = (30) (2)⁴ = 480

The number of bacteria at the end of 4th hour will be 480. 

a_{n +1} = arⁿ = (30) 2ⁿ 

Thus, number of bacteria at the end of n^th hour will be 30(2)ⁿ.

62. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Answer:

The amount deposited in the bank is Rs 500. 

At the end of first year, amount = \(Rs\,500\left(1+\frac1{10}\right)\) = Rs 500 (1.1)

At the end of 2^nd year, amount = Rs 500 (1.1) (1.1) 

At the end of 3^rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on 

Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times) = Rs 500(1.1)¹⁰

63. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Answer:

Let the root of the quadratic equation be a and b. 

According to the given condition,

The quadratic equation is given by, 

x²– x (Sum of roots) + (Product of roots) = 0 

x² – x (a + b) + (ab) = 0 

x² – 16x + 25 = 0 [Using (1) and (2)] 

Thus, the required quadratic equation is x² – 16x + 25 = 0

64. Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Answer:

The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … n^th term, 

a_n = n (n + 1)

65. Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Answer:

The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n^th term, 

a_n = n (n + 1) (n + 2) 

= (n² + n) (n + 2)

= n³ + 3n² + 2n

66. Find the sum to n terms of the series 3 × 1² + 5 × 2² + 7 × 32 + …

Answer:

The given series is 3 ×1² + 5 × 2² + 7 × 3² + … n^th term, 

a_n = (2n + 1) n² = 2n³ + n²

67. Find the sum to n terms of the series \(\frac1{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...\) 

Answer:

The given series is \(\frac1{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...\) 

n^th term, a_n = \(\frac{1}{n(n+1)}=\frac1n-\frac{1}{n+1}\)            (By partial fractions)

Adding the above terms column wise, we obtain

68. Find the sum to n terms of the series 5² + 6² + 7² +....+20²

Answer:

The given series is 5² + 6² + 7² + … + 20² n^th term, 

a_n = ( n + 4)² = n² + 8n + 16

16^th term is (16 + 4)² = 20²

69. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Answer:

The given series is 3 × 8 + 6 × 11 + 9 × 14 + … an 

= (n^th term of 3, 6, 9 …) × (n^th term of 8, 11, 14 …) 

= (3n) (3n + 5) 

= 9n² + 15n

70. Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + …

Answer:

The given series is 1² + (1² + 2²) + (1² + 2² + 3³) + … a_n 

= (1² + 2² + 3³ +…….+ n²)

71. Find the sum to n terms of the series whose n th term is given by n (n + 1) (n + 4).

Answer:

a_n = n (n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n

Answer 8

72. Find the sum to n terms of the series whose n^th terms is given by n² + 2ⁿ

Answer:

a_n = n² + 2ⁿ

Consider

The above series 2, 2², 2³ … is a G.P. with both the first term and common ratio equal to 2.

Therefore, from (1) and (2), we obtain

73. Find the sum to n terms of the series whose n^th terms is given by (2n – 1)²

Answer:

a_n = (2n – 1)² = 4n² – 4n + 1

74. Show that the sum of (m + n)^th and (m – n)^th terms of an A.P. is equal to twice the m^th term.

Answer:

Let a and d be the first term and the common difference of the A.P. respectively. It is known that the k^th term of an A. P. is given by 

a_k = a + (k –1) d

∴ a_{m + n} = a + (m + n –1) d 

a_{m – n} = a + (m – n –1) d 

a_m = a + (m –1) d

∴ a_{m + n} + a_{m – n} = a + (m + n –1) d + a + (m – n –1) d 

= 2a + (m + n –1 + m – n –1) d 

= 2a + (2m – 2) d 

= 2a + 2 (m – 1) d 

=2 [a + (m – 1) d] 

= 2a_m

Thus, the sum of (m + n)^th and (m – n)^th terms of an A.P. is equal to twice the m^th term.

75. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Answer:

Let the three numbers in A.P. be a – d, a, and a + d. 

According to the given information,

(a – d) + (a) + (a + d) = 24 … (1) 

⇒ 3a = 24 

∴ a = 8

(a – d) a (a + d) = 440 … (2) 

⇒ (8 – d) (8) (8 + d) = 440 

⇒ (8 – d) (8 + d) = 55 

⇒ 64 – d² = 55

⇒ d² = 64 – 55 = 9 

⇒ d = ± 3

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5. 

Thus, the three numbers are 5, 8, and 11.

76. Let the sum of n, 2n, 3n terms of an A.P. be S₁, S₂ and S₃, respectively, show that S₃ = 3 (S₂– S₁)

Answer:

Let a and b be the first term and the common difference of the A.P. respectively. Therefore,

From (1) and (2), we obtain

Hence, the given result is proved.

77. Find the sum of all numbers between 200 and 400 which are divisible by 7.

Answer:

The numbers lying between 200 and 400, which are divisible by 7, are 203, 210, 217 … 399

∴ First term, a = 203 

Last term, l = 399 

Common difference, d = 7

Let the number of terms of the A.P. be n. 

∴ an = 399 = a + (n –1) d 

⇒ 399 = 203 + (n –1) 7 

⇒ 7 (n –1) = 196 

⇒ n –1 = 28 

⇒ n = 29

Thus, the required sum is 8729.

78. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer:

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100. 

This forms an A.P. with both the first term and common difference equal to 2.

⇒100 = 2 + (n –1) 2 

⇒ n = 50

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100. 

This forms an A.P. with both the first term and common difference equal to 5.

∴ 100 = 5 + (n –1) 5 

⇒ 5n = 100 

⇒ n = 20

The integers, which are divisible by both 2 and 5, are 10, 20, … 100. 

This also forms an A.P. with both the first term and common difference equal to 10. 

∴ 100 = 10 + (n –1) (10) 

⇒ 100 = 10n 

⇒ n = 10

∴ Required sum = 2550 + 1050 – 550 = 3050 

Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

79. Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer:

The two-digit numbers, which when divided by 4, yield 1 as remainder, are 13, 17, … 97. 

This series forms an A.P. with first term 13 and common difference 4. 

Let n be the number of terms of the A.P. 

It is known that the n^th term of an A.P. is given by, a_n = a + (n –1) d 

∴ 97 = 13 + (n –1) (4) 

⇒ 4 (n –1) = 84 

⇒ n – 1 = 21 

⇒ n = 22

Sum of n terms of an A.P. is given by,

Thus, the required sum is 1210.

80. If f is a function satisfying f(x + y) = f(x).f(y) for all x, y ∈ N, such that f(1) = 3 and \(\sum^n_1f(x)\) = 120, find the value of n.

Answer:

It is given that, 

f (x + y) = f (x) × f (y) for all x, y ∈ N ………………. (1) 

f (1) = 3 

Taking x = y = 1 in (1), 

we obtain f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly, 

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27 

f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81 

∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.

It is known that,

\(S_n = \frac{a(r^n-1)}{r-1}\)

It is given that,

Thus, the value of n is 4.

Answer 9

81. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Answer:

Let the sum of n terms of the G.P. be 315.

It is known that,

\(S_n = \frac{a(r^n-1)}{r-1}\)

It is given that the first term a is 5 and common ratio r is 2.

∴ Last term of the G.P = 6^th term = ar^{6 – 1} = (5)(2)⁵ = (5)(32) = 160 

Thus, the last term of the G.P. is 160.

82. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Answer:

Let a and r be the first term and the common ratio of the G.P. respectively. 

a = 1   

a₃ = ar² = r² 

a₅ = ar⁴ = r⁴

∴ r² + r⁴ = 90 

⇒ r⁴ + r² – 90 = 0

Thus, the common ratio of the G.P. is ±3.

83. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Answer:

Let the three numbers in G.P. be a, ar, and ar² . 

From the given condition, 

a + ar + ar² = 56 

⇒ a (1 + r + r²) = 56 ……………… (1)

a – 1, ar – 7, ar² – 21 forms an A.P. 

∴ (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7) 

⇒ ar – a – 6 = ar² – ar – 14 

⇒ ar² – 2ar + a = 8 

⇒ ar² – ar – ar + a = 8 

⇒ a(r² + 1 – 2r) = 8 

⇒ a (r – 1)² = 8 ……………… (2)

From (1) and (2), we get 

⇒7(r² – 2r + 1) = 1 + r + r² 

⇒ 7r² – 14 r + 7 – 1 – r – r² = 0 

⇒ 6r² – 15r + 6 = 0 

⇒ 6r² – 12r – 3r + 6 = 0 

⇒ 6r (r – 2) – 3 (r – 2) = 0 

⇒ (6r – 3) (r – 2) = 0

When r = 2, a = 8 

When 

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32. 

When, r=1/2, the three numbers in G.P. are 32, 16, and 8. 

Thus, in either case, the three required numbers are 8, 16, and 32.

84. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Answer:

Let the G.P. be T₁, T₂, T₃, T₄ … T_2n. 

Number of terms = 2n 

According to the given condition,

T₁ + T₂ + T₃ + …+ T_2n = 5 [T₁ + T₃ + … +T_2n–1] 

⇒ T₁ + T₂ + T₃ + … + T_2n – 5 [T₁ + T₃ + … + T_2n–1] = 0 

⇒ T₂ + T₄ + … + T_2n = 4 [T₁ + T₃ + … + T_2n–1]

Let the G.P. be a, ar, ar² , ar³ …

Thus, the common ratio of the G.P. is 4.

85. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Answer:

Let the A.P. be a, a + d, a + 2d, a + 3d ... a + (n – 2) d, a + (n – 1)d. 

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms 

= [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d] 

= 4a + (4n – 10) d

According to the given condition, 

4a + 6d = 56 

⇒ 4(11) + 6d = 56       [Since a = 11 (given)] 

⇒ 6d = 12 ⇒ d = 2 

∴ 4a + (4n –10) d = 112 

⇒ 4(11) + (4n – 10)2 = 112 

⇒ (4n – 10)2 = 68 

⇒ 4n – 10 = 34 

⇒ 4n = 44 

⇒ n = 11 

Thus, the number of terms of the A.P. is 11.

86. If \(\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}(x\ne0)\) then show that a, b, c and d are in G.P.

Answer:

It is given that,

From (1) and (2), we obtain

\(\frac ba = \frac cb=\frac dc\)

Thus, a, b, c, and d are in G.P.

87. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ

Answer:

Let the G.P. be a, ar, ar² , ar³ … ar^{n – 1} 

According to the given information,

Hence, P² Rⁿ = Sⁿ

88. The p^th , q^th and r^th terms of an A.P. are a, b, c respectively. Show that (q - r)a + (r - p)b + (p-q)c = 0

Answer:

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

The n^th term of an A.P. is given by, a_n = t + (n – 1)d 

Therefore, 

a_p = t + (p – 1) d = a ………………. (1) 

a_q = t + (q – 1)d = b ………………. (2) 

a_r = t + (r – 1) d = c ……………….. (3)

Subtracting equation (2) from (1), we obtain 

(p – 1 – q + 1) d = a – b 

⇒ (p – q) d = a – b

\(\therefore d = \frac{a-b}{p-q}\)  ............................... (4)

Subtracting equation (3) from (2), we obtain 

(q – 1 – r + 1) d = b – c 

⇒ (q – r) d = b – c

⇒ \(d = \frac{b-c}{q-r}\)   .......................... (5)

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

Thus, the given result is proved.

89. If  \(a(\frac{1}{b}+\frac1c),b(\frac1c+\frac1a),c(\frac1a+\frac1b)\) are in A.P., prove that a, b, c are in A.P.

Answer:

It is given that \(a(\frac{1}{b}+\frac1c),b(\frac1c+\frac1a),c(\frac1a+\frac1b)\) are in A.P.

Thus, a, b, and c are in A.P.

Answer 10

90. If a, b, c, d are in G.P, prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

Answer:

It is given that a, b, c, and d are in G.P. 

\(\therefore\) b² = ac ………………… (1) 

\(\therefore\) c² = bd …………………. (2)

ad = bc …………………… (3) 

It has to be proved that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P. i.e., 

(bⁿ + cⁿ)² = (aⁿ + bⁿ) (cⁿ + dⁿ) 

Consider L.H.S.

(bⁿ + cⁿ)² = b²ⁿ + 2bⁿcⁿ + c²ⁿ 

= (b²)ⁿ+ 2bⁿcⁿ + (c²)ⁿ 

= (ac)ⁿ + 2bⁿcⁿ + (bd)ⁿ     [Using (1) and (2)] 

= aⁿcⁿ + bⁿcⁿ+ bⁿ cⁿ + bⁿ dⁿ

= aⁿ cⁿ + bⁿcⁿ+ aⁿ dⁿ + bⁿ dⁿ     [Using (3)] 

= cⁿ (aⁿ + bⁿ) + dⁿ (aⁿ + bⁿ) 

= (aⁿ + bⁿ) (cⁿ + dⁿ) = R.H.S. 

(bⁿ + cⁿ) 2 = (aⁿ + bⁿ) (cⁿ + dⁿ) 

Thus, (aⁿ + bⁿ), (bⁿ + cⁿ), and (cⁿ + dⁿ) are in G.P.

91. If a and b are the roots of x² - 3x + p = 0 and c,d are roots of x² - 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15.

Answer:

It is given that a and b are the roots of x² – 3x + p = 0 

∴ a + b = 3 and ab = p ……………………… (1)

Also, c and d are the roots of x² - 12x + q = 0

∴ c + d = 12 and cd = q ……………………… (2)

It is given that a, b, c, d are in G.P. 

Let a = x, b = xr, c = xr² , d = xr³ 

From (1) and (2), we obtain 

x + xr = 3 

⇒ x (1 + r) = 3 

xr² + xr³ =12 

⇒ xr² (1 + r) = 12 

On dividing, we obtain

Case I: 

When r = 2 and x =1, 

ab = x²r = 2 

cd = x²r⁵ = 32

Case II: 

When r = –2, x = –3, 

ab = x²r = –18 

cd = x²r⁵ = – 288

Thus, in both the cases, we obtain (q + p): (q – p) = 17:15

 92. The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that

\(a:b=(m+\sqrt{m^2-n^2}):(m -\sqrt{m^2-n^2})\)

Answer:

Let the two numbers be a and b.

A.M = \(\frac{a+b}2\) and G,M = \(\sqrt{ab}\) 

According to the given condition,

Using this in the identity (a – b)² = (a + b)² – 4ab, we obtain

Adding (1) and (2), we obtain

Substituting the value of a in (1), we obtain

93. If a, b, c are in A.P; b, c, d are in G.P and \(\frac1c,\frac1d,\frac1e\) are in A.P. prove that a, c, e are in G.P.

Answer:

It is given that a, b, c are in A.P. 

\(\therefore\) b – a = c – b ……………….. (1) 

It is given that b, c, d, are in G.P. 

\(\therefore\) c² = bd …………………… (2)

Also, \(\frac1c,\frac1d,\frac1e\) are in A.P.

It has to be proved that a, c, e are in G.P. i.e., c² = ae 

From (1), we obtain

2b = a + c

⇒ b = \(\frac{a+c}{2}\)

From (2), we obtain

d = \(\frac{c^2}{b}\)

Substituting these values in (3), we obtain

Thus, a, c, and e are in G.P.

94. Find the sum of the following series up to n terms: 

(i) 5 + 55 + 555 + … 

(ii) .6 +.66 +. 666 +…

Answer:

(i)5 + 55 + 555 + … 

Let S_n = 5 + 55 + 555 + ….. to n terms

(ii) .6 +.66 +. 666 +… 

Let S_n = 06. + 0.66 + 0.666 + … to n terms

95. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Answer:

The given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms 

\(\therefore\) n^th term = a_n = 2n × (2n + 2) = 4n² + 4n 

a₂₀ = 4 (20)² + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680 

Thus, the 20^th term of the series is 1680.

96. Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

Answer:

The given series is 3 + 7 + 13 + 21 + 31 + … 

S = 3 + 7 + 13 + 21 + 31 + …+ a_n–1 + a_n 

S = 3 + 7 + 13 + 21 + …. + a_{n – 2} + a_{n – 1} + a_n 

On subtracting both the equations, we obtain

S – S = [3 + (7 + 13 + 21 + 31 + …+ a_n–1 + a_n)] – [(3 + 7 + 13 + 21 + 31 + …+ a_n–1) + a_n]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – a_n

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – a_n

a_n = 3 + [4 + 6 + 8 + … (n –1) terms]

97. If S₁, S₂, S₃ are the sum of first n natural numbers, their squares and their cubes, respectively, show that \(9S^2_2=S_3(1+8S_1)\)

Answer:

From the given information,

Thus, from (1) and (2), we obtain \(9S^2_2=S_3(1+8S_1)\)