NCERT Solutions For Class 11 Maths Chapter 9

NCERT Solutions Class 11 Maths Sequences and Series

Ncert Solutions For Class 11 Maths Chapter 9 PDF Free Download

Class 11 is an important phase of a student’s life because the topics which are taught in class 11 are basics of the topics which will be taught in class 12. Students studying in class 11 should try to understand the chapters in a better way so that they don’t get confused while they will be learning the tough topics in class 12. NCERT Solutions for class 11 maths chapter 9 sequence and series are provided here so that students can refer to these solutions when they are facing difficulties to solve the mathematical problems. The NCERT chapter 9 Solutions for class 11 maths deals with the topics of Sequences and Series. The various problems that are asked during this topic are finding out the different problems related to the topics of sequences and series. Figuring out the initial terms of the sequences is important to understand, as it is the most commonly asked for problem type in Sequences and Series. The NCERT Solutions for Class 11 maths deals with the various topics in mathematics.

NCERT Solutions Class 11 Maths Chapter 9 Exercises

Exercise 9.1

Q1: as = s (s + 3) is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = s (s + 3)

Putting s = 1, 2, 3, 4 and 5 respectively, in as = s (s + 3)

a1 = 1 (1 + 3) = 4

a2 = 2 (2 + 3) = 10

a3 = 3 (3 + 3) = 18

a4 = 4 (4 + 3) = 28

a5 = 5 (5 + 3) = 40

The starting five terms of the sequence are 4, 10, 18, 28 and 40

 

Q2: as = \(\frac{s}{s + 2}\) is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = \(\frac{s}{s + 2}\)

Putting s = 1, 2, 3, 4 and 5 respectively, in \(\frac{s}{s + 2}\)

a1 = \(\frac{1}{1 + 2}\) = \(\frac{1}{3}\)

a2 = \(\frac{2}{2 + 2}\) = \(\frac{2}{4}\)

a3 = \(\frac{3}{3 + 2}\) = \(\frac{3}{5}\)

a4 = \(\frac{4}{4 + 2}\) = \(\frac{4}{6}\)

a5 = \(\frac{5}{5 + 2}\) = \(\frac{5}{7}\)

The starting five terms of the sequence are \(\frac{1}{2},\; \frac{2}{3},\; \frac{3}{4},\; \frac{4}{5}\; and\; \frac{5}{6}\)

 

Q3: as = 5s is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = 5s

Putting s = 1, 2, 3, 4 and 5 respectively, in as = 5s

a1 = 51 = 5

a2 = 52 = 25

a3 = 53 = 125

a4 = 54 = 625

a5 = 55 = 3125

The starting five terms of the sequence are 5, 25, 125, 625 and 3125.

 

Q4: as = \(\frac{2s + 2}{3}\) is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = \(\frac{2s + 2}{3}\)

Putting s = 1, 2, 3, 4 and 5 respectively, in \(\frac{2s + 2}{3}\)

a1 = \(\frac{2.1 + 2}{3} = \frac{4}{3}\)

a2 = \(\frac{2.2 + 2}{3} = \frac{6}{3}\)

a3 = \(\frac{2.3 + 2}{3} = \frac{8}{3}\)

a4 = \(\frac{2.4 + 2}{3} = \frac{10}{3}\)

a5 = \(\frac{2.5 + 2}{3} = \frac{12}{3}\)

The starting five terms of the sequence are \(\frac{4}{3},\; \frac{6}{3},\; \frac{8}{3},\; \frac{10}{3}\; and\; \frac{12}{3}\)

 

Q5: as = \((- 1)^{s – 1} + 3^{s + 1}\) is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = \((- 1)^{s – 1} + 3^{s + 1}\)

Putting s = 1, 2, 3, 4 and 5 respectively, in \((- 1)^{s – 1} + 3^{s + 1}\)

a1 = \((- 1)^{1 – 1} + 3^{1 + 1}\) = 1 + 9 = 10

a2 = \((- 1)^{2 – 1} + 3^{2 + 1}\) = – 1 + 27 = 26

a3 = \((- 1)^{3 – 1} + 3^{3 + 1}\) = 1 + 81 = 82

a4 = \((- 1)^{4 – 1} + 3^{4 + 1}\) = – 1 + 243 = 242

a5 = \((- 1)^{5 – 1} + 3^{5 + 1}\) = 1 + 729 = 730

The starting five terms of the sequence are 10, 26, 82, 242 and 730.

 

Q6: as = \(\frac{(- 1)^{s – 1}}{3^{s + 1}}\) is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = \(\frac{(- 1)^{s – 1}}{3^{s + 1}}\)

Putting s = 1, 2, 3, 4 and 5 respectively, in \(\frac{(- 1)^{s – 1}}{3^{s + 1}}\)

a1 = \(\frac{(- 1)^{1 – 1}}{3^{1 + 1}} = \frac{(- 1)^{0}}{3^{2}} = \frac{1}{9}\)

a2 = \(\frac{(- 1)^{2 – 1}}{3^{2 + 1}} = \frac{(- 1)^{1}}{3^{3}} = \frac{- 1}{27}\)

a3 = \(\frac{(- 1)^{3 – 1}}{3^{3 + 1}} = \frac{(- 1)^{2}}{3^{4}} = \frac{1}{81}\)

a4 = \(\frac{(- 1)^{4 – 1}}{3^{4 + 1}} = \frac{(- 1)^{3}}{3^{5}} = \frac{- 1}{243}\)

a5 = \(\frac{(- 1)^{5 – 1}}{3^{5 + 1}} = \frac{(- 1)^{4}}{3^{6}} = \frac{1}{729}\)

The starting five terms of the sequence are \(\frac{1}{9},\; \frac{- 1}{27},\; \frac{1}{81},\; \frac{- 1}{243},\; and\; \frac{1}{729}\)

 

Q7: as = s + 3s is the sth term of a sequence. Obtain the 17th and 22nd term in the required sequence.

Answer:

as = s + 3s

Putting s = 17 and 22 respectively, in as = s + 3s

a17 = 17 + 317

a22 = 22 + 322

 

Q8: as = \(\frac{9 + s}{s^{2}}\) is the sth term of a sequence. Obtain the 24th term in the required sequence.

Answer:

as = \(\frac{9 + s}{s^{2}}\)

Putting s = 24, in as = \(\frac{9 + s}{s^{2}}\)

a24 = \(\frac{9 + 24}{24^{2}} = \frac{33}{576}\)

 

Q9: as = \((- 1)^{s} \times 6^{s}\) is the sth term of a sequence. Obtain the 11th and 12th term in the required sequence.

Answer:

as = \((- 1)^{s} \times 6^{s}\)

Putting s = 11 and 12, in as = \((- 1)^{s} \times 6^{s}\)

a11 = \((- 1)^{11} \times 6^{11} = – 6^{11}\)

a12 =  \((- 1)^{12} \times 6^{12} = 6^{12}\)

 

Q10. as = \(\frac{s + 3}{2 + s}\) is the sth term of a sequence. Obtain the 21st and 82nd term in the required sequence.

Answer:

as = \(\frac{s + 3}{2 + s}\)

Putting s = 21 and 82, in as = \(\frac{s + 3}{2 + s}\)

\(a_{21} = \frac{21 + 3}{2 + 21} = \frac{24}{23}\)

\(a_{82} = \frac{82 + 3}{2 + 82} = \frac{85}{84}\)

 

Q11. Find out the starting five terms and also the following series of the required sequence given below:

a1 = 2, as = 2 as – 1 for all s > 1

Answer:

a1 = 2, as = 2 as – 1 for all s > 1

a2 = 2 a1 = 2 x 2 = 4

a3 = 2 a2 = 2 x 4 = 8

a4 = 2 a3 = 2 x 8 = 16

a5 = 2 a4 = 2 x 16 = 32

The starting five terms of the sequence are 2, 4, 8, 16 and 32

The following series is 2 + 4 + 8 + 16 + 32 + …….

 

Q12. Find out the starting five terms and also the following series of the required sequence given below:

a1 = 2, as = \(\frac{a_{s – 1} – 1}{2}\) for all s > 3

Answer:

a1 = 2, as = \(\frac{a_{s – 1} – 1}{2}\) for all s > 3

\(\frac{a_{2 – 1} – 1}{2} = \frac{a_{1} – 1}{2} = \frac{2 – 1}{2} = \frac{1}{2} \\ a_{3} = \frac{a_{3 – 1} – 1}{2} = \frac{a_{2} – 1}{2} = \frac{\frac{1}{2} – 1}{2} = – \frac{1}{4} \\ a_{4} = \frac{a_{4 – 1} – 1}{2} = \frac{a_{3} – 1}{2} = \frac{- \frac{1}{4} – 1}{2} = – \frac{5}{8} \\ a_{5} = \frac{a_{5 – 1} – 1}{2} = \frac{a_{4} – 1}{2} = \frac{- \frac{5}{8} – 1}{2} = – \frac{13}{16} \\\)

The starting five terms of the sequence are \(2,\; \frac{1}{2},\; – \frac{1}{4}, \; – \frac{5}{8}\; and\; – \frac{13}{16}\)

The following series is \(2\; + \frac{1}{2}\; + (- \frac{1}{4})\; + (- \frac{5}{8})\; +\; (- \frac{13}{16}) + ……\)

 

Q13.  Find out the starting five terms and also the following series of the required sequence given below:

Answer

a1 = a2 = 4, as = as – 1 + 1 for all s > 4

a3 = a 3 – 1 + 1 = a2 +1 = 4 + 1 = 5

a4 = a 4 – 1 + 1 = a3 + 1 = 5 + 1 = 6

a5 = a 5 – 1 + 1 = a4 + 1 = 6 + 1 = 7

The starting five terms of the sequence are 4, 4, 5, 6 and 7

The following series is 4 + 4 + 5 + 6 + 7 + ….

 

Q14. The sequence of Fibonacci is described as 1 = a1 = a2 and as = as – 1 + as – 2, n > 1. Obtain \(\frac{a_{s + 1}}{a_{s}}\)  for n = 2, 4.

Answer:

1 = a1 = a2 and as = as – 1 + as – 2, n > 1.

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

For s = 2 and 4 respectively,

\(\frac{a_{2 + 1}}{a_{2}} = \frac{a_{3}}{a_{2}} = \frac{2}{1} = 2\)

\(\frac{a_{4 + 1}}{a_{4}} = \frac{a_{5}}{a_{4}} = \frac{5}{3}\)

 

Exercise 9.2

Q1. Obtain the sum of all integers which are odd between 1 and 4001 including 1 and 4001.

Answer:

The integers which are odd between 1 and 2001 are 1, 3, 5, 7, 9, ……………. , 3997, 3999, 4001.

The sequence is in A.P form.

a = 1 [1st term]

Difference, d = 2

The standard equation of A.P,

a + (s – 1) d = 4001

1 + (s – 1) 2 = 4001

s = 2001

SS = \(\frac{s}{2} [2a + (s – 1) d] \\ S_{s} = \frac{2001}{2}[2.1 + (2001 – 1) 2] \\ = \frac{2001}{2}[2 + 4000] \\ = \frac{2001}{2} \times 4002 \\ = 2001 \times 2001 \\ = 4004001\)

Hence, the sum of all integers which are odd between 1 and 4001 is 4004001.

 

Q2. Obtain the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5.

Answer:

All natural numbers which are lying between 10 and 100, and are multiples of 5, are 15, 20, 25,  … , 95.

The sequence is in A.P form.

a = 15 [1st term]

Difference, d = 5

The standard equation of A.P,

a + (s – 1) d = 95

15 + (s – 1) 5 = 95

s = 17

SS = \(\frac{s}{2} [2a + (s – 1) d] \\ S_{s} = \frac{17}{2}[2.15 + (17 – 1) 5] \\ = \frac{17}{2}[30 + 80] \\ = \frac{17}{2} \times 110 \\ = 17 \times 55 \\ = 935\)

Hence, the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5 is 935.

 

Q3. Prove that the 21st term in the sequence is – 118 provided that the sequence is in A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms and 2 is the first term.

Answer:

Given:

2 is the first term, a = 2

The sequence is in A.P form,

In A. P = 2, 2 + d, 2 + 2d, 2 + 3d ……

In A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms,

So, according to the given condition,

10 + 10d = (1 / 4) (10 + 35d)

40 + 40d = 10 + 35d

d = – 6

a21 = a + (21 – 1) d = 2 + 20 (- 6) = – 118

Hence proved

 

Q4. Obtain the number of terms in A.P which are needed to get – 25 from the sum of – 6, (- 11 / 2), – 5, ….

Answer:

Suppose,

In, A.P the sum of s terms = – 25

a = – 6 [1st term]

Difference, d = (- 11 / 2) + 6 = (1 / 2)

\(\frac{s}{2} [2a + (s – 1) d]\)

\(- 25 = \frac{s}{2}[2.(- 6) + (n – 1) \frac{1}{2}] \\ – 50 = n [- 12 + \frac{n}{2} – \frac{1}{2}] \\ – 50 = n [\frac{- 25 + n}{2}] \\ – 100 = n (- 25 + n) \\ n^{2} – 25n + 100 = 0 \\ n^{2} – 5n – 20n + 100 = 0 \\ (n – 5)(n – 20) = 0 \\ n = 5 or 20\)

 

Q5. If m th term is 1 / n and nth term is 1 / m provided that the sequence is in A.P form, prove that the sum of the first mn terms is (1 / 2) (mn + 1), where m  n.

Answer:

Given,

m th term is 1 / n and nth term is 1 / m

So, acc. to the above condition

m th term in A.P form = am = a + (m – 1) d = 1 / m    ……. (1)

n th term in A.P form = an = a + (n – 1) d = 1 / n      ……. (2)

(2) – (1), we get,

(n – 1) d – (m – 1) d = \(\frac{1}{m} – \frac{1}{n} = \frac{n – m}{mn} \\\)

(n – m) d = \(\frac{n – m}{mn}\)

d = \(\frac{1}{mn}\)

Substituting the value of d in equation (1), we get,

a + (m – 1) \(\frac{1}{mn}\) = 1 / m

a = \(a = \frac{1}{n} – \frac{1}{n} + \frac{1}{mn} = \frac{1}{mn}\)

\(S_{mn} = \frac{mn}{2} [2a + (mn – 1) d] \\ S_{mn} = \frac{mn}{2} [2\frac{1}{mn} + (mn – 1) \frac{1}{mn}] \\ = 1 + \frac{1}{2}(mn – 1) \\ = \frac{1}{2}(mn + 1)\)

Hence proved

 

Q6. Obtain the last term if the addition of some numbers in A.P. 25, 22, 19, is 116.

Answer:

Addition of s terms in A.P be 116

Here, first term a = 25, d = – 3

\(S_{s} = \frac{s}{2} [2a + (s – 1) d] \\ 116 = \frac{s}{2}[2.(25) + (s – 1) (-3)] \\ 116 = s [50 – 3s + 3] \\ 232 = s (53 – 3s) = 53s – s^{2}\\ 3s^{2} – 53 n + 232 = 0 \\ 3s^{2} – 24 n – 29 n + 232 = 0 \\ (3s – 29)(n – 8) = 0 \\ n = 8 \;or\; n = (\frac{29}{3})\)

n = 8 is considered

a8 (last term) = a + (s – 1) d = 25 + 7 (- 3) = 25 – 21 = 4

Hence, the last term is 4

 

Q7. In A.P, obtain the total to s terms whose nth term is 5n + 1.

Answer:

Given,

In A.P, the total to s terms whose nth term is 5n + 1.

nth term = an = a + (n – 1) d

an = a + (n – 1) d = 5n + 1

a + nd – d = 5n + 1

By comparing the coefficient of n, we get

d = 5 and a – d = 1

a – 5 = 1

a = 6

\(S_{s} = \frac{s}{2} [2a + (s – 1) d] \\ = \frac{s}{2} [2.6 + (s – 1) d] \\ = \frac{s}{2} [12 + (s – 1) 5] \\ = \frac{s}{2} [12 + 5s – 5] \\ = \frac{s}{2} [5s + 7]\).

 

Q8. Obtain the common difference of a sequence in A.P, when the total of s terms is (ms + ns2), where m and n are constants.

Answer:

Given,

As mentioned in the given condition,

\(S_{s} = \frac{s}{2} [2a + (s – 1) d]\) = (ms + ns2)

\(\frac{s}{2} [2a + (s – 1) d] = ms + ns ^{2} \\ \frac{s}{2} [2a + sd – d] = ms + ns ^{2} \\ sa + s ^{2}\frac{d}{2} – s \frac{d}{2} = ms + ns ^{2} \\\)

Considering the coefficients of the of s2 , we get,

\(\frac{d}{2} = n\)

d = 2n

Hence, in A.P the difference d = 2n

 

Q9. The ratio of the total of s terms of two arithmetic progressions are 5s + 4 : 9 + 6 . Obtain the ratio of 18th term.

Answer:

Suppose the first terms are a1 and a2 respectively, and the common differences be d1 and d2 of the first two consecutive arithmetic progressions respectively.

As mentioned in the given condition,

\(\frac{Total\; addition\; of\; s\; terms\; of\; 1st\; A.P}{Total\; addition\; of\; s\; terms\; of\; 2nd\; A.P} = \frac{5 s + 4}{9 s + 6} \\ \frac{\frac{s}{2} [2a_{1} + (s – 1) d_{1}]}{\frac{s}{2} [2a_{2} + (s – 1) d_{2}] } = \frac{5 s + 4}{9 s + 6} \\ \frac{2a_{1} + (s – 1) d_{1}}{2a_{2} + (s – 1) d_{2}} = \frac{5 s + 4}{9 s + 6} \\ Putting\; s = 35 in (1),we get \\ \frac{2a_{1} + 34 d_{1}}{2a_{2} + 34 d_{2}} = \frac{5 (35) + 4}{9 (35) + 6} \\ \frac{a_{1} + 17 d_{1}}{a_{2} + 17 d_{2}} = \frac{179}{321} \\\)

18th term = \(a_{18} = \frac{18}{2} [2a + (18 – 1) d]\)

\(\frac{18^{th}\; term\; of\; 1^{st}\; A.P}{18^{th}\; term\; of\; 2^{nd}\; A.P} = a_{18} = \frac{18}{2} [2a + (18 – 1) d] = \frac{a_{1} + 17 d_{1}}{a_{2} + 17 d_{2}} = \frac{179}{321} \\\)

Hence, 179 : 321 is the required ratio of the 18th term.

 

Q10. In an A.P, the total of starting m terms is equal to the total of starting n terms. Obtain the total of starting (m + n) terms.

Answer:

m th term = Sm = \(\frac{m}{2} [2a + (m – 1) d]\)

n th term = Sn = \(\frac{n}{2} [2a + (n – 1) d]\)

As mentioned in the given condition,

\(\frac{m}{2} [2a + (m – 1) d] = \frac{n}{2} [2a + (n – 1) d] \\ m [2a + (m – 1) d] = n [2a + (n – 1) d] \\ 2am + (m – 1) md = 2an + (n – 1) nd \\ 2a (m – n) + d [m (m – 1) – n (n – 1)] = 0 \\ 2a (m – n) + d [m ^{2} – m – n ^{2} + n] = 0 \\ 2a (m – n) + d [(m + n) (m – n) – (m – n)] = 0 \\ 2a (m – n) + d [(m – n) (m + n – 1)] = 0 \\ 2a + d(m + n – 1) = 0 \\ d = \frac{- 2a}{m + n – 1} \\ S_{m + n} = \frac{m + n}{2} [2a + (m + n – 1) d] \\ S_{m + n} = \frac{m + n}{2} [2a + (m + n – 1) \frac{- 2a}{m + n – 1}] \\ S_{m + n} = \frac{m + n}{2} [2a – 2a] \\ S_{m + n} = 0\)

Hence, the total of starting (m + n) terms is 0.

 

Q11. . In an A.P, the total of starting m, n and o terms are p, q and r. Show that the \(\frac{p}{m}(n – o) + \frac{q}{n}(m – o) + \frac{r}{o}(m – n) = 0\).

Answer:

Given,

In an A.P, the total of starting m, n and o terms are p, q and r.

As mentioned in the given condition,

\(S_{m} = \frac{m}{2} [2a_{1} + (m – 1) d] = p \\ 2a_{1} + (m – 1) d = \frac{2p}{m} …… (i) \\ S_{n} = \frac{n}{2} [2a_{1} + (n – 1) d] = q \\ 2a_{1} + (n – 1) d = \frac{2q}{n} …… (ii) \\ S_{o} = \frac{o}{2} [2a_{1} + (o – 1) d] = r \\ 2a_{1} + (o – 1) d = \frac{2r}{o} …… (iii) \\ Subtract\; (i)\; -\; (ii)\;, we\; get\; \\ (m – 1) d – (n – 1) d = \frac{2p}{m} – \frac{2q}{n} \\ d (m – 1 – n + 1) = \frac{2pn – 2qm}{mn} \\ d (m – n) = \frac{2 (pn – qm)}{mn} \\ d = \frac{2 (pn – qm)}{mn(m – n)} …. (4) \\\)

\(Subtract\; (ii)\; -\; (iii)\;, we\; get\; \\ (n – 1) d – (o – 1) d = \frac{2q}{n} – \frac{2r}{o} \\ d (n – 1 – o + 1) = \frac{2qo – 2rn}{no} \\ d (n – o) = \frac{2 (qo – rn)}{no} \\ d = \frac{2 (qo – rn)}{no(n – o)} …. (5) \\\)

Now, equating the values of d in equation (4) and (5)

\(\frac{2 (pn – qm)}{mn(m – n)} = \frac{2 (qo – rn)}{no(n – o)} \\ no(n – o)(pn – qm) = mn(m – n)(qo – rn) \\ o(n – o)(pn – qm) = m(m – n)(qo – rn) \\ (n – o)(pno – qmo) = (m – n)(mqo – mrn) \\ Multiply\; both\; the\; sides\; by\; \frac{1}{mno},we\; get\;, \\ (\frac{p}{m} – \frac{q}{n})(n – o) = (\frac{q}{n} – \frac{r}{o})(m – n) \\ \frac{p}{m}(n – o) – \frac{q}{n}(n – o) = \frac{q}{n}(m – n) – \frac{r}{o}(m – n) \\ \frac{p}{m}(n – o) + \frac{q}{n}(n – o + m – n) + \frac{r}{o}(m – n) = 0 \\ \frac{p}{m}(n – o) + \frac{q}{n}(m – o) + \frac{r}{o}(m – n) = 0 \\\)

Hence proved

 

Q12. The ratio of the total of r and s terms of two arithmetic progressions is r2 : s2. Show that the ratio of r th and s th term is (2r – 1) : (2s – 1).

Answer:

Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.

As mentioned in the given condition,

\(\frac{Sum\; of\; r\; terms}{Sum\; of\; s\; terms} = \frac{r^{2}}{s^{2}} \\ \frac{\frac{r}{2} [2a + (r – 1) d]}{\frac{s}{2} [2a + (s – 1) d]} = \frac{r^{2}}{s^{2}} \\ \frac{2a + (r – 1) d}{2a + (s – 1) d} = \frac{r}{s} …..(i) \\ Substituting\; r = 2r – 1 and s = 2s – 1 in (i), we\; get, \\ \frac{2a + (2r – 1 – 1) d}{2a + (2s – 1 – 1) d} = \frac{2r – 1}{2s – 1} \\ \frac{2a + (2r – 2) d}{2a + (2s – 2) d} = \frac{2r – 1}{2s – 1} \\ \frac{a + (r – 1) d}{a + (s – 1) d} = \frac{2r – 1}{2s – 1} …. (ii)\\ \frac{r^{th} term of A.P}{s^{th} term of A.P} = \frac{a + (r – 1) d}{a + (s – 1) d} = \frac{2r – 1}{2s – 1} \\ \frac{r^{th} term of A.P}{s^{th} term of A.P} = \frac{2r – 1}{2s – 1}\)

Hence proved

 

Q13. In an A.P, the total of n terms is 3p 2 + 5p and its rth term is 164. Obtain the value of r.

Answer:

Given,

Sr = \(\frac{r}{2} [2a + (r – 1) d]\) = 164 …… (i)

Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.

As mentioned in the given condition,

p th term = Sp = \(\frac{p}{2} [2a + (p – 1) d]\) = 3 p 2 + 5p

As mentioned in the given condition,

\(pa + \frac{d}{2} p^{2} – \frac{d}{2} p = 3p^{2} + 5 p \\ \frac{d}{2} p^{2} + (a – \frac{d}{2}) p = 3p^{2} + 5 p \\ Equating, the\; coefficients\; of\; p^{2}\; on\; both\; the\; sides, we\; get\; \frac{d}{2} = 3 \\ d = 6 \\ Equating, the\; coefficients\; of\; p\; on\; both\; the\; sides, we\; get\; \\ a – \frac{d}{2} = 5 \\ a – 3 = 5 \\ a = 8\)

From equation (i), we get

8 + (r – 1) 6 = 164

(r – 1) 6 = 164 – 8

(r – 1) 6 = 156

(r – 1) = 26

r = 7

 

Q14. Place five numbers between 8 and 26 in a way such that the sequence results in an A.P form.

Answer:

Suppose Q1, Q2, Q 3, Q 4 and Q 5 be the five required numbers.

First term, a = 8,

Last term, p = 26

s = 7,

p = a + (s – 1) d

26 = 8 + (7 – 1) d

6 d = 18

d = 3

Q1 = a + d = 8 + 3 = 11

Q2 = a + 2 d = 8 + 2.3 = 14

Q 3 = a + 3 d = 8 + 3.3 = 17

Q= a + 4 d = 8 + 3.4 = 20

Q 5 = a + 5 d = 8 + 3. 5 = 23

Hence, Q1, Q2, Q 3, Q 4 and Q 5 is equal to 11, 14, 17, 20 and 23 are the required numbers in an A.P

 

Q15. Suppose in an A.M \(\frac{p^{m} + q^{m}}{p^{m – 1} + q^{m – 1}} \\\) lies between p and q terms, then obtain the value of m.

Answer:

A.M of p and q = \(\frac{p + q}{2}\)

As mentioned in the given condition,

\(\frac{p + q}{2} = \frac{p^{m} + q^{m}}{p^{m – 1} + q^{m – 1}} \\ (p + q)(p^{m – 1} + q^{m – 1}) = 2(p^{m} + q^{m}) \\ p^{m} + pq^{m – 1} + qp^{m – 1} + q^{m} = 2 p^{m} + 2 q^{m} \\ pq^{m – 1} + qp^{m – 1} = p^{m} + q^{m} \\ q^{m – 1} (p – q) = p^{m – 1} (p – q) \\ q^{m – 1} = p^{m – 1} \\ (\frac{p}{q})^{m – 1} = 1 = (\frac{p}{q})^{0} \\ m – 1 = 0 \\ m = 1\)

 

Q16. Insert n numbers between 1 and 31 in a way such that the sequence results in an A.P. The ratio is 5 : 9 of the 7th and the (n – 1)th term. Find the value of n.

Answer:

Suppose Q1, Q2, Q 3, Q 4.…..  Q m, be the required n numbers.

First term, a = 1,

Last term, p = 31

s = n + 2,

p = a + (s – 1) d

31 = 1 + (n + 2 – 1) d

30 = (n + 1) d

d = \(\frac{30}{(n + 1)}\)

Q1 = a + d

Q2 = a + 2 d

Q 3 = a + 3 d

Q= a + 4 d …….

Q 7 = a + 7 d

Q n – 1 = a + (n – 1) d

As mentioned in the given condition,

\(\frac{a + 7 d}{a + (n – 1) d} = \frac{5}{9} \\ \frac{1 + 7 (\frac{30}{n + 1})}{1 + (n – 1) (\frac{30}{n + 1})} = \frac{5}{9} \\ \frac{n + 1 + 7(30)}{n + 1 + 30 (n – 1)} = \frac{5}{9} \\ \frac{n + 211}{31m – 29} = \frac{5}{9} \\ 9n + 1899 = 155n – 145 \\ 155n – 9n = 1899 + 145 \\ 146 n = 2044 \\ n = 14\)

Hence, the value of n = 14

 

Q17. A boy will be repaying his loans and his first installment is Rs 100. What is the amount should he pay in the 30th installment if he increases the installment by Rs 5 every month?

Answer:

Given,

His first installment is Rs 100

His 2nd installment is Rs 105 and third installment is Rs 110 and so on

The sequence of money paid by the boy is in an A.P form every month is

100, 105, 110, 115 and so on …….

a = 100 [First term]

Common difference, d = 5

a 30 = a + (30 – 1) d

= 100 + 29 (5)

= 245

Hence, the amount should be paid by him in the 30th installment is Rs 245.

 

Q18.  In a polygon, 120o is the smallest angle and 5o is the difference between consecutive angles on the interior side. Obtain the number of sides the polygon has.

Answer:

Given,

120o is the smallest angle i.e., first term, a = 120

And 5o is the difference, i.e., d = 5

Total of all angles of a polygon having s sides = 180o (s – 2)

\(S_{s} = 180^{\circ} (s – 2) \\ \frac{s}{2} [2a + (s – 1) d = 180^{\circ} (s – 2) \\ s [240 + (s – 1) 5 = 360 (s – 2) \\ 240 s + s (s – 1) 5 = 360 (s – 2) \\ 240 s + 5 s^{2} – 5 s = 360 s – 720 \\ 5 s^{2} + 235 s – 360 s + 720 = 0 \\ 5 s^{2} – 125 s + 720 = 0 \\ s^{2} – 25 s + 144 = 0 \\ s^{2} – 16 s – 9 s + 144 = 0 \\ s (s – 16) – 9 (s – 16) = 0 \\ (s – 16) (s – 9) = 0 \\ s = 16 or s = 9\)

 

Exercise 9.3

Q1. Find 20th and nth term for the G.P \( \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, . . . . \)

Soln:

Given G.P = \( \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, . . . . \)

Here, first term = a = \( \frac{5}{2} \)

Common ratio = r = \( \frac{ \frac{5}{4} }{ \frac{5}{2}} = \frac{1}{2} \)

\( a_{20} = ar^{20 – 1} = \frac {5}{2} \left ( \frac{1}{2} \right )^{19} = \frac{5}{(2)(2)^{19}} = \frac{5}{(2)^{20}}\)

\( a_{n} = ar^{n – 1} = \frac {5}{2} \left ( \frac{1}{2} \right )^{n – 1} = \frac {5}{(2)(2)^{n – 1}} = \frac{5}{(2)^{n}}\)

 

Q2. Find 12th term for the G.P that has 8th term 192 and common ratio of 2.

Soln:

Given,

Common ratio = r = 2

Assume = first term  = a

\( a_{8} = ar^{8 – 1} = ar^{7} \Rightarrow ar^{7} = 192 a(2)^{7} = (2)^{6} (3) \)

\( \Rightarrow a = \frac{(2)^{6} \times 3}{(2)^{7}} =  \frac{3}{2} \)

\( a_{12} = ar^{12 – 1} = \left ( \frac{3}{2} \right ) (2)^{11} = (3)(2)^{10} = 3072 \)

 

Q3. If p is the 5th, q is the 8th and s is the 11th term of a G.P. Prove that q2 = ps

Soln:

Let the first term be ‘a’ and the common ratio be r for the G.P.

Given condition

\( a_{5} = ar^{5 – 1} = ar^{4} = p . . . .  (1) \)

\( a_{8} = ar^{8 – 1} = ar^{7} = q . . . .  (2) \)

\( a_{11} = ar^{11 – 1} = ar^{10} = s . . . .  (3) \)

Eqn (2) divided by Eqn (1), we have

\( \frac{ar^{7}}{ ar^{7}} = \frac{q}{p} \)

\( r^{3} = \frac{q}{p} \)  . . . . .  (4)

Eqn (3) divided by Eqn (2), we have

\( \frac{ar^{10}}{ ar^{7}} = \frac{s}{q} \)

\( \Rightarrow r^{3} = \frac{s}{q} \)  . . . . .  (5)

From eqn (4) and eqn (5), we get

\( \frac{q}{p} = \frac{s}{q}\)

\( \Rightarrow q^{2} = ps \)

Hence proved

 

Q4. If first term of a G.P is -3 and the 4th term is square of the 2nd term. Find the 7th term.

Soln:

Let the first term be ‘a’ and common ratio be ‘r’.

Therefore, a = -3

We know that, an = arn – 1

a4 = ar3 = (-3)r3

a2 = ar1 = (-3)r

Given condition

(-3)r3 = [(-3) r]2

\( \Rightarrow -3r^{3} = 9 r^{2} \Rightarrow r = – 3 a_{7} = ar^{7 – 1} = a \)

r6 = (-3) (-3)6 = – (3)7 = -2187

Hence, the 7th term is -2187.

 

Q5. Which term of

(a) \( 2, 2\sqrt{2}, 4,  . . . . \; is \; 128? \)

(b) \( \sqrt{3}, 3, 3\sqrt{3} . . . . \; is \; 729? \)

(c) \( \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . . . . \; is \; \frac{1}{19683}? \)

Soln:

(a) Given sequence = \( 2, 2\sqrt{2}, 4,  . . . .  \)

Here,

First term = a = 2

Common ratio = r = \( \frac{2\sqrt{2}}{2} = \sqrt{2}\)

Assume nth term = 128

\( a_{n} = a r^{n – 1} \)

\( \Rightarrow (2)(\sqrt{2})^{n – 1} = 128 \)

\( \Rightarrow (2)(2)^{\frac{n – 1}{2}} = (2)^{7}\)

\( \Rightarrow (2)^{\frac{n – 1}{2} + 1} = (2)^{7}\)

\( \frac{n – 1}{2} + 1 = 7 \)

\( ⇒ \frac{n – 1}{2} = 6 \)

\( ⇒ n – 1 = 12 \)

\( ⇒ n = 13 \)

Hence, 128 is the 13th term

(b) Given sequence = \( \sqrt{3}, 3, 3\sqrt{3} . . . . \)

Here,

First term = a = \( \sqrt{3} \)

Common ratio = r = \( \frac{3}{\sqrt{3}} = \sqrt{3}\)

Assume nth term = 729

\( a_{n} = a r^{n – 1} \)

\( a r^{n – 1} = 729 \)

\( \Rightarrow (\sqrt{3})(\sqrt{3})^{n – 1} = 729 \)

\( \Rightarrow (3)^{\frac{1}{2}}(3)^{\frac{n – 1}{2}} = (3)^{6} \)

\( \Rightarrow (3)^{\frac{1}{2} + \frac{n – 1}{2}} = (3)^{6}\)

Therefore,

\( \frac{1}{2} + frac{n – 1}{2} = 6 \)

\( \Rightarrow \frac{1 + n – 1}{2} = 6 \)

\( \Rightarrow n = 12 \)

Hence, 729 is the 12th term of the sequence.

(c) Given sequence = \( \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . . . . \)

Here,

First term = a = \( \frac{1}{3} \)

Common term = r = \( \frac{1}{9} \div \frac{1}{3} = \frac{1}{3} \)

Assume the nth term be \( \frac{1}{19683} \)

\( a_{n} = a r^{n – 1} \)

\( a r^{n – 1} = \frac{1}{19683} \)

\( \Rightarrow \left ( \frac{1}{3} \right ) \left ( \frac{1}{3} \right )^{n – 1} = \frac{1}{19683} \)

\( \Rightarrow \left ( \frac{1}{3} \right )^{n} = \left ( \frac{1}{3} \right )^{9}\)

n = 9

Hence, \( \frac{1}{19683} \) is the 9th term of the sequence.

 

Q6. If \( \frac{2}{7}, x, -\frac{7}{2} \) are in G.P then find the value of x?

Soln:

Given sequence = \( \frac{2}{7}, x, -\frac{7}{2} \)

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

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

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

\( \Rightarrow x^{2} = \frac{-2 \times 7}{-2 \times 7} = 1 \)

\( \Rightarrow x = \sqrt{1}\)

\( \Rightarrow x = \pm 1\)

Hence, the sequence is in G.P if \( x = \pm 1\)

 

Q7. The sum of the first 20 terms of the G.P 0.15, 0.015, 0.0015 . . . .

Soln:

Given G.P = 0.15, 0.015, 0.0015 . . . .

Here,

First term = a = 0.15

Common ration = r = \(\frac{0.015}{0.15} = 0.1 \)

\( S_{n} = \frac{a(1 – r^{n})}{1 – r} \)

\( = \frac{0.15[1 – (0.1)^20]}{1 – 0.1} \)

\( = \frac{0.15}{0.9}[1 – (0.1)^{20}] \)

\( = \frac{15}{90}[1 – (0.1)^{20}] \)

\( = \frac{1}{6}[1 – (0.1)^{20}] \)

 

Q8. \( \sqrt{7}, \sqrt{21}, 3\sqrt{7}, . . . . \) is a G.P series. Find sum till the nth term.

Soln:

Given series = \( \sqrt{7}, \sqrt{21}, 3\sqrt{7}, . . . . \)

Here,

First term = a = \( \sqrt{7} \)

Common ratio = r = \( \frac{\sqrt{21}}{7} = \sqrt{3} \)

\( S_{n} = \frac{a(1 – r^{n})}{1 – r} \)

\( = \frac{\sqrt{7}[1 – (\sqrt{3})^{n}]}{1 – \sqrt{3}} \)

\( = \frac{\sqrt{7}[1 – (\sqrt{3})^{n}]}{1 – \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \)

\( = \frac{\sqrt{7}(\sqrt{3} + 1)[1 – (\sqrt{3})^{n}]}{1 – 3} \)

\( = \frac{ – \sqrt{7}(\sqrt{3} + 1)[1 – (\sqrt{3})^{n}]}{2} \)

 

Q9. What is the sum of the nth term of the G.P series

1, -a, a2, -a3 . . . .  (if a ≠ -1)?

Soln:

Given series = 1, -a, a2, -a3 . . .

Here,

The first term of the G.P = a1 = 1

Common ratio of the G.P = r = -a

\( S_{n} = \frac{a_{1}(1 – r^{n})}{1 – r}\)

\( S_{n} = \frac{1[1 – (-a)^{n}]}{1 – (-a)} = \frac{[1 – (-a)^{n}]}{1 + a}\)

 

Q10. What is the sum of the nth term of the G.P series

x3, x5, x7 . . . . (if x ≠ ± 1)

Soln:

Given series = x3, x5, x7 . . . .

Here,

First term = a = x3

Common ratio = r = x2

\( S_{n} = \frac{a(1 – r^{n})}{1 – r} = \frac{x^{3}[1 – (x^{2})^{n}]}{1 – x^{2}} = \frac{x^{3}(1 – x^{2n})}{1 – x^{2}}\)

 

Q11. Evaluate \(\sum_{k = 1}^{11}(2 + 3^{k})\)

Soln:

\(\sum_{k = 1}^{11}(2 + 3^{k}) = \sum_{k = 1}^{11}(2) + \sum_{k = 1}^{11} 3^{k} = 2(11) + \sum_{k = 1}^{11} 3^{k} = 22 + \sum_{k = 1}^{11} 3^{k} . . . . . (1)\)

\(\sum_{k = 1}^{11} 3^{k} = 3^{1} + 3^{2} + 3^{3} + . . . . + 3^{11}\)

The above sequence is in a G.P 3, 32, 33, . . .

\( S_{n} = \frac{a(r^{n} – 1)}{ r – 1} \)

\( \Rightarrow S_{n} = \frac{3[(3)^{11} – 1]}{3 – 1} \)

\( \Rightarrow S_{n} = \frac{3}{2}(3^{11} – 1) \)

\(\sum_{k = 1}^{11} 3^{k} = \frac{3}{2}(3^{11} – 1) \) ….(2)

Substituting eqn (2) in eqn (1), we get

\(\sum_{k = 1}^{11} (2 + 3^{k}) = 22 + \frac{3}{2}(3^{11} – 1)\)

 

Q12. Sum of 1st 3 terms in a G.P is \( \frac{39}{10} \) and the product of the terms is 1. What is the the terms and common ratio of the sequence?

Soln:

Assume \( \frac{a}{r}, a, ar \) be the 1st 3 terms in the G.P

\(\frac{a}{r} + a + ar = \frac{39}{10}\) . . . . (1)

\(\left (\frac{a}{r} \right ) \left (a \right ) \left ( ar \right ) = 1\) . . . . (2)

From eqn (2), we get a3 = 1

a = 1

Substituting (a = 1) in eqn (1), we get

\(\frac{1}{r} + 1 + r = \frac{39}{10}\)

\(\Rightarrow 1 + r + r^{2} = \frac{39}{10}r\)

\(\Rightarrow 10 + 10 r + 10 r^{2} -39 r = 0\)

\(\Rightarrow 10 r^{2} – 29 r + 10 = 0\)

\(\Rightarrow 10 r^{2} – 25 r – 4 r + 10 = 0\)

\( ]\Rightarrow 5r (2r – 5) -2(2r – 5) = 0\)

\( ]\Rightarrow (5r – 2) (2r – 5) = 0\)

\( ]\Rightarrow r = \frac{2}{5} or \frac{5}{2}\)

Hence,

The terms are \(\frac{5}{2}, 1 \; and \; \frac{2}{5}\)

 

Q13. For the G.P 3, (3)2, (3)3 …. sum of how many terms is 120?

Soln:

Given G.P = 3, (3)2, (3)3 ….

Let n number of terms are required for obtaing the sum of 120

\(S_{n} = \frac{a(1 – r)^{n}}{1 – r}\)

Here,

First term = a = 3

Common ratio = r = 3

\(S_{n} = 120 = \frac{3(3^{n} – 1)}{3 – 1}\)

\(\Rightarrow 120 = \frac{3(3^{n} – 1)}{2}\)

\(\Rightarrow \frac{120 \times 2}{3} = 3^{n} – 1\)

\(\Rightarrow 3^{n} – 1 = 80 \)

\(\Rightarrow 3^{n} = 81 \)

\(\Rightarrow 3^{n} = 3^{4} \)

n = 4

Hence, 4 terms are required to get the sum of 120

 

Q14. In a G.P sum of the first 3 terms of are 16 where as sum for the next 3 terms are 128.

Find common ratio, first term and sum of n terms for the G.P

Soln:

Assume G.P = a, ar, a(r)2, a(r)3, . . .

Given condition =

a + a(r)2 + ar + = 16 and a(r)5 + a(r)4 + a(r)3 = 128

a (1 + (r)2 + r) = 16 …… (1)

a(r)3(1+ (r)2 + r) = 128 …….. (2)

eqn (2) divided by eqn (1), we get

\(\frac{ar^{3}(1 + r + r^{2})}{a(1 + r + r^{2})} = \frac{128}{16}\)

r3 = 8

r = 2

substituting r = 2 in eqn(1), we get

a(1 + 2 + 4) = 16

a(7) = 16

\( a = \frac{16}{7}\)

\( S_{n} = \frac{16}{7} \frac{(2^{n} – 1)}{2 – 1}  = \frac{16}{7}(2^{n} – 1)\)

 

Q15. Given first term ‘a’ = 729 and the 7th term = 64, then determine S7.

Soln:

Given a = 729 , a7 = 64

Assume common ratio = r

We know that,

an = a rn – 1

a7 = a r7 – 1 = (729) (r)6

64 = 729 (r)6

\( r^{6} = \frac{64}{729} \)

\( r^{6} = \left ( \frac{2}{3}\right )^{6} \)

\( r = \frac{2}{3} \)

Also, we know that

\(S_{n} = \frac{a(1 – r^{n})}{1 – r}\)

\(S_{7} = \frac{729 \left [ 1 – \left ( \frac{2}{3} \right )^{7} \right ] }{1 – \frac{2}{3}}\)

\(= 3 \times 729 \left [ 1 – \left ( \frac{2}{3} \right )^{7} \right ]\)

\(= (3)^{7} \left [ \frac{(3)^{7} – (2)^{7}}{(3)^{7}} \right ]\)

\(= (3)^{7} – (2)^{7} \)

= 2187 – 128

= 2059

 

Q16. The sum of first 2 terms of a G.P is -4 and 5th term is 4 times that of third term. Find the G.P

Soln:

Assume first term = a

Common ratio = r

\(S_{2} = -4 = \frac{a(1 – r^{2})}{1 – r}\) ….. (1)

\( a_{5} = 4 \times a_{3} \)

\( \Rightarrow ar^{4} = 4ar^{2} \Rightarrow r^{2} = 4\)

\( r = \pm 2 \)

From (1), we get

\(-4 = \frac{a\left [ 1 – (2)^{2} \right ]}{1 – 2} \; for \; r = 2\)

\(-4 = \frac{a(1 – 4)}{-1} \)

-4 = a(3)

\( a = \frac{-4}{3} \)

Also, \(-4 = \frac{a\left [ 1 – (-2)^{2} \right ]}{1 – (-2)} \; for \; r =  -2\)

\(-4 = \frac{a(1 – 4)}{1 + 2}\)

\(-4 = \frac{a(- 3)}{3}\)

a = 4

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

 

Q17. If p, q, and s are fourth, tenth and sixteenth terms of G.P prove that p, q, s are in G.P

Soln:

Let

First term = a

Common ratio = r

According to given condition,

a4 = ar3 = p ………. (1)

a10 = ar9 = q ………. (2)

a16 = ar15 = x ………. (3)

eqn (2) divided by eqn (1)

\(\frac{q}{p} = \frac{ar^{9}}{ar^{3}} \Rightarrow \frac{q}{p} = r^{6}\)

eqn (3) divided by eqn (2)

\(\frac{s}{q} = \frac{ar^{15}}{ar^{9}} \Rightarrow \frac{s}{q} = r^{6}\)

\(\frac{q}{p} = \frac{s}{q} \)

 

Q18. 8, 88, 888, 8888, …… are in sequence find sum of the nth terms

Soln:

Given sequence 8, 88, 888, 8888, ……

This sequence isn’t a G.P

It may be changed to a G.P if the terms are written as Sn = 8 + 88 + 888 + 8888 + …..  to n terms

\(= \frac{8}{9}\left [ 9 + 99 + 999 + 9999 + ……….\; to \; n \; terms \right ]\)

\(= \frac{8}{9}\left [ (10 – 1) + (10^{2} – 1) + (10^{3} – 1) + (10^{4} – 1) + ……….\; to \; n \; terms \right ]\)

\(= \frac{8}{9}\left [ (10 + 10^{2} + …. n \; terms) – (1 + 1 + 1 ……….\; to \; n \; terms) \right ]\)

\(= \frac{8}{9}\left [ \frac{10(10^{n} – 1)}{10 – 1} – n \right ]\)

\(= \frac{8}{9}\left [ \frac{10(10^{n} – 1)}{9} – n \right ]\)

\(= \frac{80}{81}\left ( 10^{n} – 1 \right ) – \frac{8}{9}n\)

 

Q19. If 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½ are in sequences find the sum of the products

Soln:

Required sum = \(2 \times 128 + 4 \times 32 + 8\times 8 + 16 \times 2 + 32 \times \frac{1}{2}\)

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

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

Now first term, a = 4

And common ratio, \( r = \frac{1}{2}\)

We know that,

\(S_{n} = \frac{a(1 – r^{n})}{1 – r}\)

\(S_{5} = \frac{4\left [ 1 – \left ( \frac{1}{2} \right )^{5}\right ]}{1 – \frac{1}{2}} = \frac{4\left [ 1 – \frac{1}{32} \right ]}{ \frac{1}{2}} = 8 \left ( \frac{32 – 1}{32} \right ) = \frac{31}{4}\)

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

 

Q20. Prove that product of corresponding terms from a, ar, ar2,arn-1 and A, AR, AR2,ARn-1 are in sequences find common ratio

Soln:

To show = aA, ar(AR), ar2 (AR2), … arn – 1 (ARn – 1) is a G.P series

\(\frac{Second \; term}{First \; term} = \frac{arAR}{aA} = rR\)

\(\frac{Third \; term}{Second \; term} = \frac{ar^{2}AR^{2}}{arAR} = rR\)

Hence, the sequences are in G.P and rR is common ratio.

 

Q21. If in a G.P of 4 numbers 3rd term is greater to 1st term by 9, and the 2nd term is greater to 4th by 18.

Soln:

Let

First term = a

Common ratio = r

a1 = a,  a2 = ar,  a3 = ar2,  a4 = ar3

From given conditions,

a3 = a1 + 9 => ar2 = a + 9 ………. (1)

a2 = a4 + 18 => ar = ar3 + 18 ………. (2)

From eqn (1) and eqn (2), we get

a(r2 – 1) = 9 ……… (3)

ar(1 – r2) = 18 ……..(4)

Dividing eqn (4) by eqn (3), we get

\(\frac{ar(1 – r^{2})}{a( 1 – r^{2})} = \frac{18}{9}\)

=>  – r = 2

=> r = -2

Substituting r = -2 in eqn (1), we get

4a = a + 9

=> 3a = 9

=> a = 3

Hence, first four terms of G.P are 3, 3(-2), 3(-2)2, and 3(-2)3

i.e. 3, -6, 12, -24

 

Q22. If in a G.P a, b and c are the pth, qth and rth terms. Prove that

aq – r.br – p.cp – q = 1

Soln:

Let

First term = a

Common ratio = r

From the given condition

ARp – 1 = a

ARq – 1 = b

ARr – 1 = c

aq – r.br – p.cp – q

\( = A^{q – r} \times R^{(p – 1)(q – r)} \times A^{r – p} \times R^{(q – 1)(r – p)} \times A^{p – q} \times R^{(r – 1)(p – q)}\)

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

= A0 x R0

= 1

Hence proved

 

Q23. If in a G.P ‘a’ and ‘b’ are the first and nth term and product of nth terms is P.

Show P2 = (ab)n

Soln:

Given:

First term = a

Last term = b

Therefore,

G.P = a, ar, ar2, ar3 . . . arn – 1, where r is the common ratio.

b = arn – 1…….. (1)

P = product of n terms

= (a)(ar)(ar2)…(arn – 1)

= (a x a x …. a)(r x r2 x … rn – 1)

= an r 1 + 2 + … (n – 1) …… (2)

Here, 1, 2, …..(n – 1) is an A.P

1 + 2 + …. + (n – 1)

\(= \frac{n – 1}{2}\left [ 2 + (n – 1 – 1) \times 1 \right ] = \frac{n – 1}{2}[2 + n – 2] = \frac{n(n – 1)}{2}\)

\(P = a^{n}r^{\frac{n(n -1)}{2}}\)

\(P = a^{2n}r^{n(n -1)}\)

\( = [a^{2}r^{n – 1}]^{n}\)

= (ab)^{n}

Hence proved

 

Q24. Prove that the ratio of sum of first n terms to sum of terms from (n + 1)th to (2n)th terms is 1/ rn .

Soln:

Let

First term = a

Common ratio = r

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

Since there are ‘n’ number of terms from (n + 1)th to (2n)th term,

Sum of the terms

\(S_{n} = \frac{a_{n + 1(1 – r^{n})}}{1 – r}\)

Required ratio = \(\frac{a (1 – r^{n})}{(1 – r)} \times \frac{(1 – r)}{ar^{n}(1 – r^{n})} = \frac{1}{r^{n}}\)

Thus, the ratio is 1/rn

 

Q25. Prove that:

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

If a, b, c and d are in a G.P series

Soln:

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

Therefore

bc = ad …………….. (1)

b2 = ac ……………… (2)

c2 = bd ……………… (3)

It is to be proven that,

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

R.H.S

= (ab + bc + cd)2

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

= [ab + d(a + c)]2

= a2 (b2) + 2abd(a + c) + (d2)(a + c)2

= (a2) (b2) + 2a2bd + 2acbd + (d2)(a2 + 2ac + c2)

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

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

= a2b2 + a2c2 + a2d2 + b2 x b2 + b2c2 + b2d2 + c2b2 + c2 x c2 + c2d2
[From eqn (1) and (2)]

= 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

 

Q26. Between 3 and 81 insert two numbers such that the sequence is a G.P series.

 Soln:

Let the numbers be g1 and g2

So from the given condition 3, g1, g2, 81 is in G.P

Assume first term = a

Common ratio = r

Therefore 81 = (3)(r)3

=> r3 = 27

r = 3 (taking real roots only)

For r = 3,

Q26.

G1 = ar = (3)(3) = 9

G2 = ar2 = (3)(3)2 = 27

Hence, the numbers are 9 and 27

 

Q27. If \(\frac{a^{2 + 1} + b^{n + 1}}{a^{n} + b^{n}}\) is the geometric mean for a and b. What is the value of n?

Soln:

Mean of a and b is \( \sqrt{ab}\)

From the given condition \(\frac{a^{2 + 1} + b^{n + 1}}{a^{n} + b^{n}} = \sqrt{ab} \)

Squaring both the sides, we get

\(\frac{(a^{2 + 1} + b^{n + 1})^{2}}{(a^{n} + b^{n})^{2}} = ab \)

=> a2n +1 + 2an + 1 bn + 1 + b2n + 1 = (ab)(a2n + 2anbn + b2n)

=> a2n +1 + 2an + 1 bn + 1 + b2n + 1 = a2n +1b + 2an + 1 bn + 1 + ab2n + 1

=> a2n +1 + b2n + 1 = a2n +1b + ab2n + 1

=> a2n +1 – a2n +1b = ab2n + 1 – b2n + 2

=> a2n +1 (a – b) = b2n + 1 (a – b)

=> \(\left ( \frac{a}{b} \right )^{2n + 1} = 1 = \left ( \frac{a}{b} \right )^{0}\)

=> 2n + 1 = 0

=> \( n = \frac{-1}{2}\)

 

Q28. Prove that the ratio of two numbers is  \((3 + 2\sqrt{2})(3 – 2\sqrt{2})\) if the sum of the two numbers is 6 times their geometric mean.

Soln:

Let the two numbers be a and b.

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

From given conditions,

\(a + b = 6\sqrt{ab}\) ……. (1)

=> (a + b)2 = 36(ab)

Also,

(a – b)2 = (a + b)2 – 4ab = 36ab – 4ab = 32 ab

=> \(a – b = \sqrt{32}\sqrt{ab}\)

= \( 4\sqrt{2}\sqrt{ab}\) ……. (2)

Eqn(1) + Eqn (2)

\(2a = (6 + 4 \sqrt{2})\sqrt{ab}\)

=> \(a = (3 + 2 \sqrt{2})\sqrt{ab}\)

Putting the value of a in eqn (1),we get

\(b = 6\sqrt{ab} – (3 + 2\sqrt{2})\sqrt{ab}\)

=> \(b = (3 – 2\sqrt{2})\sqrt{ab}\)

\(\frac{a}{b} = \frac{(3 + 2\sqrt{2})\sqrt{ab}}{(3 – 2\sqrt{2})\sqrt{ab}} = \frac{3 + 2\sqrt{2}}{3 – 2\sqrt{2}}\)

Hence, required ratio = \(\left ( 3 + 2\sqrt{2} \right ) : \left ( 3 – 2\sqrt{2} \right )\)

 

Q29. For two positive numbers if G and A are the G.M and A.M show that the numbers are \(A \pm \sqrt{(A + G)(A – G)}\)

Soln:

Given G and A are A.M and G.M

Assume the numbers be a and b.

\(AM = A = \frac{a + b}{2} ……… (1)\)

\(GM = G = \sqrt{ab} ……… (2)\)

From Eqn(1) and Eqn (2), we get

a + b = 2A ……. (3)

ab = G2 ……. (4)

Putting the value of a and b from (3) and (4)

(a – b)2 = (a + b)2 – 4ab, we get

(a – b)2 = 4A2 – 4G2 = 4(A2 – G2)

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

From eqn (3) and (5), we get

\(2a  = 2A + 2\sqrt{(A + G)(A – G)} \)

=> \(a = A + \sqrt{(A + G)(A – G)} \)

Putting the value of a in eqn(3), we get

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

Hence, the numbers are \(A \pm \sqrt{(A + G)(A – G)}\)

 

Q30. Bacteria over certain culture multiply twice in one hour. If in initial stage there were only 30 bacteria, what will be the amount of bacteria in the 2nd, 4th and nth hour?

Soln:

Given that amount of bacteria multiplies two times in one hour

Therefore,

Bacteria amount will form G.P

Here, first term = (a) = 30

Common ratio = (r) = 2

Therefore ar2 = a3 = 30 x (2)2 = 120

Therefore amount of bacteria after 2nd hour is 120

ar4 = a5 = 30 x (2)4 = 480

Therefore amount of bacteria after 4th hour is 480

 arn x an + 1 = 30 x (2)n

Hence, amount of bacteria after nth hour is 30 x (2)n

 

Exercise 9.4

Q1. Find sum to nth term for the series \(1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + . . . .\)

Soln:

Given series = \(1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + . . . . \) nth term, an = n (n + 1)

\(S_{n} = \sum_{k = 1}^{n} a_{1} = \sum_{k = 1}^{n} k ( k + 1)\)

\( = \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k \)

\(\frac{n (n + 1) (2n + 1)}{6} + \frac{n (n + 1)}{2}\)

\(\frac{n (n + 1)}{2} \left ( \frac{ 2n + 1}{3} + 1 \right )\)

\(\frac{n (n + 1)}{2} \left ( \frac{ 2n + 4}{3} \right )\)

\(\frac{n (n + 1) (n + 2) }{3} \)

 

Q2. Find sum to nth term of the series \(1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + . . . . .\)

Soln:

Given series = \(1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + . . . n^{th} \; term, \; a_{n} = n (n + 1) (n + 2)\)

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n

\(S_{0} = \sum_{k = 1}^{n} a_{k}\)

\(= \sum_{k = 1}^{n} k^{3} + 3 \sum_{k = 1}^{n} k^{2} + 2 \sum^{n}_{k = 1} k\)

\(= \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{3n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2}\)

\(= \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{n(n + 1)(2n + 1)}{2} + n(n + 1)\)

\(= \frac{n(n + 1)}{2} \left [ \frac{n(n + 1)}{2} + 2n + 1 + 2 \right ]\)

\(= \frac{n(n + 1)}{2} \left [ \frac{n^{2} + n + 4n + 6}{2} \right ]\)

\(= \frac{n(n + 1)}{4} \left ( n^{2} + 5n + 6 \right )\)

\(= \frac{n(n + 1)}{4} \left ( n^{2} + 2n + 3n + 6 \right )\)

\(= \frac{n(n + 1)\left [ n (n + 2) + 3 (n + 2) \right ]}{4}\)

\(= \frac{n(n + 1) (n + 2) (n + 3) }{4}\)

 

Q3. Find sum to nth term of the series \(3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + . . .\)

Soln:

Given series = \(3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + . . . n^{th} term, \)

an = (2n + 1) n2 = 2n3 + n2

\(S_{n} = \sum^{n}_{k = 1} a_{k}\)

\(= \sum^{n}_{k = 1} = (2k^{3} + k^{2}) = 2\sum^{n}_{k = 1} k^{3} + \sum^{n}_{k = 1} k^{2}\)

\(= 2 \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{n(n + 1)(2n + 1)}{6}\)

\(= \frac{n^{2} (n + 1)^{2} }{2} + \frac{n(n + 1)(2n + 1)}{6}\)

\(= \frac{n (n + 1) }{2} \left [ n(n + 1) + \frac{2n + 1}{3} \right ]\)

\(= \frac{n (n + 1) }{2} \left [ \frac{3n^{2} + 3n + 2n + 1}{3} \right ]\)

\(= \frac{n (n + 1) \left ( 3n^{2} + 5n + 1 \right ) }{6} \)

 

Q4. Find sum to nth term of the series \(\frac{1}{ 1 \times 2 } + \frac{1}{ 2 \times 3 } + \frac{1}{ 3 \times 4 } + . . . .\)

Soln:

Given series = \(\frac{1}{ 1 \times 2 } + \frac{1}{ 2 \times 3 } + \frac{1}{ 3 \times 4 } + . . . .\)

\(n^{th} \; term, a_{n} = \frac{1}{n(n + 1)} = \frac{1}{n} – \frac{1}{n + 1}\)

\(a_{1} = \frac{1}{1} – \frac{1}{2}\)

\(a_{2} = \frac{1}{2} – \frac{1}{3}\)

\(a_{3} = \frac{1}{3} – \frac{1}{4} . . . \)

\(a_{n} = \frac{1}{n} – \frac{1}{n + 1}\)

Column wise adding the above terms, we get

\(a_{1} +a_{2} + a_{3} + … + a_{n} = \left [ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n} \right ] – \left [ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{n + 1} +\right ]\)

\(S_{n} = 1 – \frac{1}{n + 1} = \frac{n + 1 – 1}{n + 1} = \frac{n}{n + 1}\)

 

Q5. Find sum to nth term of the series \(5^{2} + 6^{2} + 7^{2} + . . . + 20^{2} \; n^{th} \; term, \; a_{n} = (n + 4)^{2} = n^{2} + 8 n + 16\)

Soln:

\(S_{n} = \sum^{n}_{k = 1} a_{k} = \sum^{n}_{k = 1}(k^{2} + 8k + 16)\)

\(S_{n} = \sum^{n}_{k = 1} a_{k} = 8\sum^{n}_{k = 1}k + \sum^{n}_{k = 1}16\)

\(= \frac{n(n + 1)(2n + 1)}{6} + \frac{8n (n + 1)}{2} + 16 n\)

16th term is (16 + 4)2 = 202

\(S_{16} = \frac{16(16 + 1)(2 \times 16 + 1)}{6} + \frac{8 \times 16 \times (16 + 1)}{2} + 16 \times 16\)

\(= \frac{(16)(17)(33)}{6} + \frac{(8) \times 16 \times (16 + 1)}{2} + 16 \times 16\)

\(= \frac{(16)(17)(33)}{6} + \frac{(8) (16) (17)}{2} + 256 \)

= 1496 + 1088 + 256

= 2840

52 + 62 + 72 + …. + 202 = 2840

 

Q6. Find sum to nth term of the series \(3 \times 8 + 6 \times 11 + 9 \times 14 + …\)

Soln:

Given series = \(3 \times 8 + 6 \times 11 + 9 \times 14 + … a_{n} \)

= (nth term of 3, 6, 9, …) x (nth term of 8, 11, 14 .. .)

= (3n) (3n + 5)

= 9n2 + 15n

\(S_{n} = \sum^{n}_{k = 1}a_{k} = \sum^{n}_{k = 1} (9k^{2} + 15k)\)

\( = \sum^{n}_{k = 1}k^{2} + 15 \sum^{n}_{k = 1} k\)

\(9 \times \frac{n(n + 1)(2n + 1)}{6} + 15 \times \frac{n (n + 1)}{ 2}\)

\( = \frac{3n (n + 1)(2n + 1)}{2} + \frac{ 15n (n + 1)}{ 2}\)

\( = \frac{3n (n + 1)}{2}(2n + 1 + 5)\)

\( = \frac{3n (n + 1)}{2}(2n + 6)\)

\( = 3n (n + 1)(n + 3)\)

 

Q7. Find sum to nth term of the series \(1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . .\)

Soln:

Given series = \(1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . . a^{n} \)

\(= \left (1^{2} + 2^{2} + 3^{2} + . . . . + n^{2} \right )\)

\(= \frac{n(n + 1)(2n + 1)}{6}\)

\(= \frac{n(2n^{2} + 3n + 1)}{6}\)

\(= \frac{2n^{3} + 3n^{2} + n)}{6}\)

\(= \frac{1}{3}n^{3} + \frac{1}{2}n^{2} + \frac{1}{6}n\)

\(S_{n} = \sum_{k = 1}^{n} a_{k}\)

\(= \sum_{k = 1}^{n} \left ( \frac{1}{3} k^{3} + \frac{1}{2} k^{2} + \frac{1}{6} k \right )\)

\(= \frac{1}{3} \sum_{k = 1}^{n} k^{3} + \frac{1}{2} \sum_{k = 1}^{n} k^{2} + \frac{1}{6} \sum_{k = 1}^{n} k\)

\(= \frac{1}{3} \frac{n^{2}(n + 1)^{2}}{(2)^{2}} + \frac{1}{2} \times \frac{n(n + 1)(2n + 1)}{6} + \frac{1}{6} \times \frac{n(n + 1)}{2}\)

\(= \frac{n(n + 1)}{6} \left [ \frac{n(n + 1)}{2} + \frac{2n + 1}{2} + \frac{1}{2} \right ]\)

\(= \frac{n(n + 1)}{6} \left [ \frac{n^{2} + n + 2n + 1 + 1 }{2} \right ]\)

\(= \frac{n(n + 1)}{6} \left [ \frac{n^{2} + n + 2n + 2 }{2} \right ]\)

\(= \frac{n(n + 1)}{6} \left [ \frac{n( n + 1) + 2(n + 1) }{2} \right ]\)

\(= \frac{n(n + 1)}{6} \left [ \frac{( n + 1) (n + 2) }{2} \right ]\)

\(= \frac{n(n + 1)^{2} (n + 2) }{12}\)

 

Q8. Find sum to nth term of the series that has nth term given by n(n + 1)(n + 4)

Soln:

\(a_{n} = n(n + 1)(n + 4) = n(n^{2} + 5n + 4) = n^{3} + 5n^{2} + 4n\)

\(S_{n} = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} k^{3} + 5 \sum_{k = 1}^{n} k^{2} + 4 \sum_{k = 1}^{n} k\)

\(= \frac{n^{2} (n + 1)^{2}}{4} + \frac{5n (n + 1)(2n + 1)}{6} + \frac{4n( n + 1)}{2}\)

\(= \frac{n(n + 1)}{2} \left [ \frac{ n(n + 1)}{2} + \frac{5(2n + 1)}{6} + 4 \right ]\)

\(= \frac{n(n + 1)}{2} \left [ \frac{ 3n^{2} + 3n + 20n + 10 + 24 }{6} \right ]\)

\(= \frac{n(n + 1)}{2} \left [ \frac{ 3n^{2} + 23n  + 34 }{6} \right ]\)

\(\frac{n(n + 1) \left ( 3n^{2} + 23n + 34 \right ) }{12}\)

 

Q9. Find sum to nth term of the series that has nth term given by n2 + 2n

Soln:
an = n2 + 2n

\(S_{n} = \sum_{k = 1}^{n} k^{2} + 2^{k} = \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} 2^{k}\) ……………(1)

Take \(\sum_{k = 1}^{n} 2^{k} = 2^{1} + 2^{2} + 2^{2} + . . . .\)

The series 2, 22, 23, …. Is a G.P with 2 as first term as well as common ratio.

\(\sum_{k = 1}^{n} 2^{k} = \frac{(2)\left [ (2)^{n} – 1 \right ]}{2 – 1} = 2 (2^{n} – 1)\) ……………… (2)

From (1) and (2), we get

\(S_{n} = \sum_{k = 1}^{n} k^{2} + 2(2^{n} – 1) = \frac{n (n + 1)(2n + 1)}{6} + 2 (2^{n} – 1)\)

 

Q10. Find sum to nth term of the series that has nth term given by (2n – 1)2

Soln:

an = (2n – 1)2 = 4n2 – 4n + 1

\(S_{n} = 4 \sum_{k = 1}^{n} a_{k} – 4 \sum_{k = 1}^{n} (4k^{2} – 4k + 1)\)

\( = 4\sum_{k = 1}^{n} k^{2} – 4 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1\)

\( = \frac{4n (n + 1) (2n + 1)}{6} – \frac{4n (n + 1)}{2} + n\)

\( = \frac{2n (n + 1) (2n + 1)}{3} – 2n (n + 1) + n\)

\(= n \left [ \frac{2 (2n^{2} + 3n + 1)} {3} – 2 (n + 1) + 1 \right ]\)

\(= n \left [ \frac{4n^{2} + 6n + 2 – 6n – 6 + 3} {3} \right ]\)

\(= n \left [ \frac{4n^{2} – 1} {3} \right ]\)

\(= \frac{n (2n + 1) (2n – 1)}{3}\)

Thus, the above NCERT Solutions for class 11 maths chapter 9 is displayed above. It is important to be aware of the various types of problems that can be asked for during the examination. It is important to be aware of the different solutions for the many problems that one may get asked. The NCERT solutions for class 11 deals with the various types of problems and figuring out the best solutions for class 11. Some of the other major topics in mathematics are Sets, Relations and Functions, Binomial Theorem, Conic Sections etc. These are some of the major different types of exercises for the chapter of mathematics. There are quite a few different topics that students for class 11 need to study for. Thus, these were the major problems of NCERT solutions class 11 maths chapter 9.


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If you were given an angle of 60, it is possible to get 30.