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## Karnataka 1st PUC Basic Maths Question Bank Chapter 5 Progressions

Arithmetic Progression (A.P)

Question 1.

Which term of A.P 4\(\left(5 \frac{1}{3}-4\right), 6 \frac{2}{3} \ldots \ldots\) is 104?

Answer:

In given A.P: a = 4, d = 5^{1/3} – 4 = \(\frac { 16 }{ 3 }\) – 4 = \(\frac { 4 }{ 3 }\)

Let n^{th} term be 104 = T_{n} = a + (n – 1) d

⇒ 4 + (n – 1) \(\frac { 4 }{ 3 }\) = 104

⇒ 12 + 4n – 4 = 312 ⇒ 4n – 4 = 312

⇒ 4n = 312 – 8

⇒ 4n = 304

∴ n = 76

Question 2.

How many terms of series 24 + 20 + 16 + ……. must be taken so that their sum is 72.

Answer:

In the given A.P; a = 24, d = 4, S_{n} = 72

we have S_{n} = \(\frac { n }{ 2 }\){2a + (n – 1)d}

72 = \(\frac { n }{ 2 }\){2(24) + (n – 1)(-4)}

= 144 = n{48 – 4n + 4}

⇒ 144 = n{52 – 4n}

⇒ 144 = 52n – 4n^{2
}⇒ 4n^{2} – 52n + 144 = 0

⇒ n^{2} – 13n + 36 = 0

⇒ (n – 9)(n – 4) = 0

∴ n = 9,4

Question 3.

The first and the last term of an A.P are -4 and 146 and the sum of the A.P is 7171. Find the number of terms in the A.P and the A.P.

Answer:

Given a = -4, l = 146.

We have l = a + (n – 1 )d = 146 = -4 + (n – 1) d

⇒ (n – 1)d = 150

Also we have Sn = \(\frac { n }{ 2 }\){2a + (n – 1)d}

7171 = \(\frac { n }{ 2 }\){2(-4) + 150}

14342 = n{-8 + 150}

⇒ 142n = 14342

∴ n = \(\frac{14342}{142}\) = 101

So number of terms is A.P = 101

consider (n – 1)d = 150.

Question 4.

Find the sum of all numbers between 100 and 1000 which are divisible by 11.

Answer:

The 1^{st} greatest No. greater than 100 is 110. the last which is less than 1000 is, 990.

110,121, ………990 be the number

T_{n} = 90 ⇒ 110 + (n – 1) 11 = 990

11n – 11 = 990 – 110

11n = 880 + 11 = 891

∴ n = \(\frac { 891 }{ 11 }\) = 81

Now S_{81} = 8\(\frac { 1 }{ 2 }\)(110 + 990) = \(\frac { 81 }{ 2 }\)(550) = 4450

Question 5.

The first, second and the last terms of an A.P are a, b, c respectively prove that the sum is \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\)

Answer:

1^{st} term is a, common difference = d = b – a.

Let ‘n’ be the number of term in the series.

∴ c = a + (n – 1)d

⇒ (n – 1)d = c – a

⇒ (n – 1 )(b – a) = c – a

⇒ n – 1 = \(\frac{c-a}{b-a}\)

⇒ n = \(\frac{c-a}{b-a}\) + 1

= \(\frac{n}{2}\){2a + c – a} = \(\frac{n}{2}\){a + c} = \(\left(\frac{b+c-2 a}{b-a}\right)\left(\frac{a+c}{2}\right)\)

Geometric Progression (G.P)

Question 1.

Find the 9^{th} term and sum of 6 term of the geometric sequence \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27} \ldots \ldots\)

Answer:

a = \(\frac{1}{3}\) r = \(\frac{1}{3}\)

Question 2.

The second and fifth term of GP are 3 and \(\frac{81}{8}\) respectively. Find the GP

Answer:

Given T_{2} = ar = 3, T_{5} = ar^{4} = \(\frac{81}{8}\)

∴ r = \(\sqrt[3]{\frac{33}{23}}=\frac{3}{2}\)

If ar = 3 ⇒ a. \(\frac{3}{2}\) = 3 ⇒ a = 2

G.p is 2, 3, \(\frac{9}{2}\), …….

Question 3.

The sum of three numbers which are in G.P is \(\frac{39}{10}\) and their product is 1. Find the numbers.

Answers:

Let the number be \(\frac{a}{2}\), a, ar. which are in G.P.

i.e., \(\frac{a}{2}\) + a + ar = \(\frac{39}{10}\)

⇒ a^{3} = 1

∴ a = 1

put a = 1 in (1),

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

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

⇒ 10 + 10r + 10r^{3 }= 39

r ⇒ 10r^{2} – 29r + 10 = 0

⇒ r = \(\frac{5}{2}\) and \(\frac{2}{5}\)

Question 4.

Find three numbers in GP. Whose sum is 35 and sum of whose squares is 525.

Answer:

Let the three numbers be a, ar, ar^{2} which are is GP.

Given a + ar + ar^{2} = 35 ⇒ a( 1 + r + r^{2}) = 35

and a^{2} + a^{2}r^{2} + a^{2}r^{4} = 525

⇒ a^{2}( 1 + r^{2} + r^{4}) = 525

\(\frac{a^{2}\left(1+r^{2}+r^{4}\right)}{a\left(1+r+r^{2}\right)}=\frac{525}{35}=15\)

a(1 + r^{2} + r^{4}) = 15(1 + r + r^{2})

⇒ a(1 – r + r^{2}) = 15 …….. (2)

Add (1) and (2)

2a + 2ar^{2} = 50

all (2) – (1) ⇒ 2ar = -20

\(\frac{1}{r}\) + r = \(\frac{-5}{2}\)

⇒ 2(1 + r^{2}) = -5r

⇒ 2r^{2} + 5r + 2 = 0

⇒ r = \(\frac{-5 \pm \sqrt{25-4(2)(2)}}{4}\)

⇒ \(\frac{-5 \pm 3}{4}=\frac{-1}{2},-2\)

when r = -2 2ar = -20 ⇒ -4a = -20 [∴ a = 5]

when r = \(-\frac{1}{2}\) , 2ar = -20 ⇒ -a = -20 ⇒ [a = 20]

Hence the three number which are in G.P are 5, -10, 20 (or) 20, – 10, 5

Question 5.

Find the sum of n terms of 5 + 55 + 555 + ……..

Answer:

Let s = 5 + 55 + 555 + …….. n terms

⇒ s = 5[1 + 11 + 111 + …… to term

= \(\frac{5}{9}\)(9 + 99 + 999 + ….. to n term) \(\frac{5}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) + …… to terms]

= \(\frac{5}{9}\)[10 + 10^{2} + 10^{3} + …… to n terms] + (1 + 1 + ……. to n terms)

Harmonic Progression (H.P)

Question 1.

If the 12^{th} element of H.P is \(\frac { 1 }{ 5 }\) and the 19th term is \(\frac { 3 }{ 32 }\). Find the 10^{th} term of H.P.

Answer:

12^{th} term = \(\frac { 1 }{ 15 }\)

19^{th} term = \(\frac{3}{2^{2}}\)

∴ 12^{th} term, AP is 5 and

19^{th} termof A.P is \(\frac { 22 }{ 3 }\)

Question 2.

If 1, x, 3 are in H.P . Find x.

Answer:

Given 1, x, 3 are in H.P

Question 3.

If the m^{th} term of H.P is ‘n’ and n^{th} term is m, prove that (m + n)^{th} is \(\frac{m n}{m+n}\)

Answer:

Given m^{th} term of H.P is n,

∴ m^{th} term of A.P = \(\frac{1}{n}\) = a + (m – 1) d

Also n^{th} term of HP is m.

∴ n^{th} term of A.P = \(\frac{1}{m}\) = a + (n – 1)d

a + (m – 1)d = \(\frac{1}{n}\) …… (1)

a + (n – 1)d = \(\frac{1}{m}\) …… (2)

Solving (1) and (2)

Arithmetic, Geometric and Hormonic Means.

Question 1.

Find six geometrical mean between 27 and \(\frac { 1 }{ 81 }\)

Answer:

Let the required GM is g_{1}, g_{2}, g_{3}, g_{4}, g_{5}, g_{6}

= 27, g_{1}, g_{2}, g_{3}, g_{4}, g_{5}, g_{6}, \(\frac{1}{81}\) are in GP

Now a = 27 and \(\frac{1}{81}\) = 8^{th }element

⇒ \(\frac{1}{81}\) = ar^{7} ⇒ \(\frac{1}{81}\) = 27.r^{7}

⇒ \(\frac{1}{81 \times 27}\) = r^{7}

⇒ r^{7} = \(\frac{1}{3^{7}}\)

⇒ r = \(\frac{1}{3}\)

∴ the six GM are 9, 3, 1,\(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\)

Question 2.

If a, b, c are in A.P and a, mb, c are in GP then show that a, m^{2}b, c are in H.P

Answer:

a, b, c, are in AP .

⇒ b = \(\frac{a+c}{2}\)

a, mb, c are in GP,

⇒ mb = \(\sqrt{a c}\)

⇒ m^{2}b^{2} = ac

⇒ m^{2}b. b. = ac

⇒ m^{2}b \(\left(\frac{a+c}{2}\right)\) = ac \(\left(\because b=\frac{a+c}{2}\right)\)

⇒ m^{2}b = \(\frac{2 a c}{a+c}\)

⇒ a, m^{2}b, c are in H.P.

Summation of Series

Question 1.

1^{2} + 3^{2} + 5^{2} + ……… to ’n’ terms.

Answer:

1, 3, 5, ………. are in A.P

T_{n} = a + (n – 1)d = 1 + (n – 1)2 = 1 + 2x – 2 = 2x – 1

∴ n^{th} term = (2n – 1)^{2}

T_{n} = (2x – 1)^{2} = 4n^{2} – 4x + 1

required sum = S_{n} = ΣT_{n} = Σ4n^{2} – 4x + 1 = 4Σn^{2} – 4Σn + Σ1.

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

= \(\frac{n}{3}\) [2(n + 1)(2n + 1) – 6(n + 1) + 3] = \(\frac{n}{3}\)[4n^{2} – 1]

Question 2.

Find the sum to ‘n’ terms the series 1.2 + 4.5 + 7.8 + ….. to term.

Answer:

1st factors are 1, 4, 7,

∴ n^{th}term = 1 + (n – 1) 3 = 3n – 2

2nd factor are 2, 5, 8,

∴ n^{th} term = 2 + (n – 1) 3 = 3n – 1

⇒ n^{th} term of the series is (3n – 2)(3n – 1)

∴ The sum = Σ(3n – 2)(3n – 1)

= Σ(9x^{2} – 9x + 2) = 9Σn^{2} -9Σn + 2Σ1

\(=9 \frac{n(n+1)(2 n+1)}{6}-9 \frac{n(n+1)}{2}+2 n\)

= \(\frac{1}{2}\)n(n + 1)[3(2n + 1) – 9] + 2n = n(3n^{2} – 1).

Question 3.

Sum the series \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots \ldots .16 \text { terms }\) 16terms

Answer: