1st PUC Basic Maths Question Bank Chapter 5 Progressions

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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 – 4n²
⇒ 4n² – 52n + 144 = 0
⇒ n² – 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 S₈₁ = 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 T₂ = ar = 3, T₅ = ar⁴ = \(\frac{81}{8}\)

∴ 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 = \(\frac{39}{10}\)

⇒ a³ = 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 + 10r³ = 39
r ⇒ 10r² – 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, ar² which are is GP.
Given a + ar + ar² = 35 ⇒ a( 1 + r + r²) = 35
and a² + a²r² + a²r⁴ = 525
⇒ a²( 1 + r² + r⁴) = 525
\(\frac{a^{2}\left(1+r^{2}+r^{4}\right)}{a\left(1+r+r^{2}\right)}=\frac{525}{35}=15\)
a(1 + r² + r⁴) = 15(1 + r + r²)

⇒ a(1 – r + r²) = 15 …….. (2)
Add (1) and (2)
2a + 2ar² = 50
all (2) – (1) ⇒ 2ar = -20

\(\frac{1}{r}\) + r = \(\frac{-5}{2}\)
⇒ 2(1 + r²) = -5r
⇒ 2r² + 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 + 10² + 10³ + …… 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 g₁, g₂, g₃, g₄, g₅, g₆
= 27, g₁, g₂, g₃, g₄, g₅, g₆, \(\frac{1}{81}\) are in GP
Now a = 27 and \(\frac{1}{81}\) = 8^th element
⇒ \(\frac{1}{81}\) = ar⁷ ⇒ \(\frac{1}{81}\) = 27.r⁷
⇒ \(\frac{1}{81 \times 27}\) = r⁷
⇒ r⁷ = \(\frac{1}{3^{7}}\)
⇒ 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, m²b, c are in H.P
Answer:
a, b, c, are in AP .
⇒ b = \(\frac{a+c}{2}\)
a, mb, c are in GP,
⇒ mb = \(\sqrt{a c}\)
⇒ m²b² = ac
⇒ m²b. b. = ac
⇒ m²b \(\left(\frac{a+c}{2}\right)\) = ac \(\left(\because b=\frac{a+c}{2}\right)\)
⇒ m²b = \(\frac{2 a c}{a+c}\)
⇒ a, m²b, c are in H.P.

Summation of Series

Question 1.
1² + 3² + 5² + ………  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 = 4n² – 4x + 1
required sum = S_n = ΣT_n = Σ4n² – 4x + 1 = 4Σn² – 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] = \(\frac{n}{3}\)[4n² – 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^thterm = 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(9x² – 9x + 2) = 9Σn² -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(3n² – 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: