Write the first five terms of each of the sequences in Exercises 1 to 6 whose \( n^{th} \) terms are:
Q1
\( a_n = n(n + 2) \)
Answer:
Substitute \( n = 1, 2, 3, 4, 5 \):
- \( a_1 = 1(1 + 2) = 3 \)
- \( a_2 = 2(2 + 2) = 8 \)
- \( a_3 = 3(3 + 2) = 15 \)
- \( a_4 = 4(4 + 2) = 24 \)
- \( a_5 = 5(5 + 2) = 35 \)
The first five terms are 3, 8, 15, 24, 35.
Q2
\( a_n = \frac{n}{n+1} \)
Answer:
- \( a_1 = \frac{1}{1+1} = \frac{1}{2} \)
- \( a_2 = \frac{2}{2+1} = \frac{2}{3} \)
- \( a_3 = \frac{3}{3+1} = \frac{3}{4} \)
- \( a_4 = \frac{4}{4+1} = \frac{4}{5} \)
- \( a_5 = \frac{5}{5+1} = \frac{5}{6} \)
The terms are \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \).
Q3
\( a_n = 2^n \)
Answer:
- \( a_1 = 2^1 = 2 \)
- \( a_2 = 2^2 = 4 \)
- \( a_3 = 2^3 = 8 \)
- \( a_4 = 2^4 = 16 \)
- \( a_5 = 2^5 = 32 \)
The terms are 2, 4, 8, 16, 32.
Q4
\( a_n = \frac{2n - 3}{6} \)
Answer:
- \( a_1 = \frac{2(1)-3}{6} = -\frac{1}{6} \)
- \( a_2 = \frac{2(2)-3}{6} = \frac{1}{6} \)
- \( a_3 = \frac{2(3)-3}{6} = \frac{3}{6} = \frac{1}{2} \)
- \( a_4 = \frac{2(4)-3}{6} = \frac{5}{6} \)
- \( a_5 = \frac{2(5)-3}{6} = \frac{7}{6} \)
The terms are \( -\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6} \).
Q5
\( a_n = (-1)^{n-1} 5^{n+1} \)
Answer:
- \( a_1 = (-1)^{0} 5^{2} = 25 \)
- \( a_2 = (-1)^{1} 5^{3} = -125 \)
- \( a_3 = (-1)^{2} 5^{4} = 625 \)
- \( a_4 = (-1)^{3} 5^{5} = -3125 \)
- \( a_5 = (-1)^{4} 5^{6} = 15625 \)
The terms are 25, -125, 625, -3125, 15625.
Q6
\( a_n = n \frac{n^2 + 5}{4} \)
Answer:
- \( a_1 = 1 \cdot \frac{1+5}{4} = \frac{6}{4} = \frac{3}{2} \)
- \( a_2 = 2 \cdot \frac{4+5}{4} = 2 \cdot \frac{9}{4} = \frac{9}{2} \)
- \( a_3 = 3 \cdot \frac{9+5}{4} = 3 \cdot \frac{14}{4} = \frac{21}{2} \)
- \( a_4 = 4 \cdot \frac{16+5}{4} = 21 \)
- \( a_5 = 5 \cdot \frac{25+5}{4} = 5 \cdot \frac{30}{4} = \frac{75}{2} \)
The terms are \( \frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \frac{75}{2} \).
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose \( n^{th} \) terms are:
Q7
\( a_n = 4n - 3; \quad a_{17}, a_{24} \)
Answer:
\( a_{17} = 4(17) - 3 = 68 - 3 = 65 \)
\( a_{24} = 4(24) - 3 = 96 - 3 = 93 \)
Q8
\( a_n = \frac{n^2}{2^n}; \quad a_7 \)
Answer:
\( a_7 = \frac{7^2}{2^7} = \frac{49}{128} \)
Q9
\( a_n = (-1)^{n-1} n^3; \quad a_9 \)
Answer:
\( a_9 = (-1)^{9-1} 9^3 = (-1)^8 (729) = 729 \)
Q10
\( a_n = \frac{n(n-2)}{n+3}; \quad a_{20} \)
Answer:
\( a_{20} = \frac{20(20-2)}{20+3} = \frac{20(18)}{23} = \frac{360}{23} \)
Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:
Q11
\( a_1 = 3, \quad a_n = 3a_{n-1} + 2 \) for all \( n > 1 \)
Answer:
- \( a_1 = 3 \)
- \( a_2 = 3a_1 + 2 = 3(3) + 2 = 11 \)
- \( a_3 = 3a_2 + 2 = 3(11) + 2 = 35 \)
- \( a_4 = 3a_3 + 2 = 3(35) + 2 = 107 \)
- \( a_5 = 3a_4 + 2 = 3(107) + 2 = 323 \)
Series: \( 3 + 11 + 35 + 107 + 323 + \dots \)
Q12
\( a_1 = -1, \quad a_n = \frac{a_{n-1}}{n}, \quad n \ge 2 \)
Answer:
- \( a_1 = -1 \)
- \( a_2 = \frac{a_1}{2} = -\frac{1}{2} \)
- \( a_3 = \frac{a_2}{3} = \frac{-1/2}{3} = -\frac{1}{6} \)
- \( a_4 = \frac{a_3}{4} = \frac{-1/6}{4} = -\frac{1}{24} \)
- \( a_5 = \frac{a_4}{5} = \frac{-1/24}{5} = -\frac{1}{120} \)
Series: \( -1 + (-\frac{1}{2}) + (-\frac{1}{6}) + (-\frac{1}{24}) + (-\frac{1}{120}) + \dots \)
Q13
\( a_1 = a_2 = 2, \quad a_n = a_{n-1} - 1, \quad n > 2 \)
Answer:
- \( a_1 = 2 \)
- \( a_2 = 2 \)
- \( a_3 = a_2 - 1 = 2 - 1 = 1 \)
- \( a_4 = a_3 - 1 = 1 - 1 = 0 \)
- \( a_5 = a_4 - 1 = 0 - 1 = -1 \)
Series: \( 2 + 2 + 1 + 0 + (-1) + \dots \)
Q14
The Fibonacci sequence is defined by \( 1 = a_1 = a_2 \) 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:
First, calculate the terms of the Fibonacci sequence:
- \( a_1 = 1 \)
- \( a_2 = 1 \)
- \( a_3 = 1 + 1 = 2 \)
- \( a_4 = 2 + 1 = 3 \)
- \( a_5 = 3 + 2 = 5 \)
- \( a_6 = 5 + 3 = 8 \)
Now find the ratios \( \frac{a_{n+1}}{a_n} \):
- For \( n=1: \frac{a_2}{a_1} = \frac{1}{1} = 1 \)
- For \( n=2: \frac{a_3}{a_2} = \frac{2}{1} = 2 \)
- For \( n=3: \frac{a_4}{a_3} = \frac{3}{2} = 1.5 \)
- For \( n=4: \frac{a_5}{a_4} = \frac{5}{3} \)
- For \( n=5: \frac{a_6}{a_5} = \frac{8}{5} \)
Q1
Find the 20th and \( n^{th} \) terms of the G.P. \( \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots \)
Answer:
First term \( a = \frac{5}{2} \).
Common ratio \( r = \frac{5/4}{5/2} = \frac{1}{2} \).
\( a_{20} = ar^{19} = \frac{5}{2} \left( \frac{1}{2} \right)^{19} = \frac{5}{2^{20}} \).
\( a_n = ar^{n-1} = \frac{5}{2} \left( \frac{1}{2} \right)^{n-1} = \frac{5}{2^n} \).
Q2
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Answer:
Given \( a_8 = 192, r = 2 \).
\( a r^7 = 192 \Rightarrow a(2^7) = 192 \Rightarrow a(128) = 192 \Rightarrow a = \frac{192}{128} = \frac{3}{2} \).
\( a_{12} = ar^{11} = \frac{3}{2} (2)^{11} = 3 \cdot 2^{10} = 3(1024) = 3072 \).
Q3
The 5th, 8th and 11th terms of a G.P. are \( p, q \) and \( s \), respectively. Show that \( q^2 = ps \).
Answer:
\( p = ar^4 \), \( q = ar^7 \), \( s = ar^{10} \).
LHS: \( q^2 = (ar^7)^2 = a^2 r^{14} \).
RHS: \( ps = (ar^4)(ar^{10}) = a^2 r^{14} \).
Since LHS = RHS, \( q^2 = ps \). Hence Proved.
Q4
The 4th term of a G.P. is square of its second term, and the first term is -3. Determine its 7th term.
Answer:
Given \( a = -3 \).
\( a_4 = (a_2)^2 \Rightarrow ar^3 = (ar)^2 \Rightarrow ar^3 = a^2 r^2 \).
Since \( a \ne 0, r \ne 0 \), dividing by \( ar^2 \): \( r = a = -3 \).
\( a_7 = ar^6 = (-3)(-3)^6 = (-3)^7 = -2187 \).
Q5
Which term of the following sequences:
(a) \( 2, 2\sqrt{2}, 4, \dots \) is 128?
(b) \( \sqrt{3}, 3, 3\sqrt{3}, \dots \) is 729?
(c) \( \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots \) is \( \frac{1}{19683} \)?
(a) \( 2, 2\sqrt{2}, 4, \dots \) is 128?
(b) \( \sqrt{3}, 3, 3\sqrt{3}, \dots \) is 729?
(c) \( \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots \) is \( \frac{1}{19683} \)?
Answer:
(a) \( a=2, r=\sqrt{2} \). \( a_n = 2(\sqrt{2})^{n-1} = 128 \).
\( (\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12} \Rightarrow n-1 = 12 \Rightarrow n = 13 \).
(b) \( a=\sqrt{3}, r=\sqrt{3} \). \( a_n = \sqrt{3}(\sqrt{3})^{n-1} = (\sqrt{3})^n = 729 \).
\( 729 = 3^6 = (\sqrt{3})^{12} \Rightarrow n = 12 \).
(c) \( a=\frac{1}{3}, r=\frac{1}{3} \). \( a_n = (\frac{1}{3})^n = \frac{1}{19683} \).
\( 19683 = 3^9 \), so \( n = 9 \).
Q6
For what values of \( x \), the numbers \( -\frac{2}{7}, x, -\frac{7}{2} \) are in G.P.?
Answer:
For terms in GP, \( b^2 = ac \).
\( x^2 = \left( -\frac{2}{7} \right) \left( -\frac{7}{2} \right) = 1 \).
\( x = \pm 1 \).
Q7
Find the sum to 20 terms: \( 0.15, 0.015, 0.0015, \dots \)
Answer:
\( a = 0.15, r = 0.1 = \frac{1}{10} \).
\( S_{20} = \frac{a(1-r^{20})}{1-r} = \frac{0.15(1 - 0.1^{20})}{1 - 0.1} = \frac{0.15}{0.9}(1 - 0.1^{20}) \).
\( = \frac{1}{6} (1 - (0.1)^{20}) \).
Q8
Find sum to \( n \) terms: \( \sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots \)
Answer:
\( a = \sqrt{7}, r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3} \).
\( S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1} \).
Rationalizing the denominator:
\( S_n = \frac{\sqrt{7}(3^{n/2} - 1)(\sqrt{3} + 1)}{2} \).
Q9
Find sum to \( n \) terms: \( 1, -a, a^2, -a^3, \dots \) (if \( a \ne -1 \))
Answer:
\( a = 1, r = -a \).
\( S_n = \frac{1(1 - (-a)^n)}{1 - (-a)} = \frac{1 - (-a)^n}{1 + a} \).
Q10
Find sum to \( n \) terms: \( x^3, x^5, x^7, \dots \) (if \( x \ne \pm 1 \))
Answer:
\( a = x^3, r = x^2 \).
\( S_n = \frac{x^3((x^2)^n - 1)}{x^2 - 1} = \frac{x^3(x^{2n} - 1)}{x^2 - 1} \).
Q11
Evaluate \( \sum_{k=1}^{11} (2 + 3^k) \)
Answer:
\( \sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^k \)
\( = (2 \times 11) + (3 + 3^2 + \dots + 3^{11}) \)
\( = 22 + \frac{3(3^{11} - 1)}{3 - 1} = 22 + \frac{3}{2}(3^{11} - 1) \).
Q12
The sum of first three terms of a G.P. is \( \frac{39}{10} \) and their product is 1. Find the terms.
Answer:
Let terms be \( \frac{a}{r}, a, ar \).
Product: \( a^3 = 1 \implies a = 1 \).
Sum: \( \frac{1}{r} + 1 + r = \frac{39}{10} \)
\( \frac{1+r+r^2}{r} = \frac{39}{10} \Rightarrow 10 + 10r + 10r^2 = 39r \)
\( 10r^2 - 29r + 10 = 0 \).
\( (2r-5)(5r-2) = 0 \Rightarrow r = \frac{5}{2} \) or \( \frac{2}{5} \).
Terms are: \( \frac{2}{5}, 1, \frac{5}{2} \) or \( \frac{5}{2}, 1, \frac{2}{5} \).
Q13
How many terms of G.P. \( 3, 3^2, 3^3, \dots \) are needed to give the sum 120?
Answer:
\( a=3, r=3 \).
\( S_n = \frac{3(3^n - 1)}{3-1} = 120 \Rightarrow \frac{3}{2}(3^n - 1) = 120 \)
\( 3^n - 1 = \frac{240}{3} = 80 \Rightarrow 3^n = 81 \Rightarrow n = 4 \).
Q14
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.
Answer:
\( S_3 = a(1+r+r^2) = 16 \).
Sum of next three (terms 4, 5, 6) \( = ar^3(1+r+r^2) = 128 \).
Ratio: \( \frac{ar^3(1+r+r^2)}{a(1+r+r^2)} = \frac{128}{16} \Rightarrow r^3 = 8 \Rightarrow r = 2 \).
\( a(1+2+4) = 16 \Rightarrow 7a = 16 \Rightarrow a = \frac{16}{7} \).
\( S_n = \frac{16}{7} \frac{(2^n - 1)}{2-1} = \frac{16}{7}(2^n - 1) \).
Q15
Given a G.P. with \( a = 729 \) and 7th term 64, determine \( S_7 \).
Answer:
\( a_7 = ar^6 = 729 r^6 = 64 \Rightarrow r^6 = \frac{64}{729} = \left( \frac{2}{3} \right)^6 \Rightarrow r = \frac{2}{3} \).
\( S_7 = \frac{729(1 - (2/3)^7)}{1 - 2/3} = \frac{729(1 - \frac{128}{2187})}{1/3} \)
\( = 3 \times 729 \times \frac{2059}{2187} = 2187 \times \frac{2059}{2187} = 2059 \).
Q16
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:
\( a_5 = 4a_3 \Rightarrow ar^4 = 4ar^2 \Rightarrow r^2 = 4 \Rightarrow r = \pm 2 \).
Sum of first two: \( a + ar = -4 \).
Case 1 (\( r=2 \)): \( a(1+2) = -4 \Rightarrow 3a = -4 \Rightarrow a = -4/3 \). GP: \( -4/3, -8/3, -16/3, \dots \)
Case 2 (\( r=-2 \)): \( a(1-2) = -4 \Rightarrow -a = -4 \Rightarrow a = 4 \). GP: \( 4, -8, 16, -32, \dots \)
Q17
If the 4th, 10th and 16th terms of a G.P. are \( x, y, z \) respectively. Prove that \( x, y, z \) are in G.P.
Answer:
\( x = ar^3, y = ar^9, z = ar^{15} \).
\( y^2 = (ar^9)^2 = a^2 r^{18} \).
\( xz = (ar^3)(ar^{15}) = a^2 r^{18} \).
Since \( y^2 = xz \), terms are in G.P.
Q18
Find the sum to \( n \) terms of the sequence \( 8, 88, 888, 8888 \dots \)
Answer:
\( S_n = 8 + 88 + 888 + \dots = \frac{8}{9}(9 + 99 + 999 + \dots) \)
\( = \frac{8}{9} [ (10-1) + (100-1) + (1000-1) + \dots ] \)
\( = \frac{8}{9} [ (10 + 10^2 + \dots + 10^n) - n ] \)
\( = \frac{8}{9} \left[ \frac{10(10^n - 1)}{9} - n \right] \).
Q19
Find the sum of the products of the corresponding terms of the sequences \( 2, 4, 8, 16, 32 \) and \( 128, 32, 8, 2, \frac{1}{2} \).
Answer:
Product terms: \( 2(128), 4(32), 8(8), 16(2), 32(1/2) \)
\( 256, 128, 64, 32, 16 \). This is a GP with \( a=256, r=1/2 \).
\( S_5 = \frac{256(1 - (1/2)^5)}{1 - 1/2} = \frac{256(31/32)}{1/2} = 512 \times \frac{31}{32} = 16 \times 31 = 496 \).
Q20
Show that the products of the corresponding terms of the sequences \( a, ar, ar^2, \dots \) and \( A, AR, AR^2, \dots \) form a G.P.
Answer:
Products: \( aA, (ar)(AR), (ar^2)(AR^2), \dots \)
\( = aA, (aA)(rR), (aA)(rR)^2, \dots \)
This is clearly a GP with first term \( aA \) and common ratio \( rR \).
Q21
Find four numbers forming a G.P. in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Answer:
\( ar^2 = a + 9 \implies a(r^2 - 1) = 9 \)
\( ar = ar^3 + 18 \implies ar(1 - r^2) = 18 \)
Divide second by first: \( \frac{ar(1-r^2)}{a(r^2-1)} = \frac{18}{9} \)
\( \frac{-ar(r^2-1)}{a(r^2-1)} = 2 \implies -r = 2 \implies r = -2 \).
Substitute \( r = -2 \): \( a(4-1) = 9 \Rightarrow 3a = 9 \Rightarrow a = 3 \).
Numbers: 3, -6, 12, -24.
Q22
If the \( p^{th}, q^{th} \) and \( r^{th} \) terms of a G.P. are \( a, b \) and \( c \), prove that \( a^{q-r} b^{r-p} c^{p-q} = 1 \).
Answer:
Let first term be \( A \) and ratio \( R \).
\( a = AR^{p-1}, b = AR^{q-1}, c = AR^{r-1} \).
\( a^{q-r} b^{r-p} c^{p-q} = (AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q} \)
Power of A: \( (q-r) + (r-p) + (p-q) = 0 \).
Power of R: \( (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) \)
Expanding and summing exponents gives 0.
Result \( A^0 R^0 = 1 \). Proved.
Q23
If the first and the \( n^{th} \) term of a G.P. are \( a \) and \( b \), and \( P \) is the product of \( n \) terms, prove \( P^2 = (ab)^n \).
Answer:
\( b = ar^{n-1} \).
\( P = a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{1+2+\dots+(n-1)} = a^n r^{\frac{n(n-1)}{2}} \).
\( P^2 = a^{2n} r^{n(n-1)} = (a^2 r^{n-1})^n = (a \cdot ar^{n-1})^n = (ab)^n \). Proved.
Q24
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 \( \frac{1}{r^n} \).
Answer:
Sum 1: \( S_n = \frac{a(1-r^n)}{1-r} \).
Sum 2: First term is \( ar^n \). Sum is \( \frac{ar^n(1-r^n)}{1-r} \).
Ratio: \( \frac{\frac{a(1-r^n)}{1-r}}{\frac{ar^n(1-r^n)}{1-r}} = \frac{1}{r^n} \).
Q25
If \( a, b, c, d \) are in G.P. show that \( (a^2+b^2+c^2)(b^2+c^2+d^2) = (ab+bc+cd)^2 \).
Answer:
Let \( b=ar, c=ar^2, d=ar^3 \).
LHS \( = (a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6) = a^2(1+r^2+r^4) \cdot a^2r^2(1+r^2+r^4) = a^4r^2(1+r^2+r^4)^2 \).
RHS \( = (a(ar) + ar(ar^2) + ar^2(ar^3))^2 = (a^2r + a^2r^3 + a^2r^5)^2 = [a^2r(1+r^2+r^4)]^2 = a^4r^2(1+r^2+r^4)^2 \).
LHS = RHS. Proved.
Q26
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Answer:
\( 3, G_1, G_2, 81 \). \( a=3, a_4 = ar^3 = 81 \).
\( 3r^3 = 81 \Rightarrow r^3 = 27 \Rightarrow r = 3 \).
\( G_1 = 3(3) = 9 \).
\( G_2 = 9(3) = 27 \).
Numbers are 9 and 27.
Q27
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:
Geometric Mean = \( \sqrt{ab} = a^{1/2}b^{1/2} \).
\( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = a^{1/2}b^{1/2} \)
\( a^{n+1} + b^{n+1} = a^{n+1/2}b^{1/2} + a^{n}b^{1/2}b^{n} \text{ (Wait, expand: } (a^n+b^n)(a^{1/2}b^{1/2}) = a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2} ) \)
\( a^{n+1} - a^{n+1/2}b^{1/2} = a^{1/2}b^{n+1/2} - b^{n+1} \)
\( a^{n+1/2}(a^{1/2} - b^{1/2}) = b^{n+1/2}(a^{1/2} - b^{1/2}) \)
\( a^{n+1/2} = b^{n+1/2} \Rightarrow \left( \frac{a}{b} \right)^{n+1/2} = 1 = \left( \frac{a}{b} \right)^0 \).
\( n + \frac{1}{2} = 0 \Rightarrow n = -\frac{1}{2} \).
Q28
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio \( (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \).
Answer:
\( a+b = 6\sqrt{ab} \).
\( \frac{a+b}{2\sqrt{ab}} = 3 \). Using componendo and dividend:
\( \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{3+1}{3-1} = \frac{4}{2} = 2 \)
\( \left( \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} \right)^2 = 2 \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{2} \).
Again apply C&D: \( \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1} \).
Square both sides: \( \frac{a}{b} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}} \). Proved.
Q29
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:
Let numbers be \( a, b \). \( A = \frac{a+b}{2}, G = \sqrt{ab} \).
Quadratic eq with roots \( a, b \): \( x^2 - (a+b)x + ab = 0 \).
\( x^2 - 2Ax + G^2 = 0 \).
\( x = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} = A \pm \sqrt{A^2 - G^2} = A \pm \sqrt{(A+G)(A-G)} \). Proved.
Q30
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 2nd hour, 4th hour and \( n^{th} \) hour?
Answer:
Sequence: 30, 60, 120... (GP with \( a=30, r=2 \)).
After 2nd hour (3rd term): \( a_3 = 30(2)^2 = 120 \).
After 4th hour (5th term): \( a_5 = 30(2)^4 = 480 \).
After \( n^{th} \) hour (\((n+1)^{th}\) term): \( 30(2^n) \).
Q31
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:
Formula \( A = P(1 + \frac{R}{100})^n \).
\( P = 500, R = 10, n = 10 \).
\( A = 500(1 + 0.1)^{10} = 500(1.1)^{10} \).
Q32
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Answer:
Let roots be \( a, b \).
\( \frac{a+b}{2} = 8 \Rightarrow a+b = 16 \).
\( \sqrt{ab} = 5 \Rightarrow ab = 25 \).
Quadratic Equation: \( x^2 - (a+b)x + ab = 0 \).
\( x^2 - 16x + 25 = 0 \).
Q1
If \( f \) is a function satisfying \( f(x+y) = f(x)f(y) \) for all \( x, y \in N \) such that \( f(1) = 3 \) and \( \sum_{x=1}^{n} f(x) = 120 \), find the value of \( n \).
Answer:
Given \( f(x+y) = f(x)f(y) \) and \( f(1) = 3 \).
\( f(2) = f(1+1) = f(1)f(1) = 3 \times 3 = 3^2 = 9 \).
\( f(3) = f(2+1) = f(2)f(1) = 3^2 \times 3 = 3^3 = 27 \).
In general, \( f(x) = 3^x \).
The sequence \( f(1), f(2), f(3), \dots \) is a G.P. with \( a=3, r=3 \).
Given sum \( S_n = 120 \).
\( \frac{a(r^n - 1)}{r - 1} = 120 \)
\( \frac{3(3^n - 1)}{3 - 1} = 120 \)
\( \frac{3}{2}(3^n - 1) = 120 \)
\( 3^n - 1 = \frac{120 \times 2}{3} = 80 \)
\( 3^n = 81 = 3^4 \). Thus, \( n = 4 \).
Q2
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:
\( a = 5, r = 2, S_n = 315 \).
\( S_n = \frac{a(r^n - 1)}{r - 1} \Rightarrow 315 = \frac{5(2^n - 1)}{2 - 1} \).
\( 315 = 5(2^n - 1) \Rightarrow 63 = 2^n - 1 \Rightarrow 2^n = 64 \).
\( 2^n = 2^6 \Rightarrow n = 6 \). Number of terms is 6.
Last term \( l = a_6 = ar^5 = 5(2)^5 = 5(32) = 160 \).
Q3
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:
\( a = 1 \).
\( a_3 + a_5 = 90 \Rightarrow ar^2 + ar^4 = 90 \).
\( r^2 + r^4 = 90 \). Let \( r^2 = x \).
\( x^2 + x - 90 = 0 \Rightarrow (x+10)(x-9) = 0 \).
\( x = 9 \) or \( x = -10 \). Since \( r^2 \) cannot be negative, \( r^2 = 9 \).
\( r = \pm 3 \).
Q4
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 numbers in G.P. be \( a, ar, ar^2 \).
Sum: \( a(1+r+r^2) = 56 \) ... (i)
New numbers: \( a-1, ar-7, ar^2-21 \) are in A.P.
\( 2(ar-7) = (a-1) + (ar^2-21) \)
\( 2ar - 14 = a + ar^2 - 22 \Rightarrow a + ar^2 - 2ar = 8 \)
\( a(r^2 - 2r + 1) = 8 \Rightarrow a(r-1)^2 = 8 \) ... (ii)
Divide (i) by (ii): \( \frac{1+r+r^2}{r^2-2r+1} = \frac{56}{8} = 7 \).
\( 1+r+r^2 = 7(r^2-2r+1) \Rightarrow 6r^2 - 15r + 6 = 0 \).
\( 2r^2 - 5r + 2 = 0 \Rightarrow (2r-1)(r-2) = 0 \).
\( r = 2 \) or \( r = \frac{1}{2} \).
If \( r=2, a(1)^2 = 8 \Rightarrow a=8 \). Numbers: 8, 16, 32.
If \( r=1/2, a(-1/2)^2 = 8 \Rightarrow a/4 = 8 \Rightarrow a=32 \). Numbers: 32, 16, 8.
Q5
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 number of terms be \( 2n \). Sequence: \( a, ar, ar^2, \dots, ar^{2n-1} \).
Sum of all terms \( S_{total} = \frac{a(r^{2n}-1)}{r-1} \).
Odd terms: \( a, ar^2, ar^4, \dots \) (n terms, common ratio \( r^2 \)).
Sum of odd terms \( S_{odd} = \frac{a((r^2)^n - 1)}{r^2 - 1} = \frac{a(r^{2n}-1)}{(r-1)(r+1)} \).
Given \( S_{total} = 5 S_{odd} \).
\( \frac{a(r^{2n}-1)}{r-1} = 5 \times \frac{a(r^{2n}-1)}{(r-1)(r+1)} \).
\( 1 = \frac{5}{r+1} \Rightarrow r+1 = 5 \Rightarrow r = 4 \).
Q6
If \( \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} (x \ne 0) \), then show that \( a, b, c \) and \( d \) are in G.P.
Answer:
Using componendo and dividend rule on each part:
\( \frac{(a+bx)+(a-bx)}{(a+bx)-(a-bx)} = \frac{(b+cx)+(b-cx)}{(b+cx)-(b-cx)} = \frac{(c+dx)+(c-dx)}{(c+dx)-(c-dx)} \)
\( \frac{2a}{2bx} = \frac{2b}{2cx} = \frac{2c}{2dx} \)
\( \frac{a}{b} = \frac{b}{c} = \frac{c}{d} \).
This implies \( b^2 = ac \) and \( c^2 = bd \), which means \( a, b, c, d \) are in G.P.
Q7
Let S be the sum, P the product and R the sum of reciprocals of \( n \) terms in a G.P. Prove that \( P^2R^n = S^n \).
Answer:
\( S = \frac{a(r^n-1)}{r-1} \).
\( P = a^n r^{\frac{n(n-1)}{2}} \).
\( R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}} = \frac{1}{a} \left( \frac{1 - (1/r)^n}{1 - 1/r} \right) = \frac{1}{a} \frac{(r^n-1)/r^n}{(r-1)/r} = \frac{r^n-1}{a r^{n-1} (r-1)} \).
\( R = \frac{S}{a^2 r^{n-1}} \). So \( S = R a^2 r^{n-1} \).
LHS \( P^2 R^n = (a^n r^{\frac{n(n-1)}{2}})^2 R^n = a^{2n} r^{n(n-1)} R^n = (a^2 r^{n-1})^n R^n = (R a^2 r^{n-1})^n = S^n \). Proved.
Q8
If \( a, b, c, d \) are in G.P, prove that \( (a^n + b^n), (b^n + c^n), (c^n + d^n) \) are in G.P.
Answer:
Let \( b=ar, c=ar^2, d=ar^3 \).
Terms: \( a^n(1+r^n), a^n r^n(1+r^n), a^n r^{2n}(1+r^n) \).
Ratio 1: \( \frac{a^n r^n(1+r^n)}{a^n(1+r^n)} = r^n \).
Ratio 2: \( \frac{a^n r^{2n}(1+r^n)}{a^n r^n(1+r^n)} = r^n \).
Since ratios are equal, they are in G.P.
Q9
If \( a \) and \( b \) are the roots of \( x^2 - 3x + p = 0 \) and \( c, d \) are roots of \( x^2 - 12x + q = 0 \), where \( a, b, c, d \) form a G.P. Prove that \( (q+p):(q-p) = 17:15 \).
Answer:
\( a+b=3, ab=p \). \( c+d=12, cd=q \).
Let terms be \( a, ar, ar^2, ar^3 \).
\( a(1+r) = 3 \) and \( ar^2(1+r) = 12 \).
Dividing: \( r^2 = 4 \Rightarrow r = 2 \) (assuming increasing GP).
\( a(3) = 3 \Rightarrow a = 1 \). So \( a=1, b=2, c=4, d=8 \).
\( p = ab = 2 \), \( q = cd = 32 \).
\( \frac{q+p}{q-p} = \frac{32+2}{32-2} = \frac{34}{30} = \frac{17}{15} \). Proved.
Q10
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:
\( \frac{A}{G} = \frac{a+b}{2\sqrt{ab}} = \frac{m}{n} \).
Using componendo and dividend:
\( \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{2m+2n}{2m-2n} \text{ (Wait, standard way)} \).
\( \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = \frac{m+n}{m-n} \).
\( \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{\frac{m+n}{m-n}} \).
Again C&D: \( \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \).
Square both sides: \( \frac{a}{b} = \frac{m+n + m-n + 2\sqrt{m^2-n^2}}{m+n + m-n - 2\sqrt{m^2-n^2}} = \frac{2m + 2\sqrt{m^2-n^2}}{2m - 2\sqrt{m^2-n^2}} \).
\( \frac{a}{b} = \frac{m + \sqrt{m^2-n^2}}{m - \sqrt{m^2-n^2}} \). Proved.
Q11
Find the sum of the following series up to \( n \) terms:
(i) \( 5 + 55 + 555 + \dots \)
(ii) \( .6 + .66 + .666 + \dots \)
(i) \( 5 + 55 + 555 + \dots \)
(ii) \( .6 + .66 + .666 + \dots \)
Answer:
(i) \( S = 5(1+11+111+\dots) = \frac{5}{9}(9+99+999+\dots) \)
\( = \frac{5}{9}[(10-1) + (100-1) + \dots] = \frac{5}{9} [ \frac{10(10^n-1)}{9} - n ] \).
(ii) \( S = 6(.1+.11+.111+\dots) = \frac{6}{9}(.9+.99+.999+\dots) \)
\( = \frac{2}{3} [ (1-0.1) + (1-0.01) + \dots ] = \frac{2}{3} [ n - \frac{1/10(1-(1/10)^n)}{1-1/10} ] \)
\( = \frac{2}{3} [ n - \frac{1}{9}(1 - 10^{-n}) ] \).
Q12
Find the \( 20^{th} \) term of the series \( 2 \times 4 + 4 \times 6 + 6 \times 8 + \dots + n \) terms.
Answer:
\( n^{th} \) term \( a_n = (2n) \times (2n+2) \).
\( a_{20} = (2 \times 20) \times (2 \times 20 + 2) = 40 \times 42 = 1680 \).
Q13
A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
Answer:
Balance = 6000. Installment = 500. Number of installments = 12.
Interest paid:
Year 1: \( 12\% \) of 6000 = 720
Year 2: \( 12\% \) of 5500 = 660
...
Year 12: \( 12\% \) of 500 = 60.
Total Interest = \( 720 + 660 + \dots + 60 \) (AP with \( a=720, d=-60, n=12 \)).
\( S = \frac{12}{2}(720 + 60) = 6(780) = 4680 \).
Total Cost = \( 12000 + 4680 = \) Rs 16680.
Q14
Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
Answer:
Balance = 18000. Installment = 1000. Number of installments = 18.
Interest: \( 10\% \) of (18000, 17000, ..., 1000).
Sum = \( 1800 + 1700 + \dots + 100 \).
\( S = \frac{18}{2}(1800 + 100) = 9(1900) = 17100 \).
Total Cost = \( 22000 + 17100 = \) Rs 39100.
Q15
A person writes a letter to four of his friends... cost 50 paise... Find the total amount spent when 8th set of letter is mailed.
Answer:
Number of letters at steps: 4, 16, 64, ... (GP with \( a=4, r=4 \)).
Total letters after 8th set = \( S_8 = \frac{4(4^8 - 1)}{4 - 1} = \frac{4}{3}(65536 - 1) = \frac{4}{3}(65535) = 4(21845) = 87380 \).
Cost = \( 87380 \times 0.50 \) Rs = Rs 43690.
Q16
A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year and after 20 years.
Answer:
Interest per year = \( 5\% \) of 10000 = 500.
Amount forms AP: 10000, 10500, 11000...
(i) Amount in 15th year (\( a_{15} \)): \( 10000 + (14 \times 500) = 10000 + 7000 = 17000 \).
(ii) Amount after 20 years (start of 21st): \( 10000 + (20 \times 500) = 10000 + 10000 = 20000 \).
Q17
A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Answer:
Depreciation rate = 20%. Value becomes 80% or \( \frac{4}{5} \) of previous year.
Value after 5 years = \( 15625 \times (\frac{4}{5})^5 \).
\( = 15625 \times \frac{1024}{3125} = 5 \times 1024 = \) Rs 5120.
Q18
150 workers were engaged to finish a job... 4 workers dropped out on second day, 4 more on third day... It took 8 more days to finish the work. Find number of days.
Answer:
Let original days be \( x \). Total work = \( 150x \).
Actual days \( n = x + 8 \).
Workers sequence: 150, 146, 142... (AP with \( a=150, d=-4 \)).
Sum of work done = \( \frac{n}{2}[2(150) + (n-1)(-4)] = 150(n-8) \).
\( \frac{n}{2}[300 - 4n + 4] = 150n - 1200 \)
\( n(152 - 2n) = 150n - 1200 \)
\( 152n - 2n^2 = 150n - 1200 \Rightarrow 2n^2 - 2n - 1200 = 0 \).
\( n^2 - n - 600 = 0 \Rightarrow (n-25)(n+24) = 0 \).
\( n = 25 \). The work was completed in 25 days.