Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.7

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.7

 

Question 1. Which of the following sequences are in G.P?
(i) 3,9,27,81,...
(ii) 4,44,444,4444,...
(iii) 0.5,0.05,0.005,...
(iv) \( \frac { 1 }{ 3 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \),.........
(v) -5, 25, -125,...
(vi) 120, 60, 30, 18,...
(vii) 16, 4, 1, \( \frac { 1 }{ 4 } \), ..........
Answer:
A sequence is a Geometric Progression (G.P.) if the ratio of any term to its previous term is constant. This constant is called the common ratio (r).
(i) Given sequence: 3, 9, 27, 81,... \[ \frac { t_2 }{ t_1 } = \frac { 9 }{ 3 } = 3 \] \[ \frac { t_3 }{ t_2 } = \frac { 27 }{ 9 } = 3 \] \[ \frac { t_4 }{ t_3 } = \frac { 81 }{ 27 } = 3 \] Since the common ratio is 3 and it is constant, this sequence is a G.P.
(ii) Given sequence: 4, 44, 444, 4444,... \[ \frac { t_2 }{ t_1 } = \frac { 44 }{ 4 } = 11 \] \[ \frac { t_3 }{ t_2 } = \frac { 444 }{ 44 } = \frac { 111 }{ 11 } \approx 10.09 \] Since the ratios are not equal (\( 11 \neq 10.09 \)), this sequence is not a G.P.
(iii) Given sequence: 0.5, 0.05, 0.005,... \[ \frac { t_2 }{ t_1 } = \frac { 0.05 }{ 0.5 } = \frac { 5 }{ 50 } = \frac { 1 }{ 10 } \] \[ \frac { t_3 }{ t_2 } = \frac { 0.005 }{ 0.05 } = \frac { 5 }{ 50 } = \frac { 1 }{ 10 } \] Since the common ratio is \( \frac { 1 }{ 10 } \) and it is constant, this sequence is a G.P.
(iv) Given sequence: \( \frac { 1 }{ 3 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \),......... \[ \frac { t_2 }{ t_1 } = \frac { \frac { 1 }{ 6 } }{ \frac { 1 }{ 3 } } = \frac { 1 }{ 6 } \times \frac { 3 }{ 1 } = \frac { 1 }{ 2 } \] \[ \frac { t_3 }{ t_2 } = \frac { \frac { 1 }{ 12 } }{ \frac { 1 }{ 6 } } = \frac { 1 }{ 12 } \times \frac { 6 }{ 1 } = \frac { 1 }{ 2 } \] Since the common ratio is \( \frac { 1 }{ 2 } \) and it is constant, this sequence is a G.P.
(v) Given sequence: -5, 25, -125,... \[ \frac { t_2 }{ t_1 } = \frac { 25 }{ -5 } = -5 \] \[ \frac { t_3 }{ t_2 } = \frac { -125 }{ 25 } = -5 \] Since the common ratio is -5 and it is constant, this sequence is a G.P.
(vi) Given sequence: 120, 60, 30, 18,... \[ \frac { t_2 }{ t_1 } = \frac { 60 }{ 120 } = \frac { 1 }{ 2 } \] \[ \frac { t_3 }{ t_2 } = \frac { 30 }{ 60 } = \frac { 1 }{ 2 } \] \[ \frac { t_4 }{ t_3 } = \frac { 18 }{ 30 } = \frac { 3 }{ 5 } \] Since the ratios are not equal (\( \frac { 1 }{ 2 } \neq \frac { 3 }{ 5 } \)), this sequence is not a G.P.
(vii) Given sequence: 16, 4, 1, \( \frac { 1 }{ 4 } \), .......... \[ \frac { t_2 }{ t_1 } = \frac { 4 }{ 16 } = \frac { 1 }{ 4 } \] \[ \frac { t_3 }{ t_2 } = \frac { 1 }{ 4 } \] \[ \frac { t_4 }{ t_3 } = \frac { \frac { 1 }{ 4 } }{ 1 } = \frac { 1 }{ 4 } \] Since the common ratio is \( \frac { 1 }{ 4 } \) and it is constant, this sequence is a G.P.
In simple words: To find if a list of numbers is a G.P., divide each number by the one before it. If the answer is always the same, then it is a G.P. The number you get is the common ratio.

🎯 Exam Tip: Always show at least two ratios of consecutive terms (\( \frac{t_2}{t_1} \) and \( \frac{t_3}{t_2} \)) to clearly demonstrate if a common ratio exists or not.

 

Question 2. Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) a = 6, r = 3
(ii) a = \( \sqrt { 2 } \), r = \( \sqrt { 2 } \).
(iii) a = 1000, r = \( \frac { 2 }{ 5 } \)
Answer:
The general form of a G.P. is \( a, ar, ar^2, ... \), where 'a' is the first term and 'r' is the common ratio. We use this to find the terms.
(i) Given: \( a = 6, r = 3 \) First term \( = a = 6 \) Second term \( = ar = 6 \times 3 = 18 \) Third term \( = ar^2 = 6 \times 3^2 = 6 \times 9 = 54 \) The first three terms are 6, 18, and 54.
(ii) Given: \( a = \sqrt { 2 }, r = \sqrt { 2 } \) First term \( = a = \sqrt { 2 } \) Second term \( = ar = \sqrt { 2 } \times \sqrt { 2 } = 2 \) Third term \( = ar^2 = \sqrt { 2 } \times ( \sqrt { 2 } )^2 = \sqrt { 2 } \times 2 = 2\sqrt { 2 } \) The first three terms are \( \sqrt { 2 } \), 2, and \( 2\sqrt { 2 } \).
(iii) Given: \( a = 1000, r = \frac { 2 }{ 5 } \) First term \( = a = 1000 \) Second term \( = ar = 1000 \times \frac { 2 }{ 5 } = 200 \times 2 = 400 \) Third term \( = ar^2 = 1000 \times ( \frac { 2 }{ 5 } )^2 = 1000 \times \frac { 4 }{ 25 } = 40 \times 4 = 160 \) The first three terms are 1000, 400, and 160.
In simple words: To find the terms of a G.P., start with the first term 'a'. Then multiply 'a' by the common ratio 'r' to get the second term. Multiply the second term by 'r' to get the third term, and so on.

🎯 Exam Tip: Remember the formula for the nth term of a G.P. is \( t_n = ar^{n-1} \), which is useful for calculating any specific term.

 

Question 3. In a G.P. 729, 243, 81,... find \( t_7 \).
Answer:
Given G.P. is 729, 243, 81,...
First term \( a = 729 \)
Common ratio \( r = \frac { t_2 }{ t_1 } = \frac { 243 }{ 729 } = \frac { 1 }{ 3 } \)
Alternatively, \( r = \frac { t_3 }{ t_2 } = \frac { 81 }{ 243 } = \frac { 1 }{ 3 } \). The formula for the nth term of a G.P. is \( t_n = ar^{n-1} \). We need to find the 7th term, so \( n = 7 \). \( t_7 = a r^{7-1} = a r^6 \) \( t_7 = 729 \times ( \frac { 1 }{ 3 } )^6 \) We know that \( 729 = 3^6 \).
\( \implies \) \( t_7 = 3^6 \times \frac { 1 }{ 3^6 } \)
\( \implies \) \( t_7 = 1 \) Therefore, the 7th term of the G.P. is 1.
In simple words: First, find the starting number (a) and how much it changes each time (r). Then use the formula \( t_n = ar^{n-1} \) to find the number at the 7th spot.

🎯 Exam Tip: It is often helpful to express the first term 'a' as a power of the common ratio 'r' (if possible) to simplify calculations when finding higher terms.

 

Question 4. Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.
Answer:
If x + 6, x + 12, and x + 15 are consecutive terms in a G.P., then the common ratio between them must be the same. So, \( \frac { t_2 }{ t_1 } = \frac { t_3 }{ t_2 } \)
\( \implies \) \( \frac { x + 12 }{ x + 6 } = \frac { x + 15 }{ x + 12 } \) Now, we cross-multiply to solve for x:
\( \implies \) \( (x + 12)^2 = (x + 6)(x + 15) \) Expand both sides of the equation:
\( \implies \) \( x^2 + 24x + 144 = x^2 + 15x + 6x + 90 \)
\( \implies \) \( x^2 + 24x + 144 = x^2 + 21x + 90 \) Subtract \( x^2 \) from both sides and gather like terms:
\( \implies \) \( 24x - 21x = 90 - 144 \)
\( \implies \) \( 3x = -54 \) Divide by 3 to find x:
\( \implies \) \( x = \frac { -54 }{ 3 } \)
\( \implies \) \( x = -18 \) Thus, the value of x is -18. This makes the terms -12, -6, -3, forming a G.P. with a common ratio of 1/2.
In simple words: For three numbers to be in a G.P., the ratio of the second to the first number must be the same as the ratio of the third to the second. Use this rule to set up an equation and find x.

🎯 Exam Tip: Always remember the property of G.P. that \( t_2^2 = t_1 \times t_3 \) (the square of the middle term equals the product of the other two) for three consecutive terms, which simplifies the equation setup.

 

Question 5. Find the number of terms in the following G.P.
(i) 4,8,16,...,8192?
(ii) \( \frac { 1 }{ 3 } \), \( \frac { 1 }{9} \), \( \frac { 1 }{ 27 } \), ..............., \( \frac { 1 }{ 2187 } \)
Answer:
(i) Given G.P.: 4, 8, 16,..., 8192.
First term \( a = 4 \)
Common ratio \( r = \frac { t_2 }{ t_1 } = \frac { 8 }{ 4 } = 2 \)
The last term \( t_n = 8192 \). The formula for the nth term is \( t_n = ar^{n-1} \).
\( \implies \) \( 8192 = 4 \times 2^{n-1} \) Divide both sides by 4:
\( \implies \) \( \frac { 8192 }{ 4 } = 2^{n-1} \)
\( \implies \) \( 2048 = 2^{n-1} \) To find 'n-1', we express 2048 as a power of 2. We can do this by repeatedly dividing 2048 by 2:

We find that \( 2048 = 2^{11} \). So, \( 2^{n-1} = 2^{11} \) Equating the exponents:
\( \implies \) \( n - 1 = 11 \)
\( \implies \) \( n = 11 + 1 \)
\( \implies \) \( n = 12 \) The number of terms in the G.P. is 12.
(ii) Given G.P.: \( \frac { 1 }{ 3 } \), \( \frac { 1 }{9} \), \( \frac { 1 }{ 27 } \), ..............., \( \frac { 1 }{ 2187 } \)
First term \( a = \frac { 1 }{ 3 } \)
Common ratio \( r = \frac { t_2 }{ t_1 } = \frac { \frac { 1 }{ 9 } }{ \frac { 1 }{ 3 } } = \frac { 1 }{ 9 } \times \frac { 3 }{ 1 } = \frac { 1 }{ 3 } \) The last term \( t_n = \frac { 1 }{ 2187 } \). The formula for the nth term is \( t_n = ar^{n-1} \).
\( \implies \) \( \frac { 1 }{ 2187 } = \frac { 1 }{ 3 } \times ( \frac { 1 }{ 3 } )^{n-1} \)
\( \implies \) \( \frac { 1 }{ 2187 } = ( \frac { 1 }{ 3 } )^{1} \times ( \frac { 1 }{ 3 } )^{n-1} \) Using the rule \( a^m \times a^n = a^{m+n} \):
\( \implies \) \( \frac { 1 }{ 2187 } = ( \frac { 1 }{ 3 } )^{1 + n - 1} \)
\( \implies \) \( \frac { 1 }{ 2187 } = ( \frac { 1 }{ 3 } )^{n} \) Now, express 2187 as a power of 3:We find that \( 2187 = 3^7 \). So, \( \frac { 1 }{ 3^7 } = ( \frac { 1 }{ 3 } )^{n} \)
\( \implies \) \( ( \frac { 1 }{ 3 } )^7 = ( \frac { 1 }{ 3 } )^{n} \) Equating the exponents:
\( \implies \) \( n = 7 \) The number of terms in the G.P. is 7.
In simple words: To find how many terms are in a G.P., use the first term, the common ratio, and the last term in the G.P. formula. Then, solve the equation by making the bases of the powers the same.

🎯 Exam Tip: When solving for 'n' in \( t_n = ar^{n-1} \), always try to express numbers as powers of the common ratio 'r' to simplify the exponential equation.

 

Question 6. In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.
Answer:
Let the first term of the G.P. be 'a' and the common ratio be 'r'. The formula for the nth term of a G.P. is \( t_n = ar^{n-1} \). Given: 9th term \( t_9 = 32805 \)
\( \implies \) \( ar^{9-1} = ar^8 = 32805 \) .....(1) 6th term \( t_6 = 1215 \)
\( \implies \) \( ar^{6-1} = ar^5 = 1215 \) .....(2) To find 'r', divide equation (1) by equation (2): \[ \frac { ar^8 }{ ar^5 } = \frac { 32805 }{ 1215 } \]
\( \implies \) \( r^{8-5} = \frac { 6561 }{ 243 } \)
\( \implies \) \( r^3 = \frac { 2187 }{ 81 } = \frac { 729 }{ 27 } = \frac { 243 }{ 9 } = \frac { 81 }{ 3 } \)
\( \implies \) \( r^3 = 27 \) We know that \( 27 = 3^3 \). So, \( r^3 = 3^3 \)
\( \implies \) \( r = 3 \) Now, substitute the value of \( r = 3 \) into equation (2) to find 'a': \( a r^5 = 1215 \)
\( \implies \) \( a \times 3^5 = 1215 \)
\( \implies \) \( a \times 243 = 1215 \)
\( \implies \) \( a = \frac { 1215 }{ 243 } \)
\( \implies \) \( a = 5 \) So, the first term is 5 and the common ratio is 3. Now we need to find the 12th term, \( t_{12} \). Using the formula \( t_n = ar^{n-1} \): \( t_{12} = a r^{12-1} = a r^{11} \)
\( \implies \) \( t_{12} = 5 \times 3^{11} \) To calculate \( 3^{11} \): \( 3^{11} = 3^5 \times 3^5 \times 3^1 = 243 \times 243 \times 3 = 59049 \times 3 = 177147 \)
\( \implies \) \( t_{12} = 5 \times 177147 \)
\( \implies \) \( t_{12} = 885735 \) The 12th term of the G.P. is 885735.
In simple words: First, use the given terms to find the common ratio (r) and the first term (a) of the G.P. Then, use these values in the G.P. formula to calculate the desired 12th term.

🎯 Exam Tip: When given two terms of a G.P., dividing the equation for the higher term by the equation for the lower term is an efficient way to find the common ratio 'r'.

 

Question 7. Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.
Answer:
Let the first term be 'a' and the common ratio be 'r'. Given: 8th term \( t_8 = 768 \) Common ratio \( r = 2 \) The formula for the nth term is \( t_n = ar^{n-1} \). So, \( t_8 = ar^{8-1} = ar^7 \)
\( \implies \) \( ar^7 = 768 \) We need to find the 10th term, \( t_{10} \). \( t_{10} = ar^{10-1} = ar^9 \) We can write \( ar^9 \) as \( ar^7 \times r^2 \). Since we know \( ar^7 = 768 \) and \( r = 2 \), we can substitute these values:
\( \implies \) \( t_{10} = 768 \times 2^2 \)
\( \implies \) \( t_{10} = 768 \times 4 \)
\( \implies \) \( t_{10} = 3072 \) The 10th term of the G.P. is 3072.
In simple words: Since we know the 8th term and the common ratio, we can find the 10th term by multiplying the 8th term by the common ratio twice.

🎯 Exam Tip: If you need to find a term not far from a known term, you can directly multiply or divide by the common ratio 'r' the required number of times instead of recalculating 'a'.

 

Question 8. If a, b, c are in A.P. then show that \( 3^a, 3^b, 3^c \) are in G.P.
Answer:
If a, b, c are in an Arithmetic Progression (A.P.), then the common difference between consecutive terms is equal.
\( \implies \) \( t_2 - t_1 = t_3 - t_2 \)
\( \implies \) \( b - a = c - b \) Rearranging the terms, we get:
\( \implies \) \( 2b = a + c \) .....(1) Now, we need to show that \( 3^a, 3^b, 3^c \) are in a Geometric Progression (G.P.). For numbers to be in G.P., the ratio of consecutive terms must be equal. So, we need to show that \( \frac { 3^b }{ 3^a } = \frac { 3^c }{ 3^b } \) Using the rule of exponents \( \frac { x^m }{ x^n } = x^{m-n} \):
\( \implies \) \( 3^{b-a} = 3^{c-b} \) If the bases are equal, then the exponents must be equal:
\( \implies \) \( b - a = c - b \) Rearranging the terms:
\( \implies \) \( 2b = a + c \) .....(2) From equations (1) and (2), we see that the condition for \( 3^a, 3^b, 3^c \) to be in G.P. (which is \( 2b = a + c \)) is the same as the condition for a, b, c to be in A.P. Since \( 2b = a + c \) is true, then \( 3^a, 3^b, 3^c \) are in G.P.
In simple words: If three numbers are in A.P., it means the middle number is the average of the first and last. If you use these numbers as powers of 3, the new numbers will form a G.P., where the middle term squared equals the product of the other two.

🎯 Exam Tip: This type of proof often requires using the definitions of A.P. and G.P. and algebraic manipulation of exponents to show equivalence between the conditions.

 

Question 9. In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is \( \frac { 57 }{ 2 } \). Find the three terms.
Answer:
Let the three consecutive terms of the G.P. be \( \frac { a }{ r }, a, ar \). This choice simplifies calculations involving products.
Given: Product of three consecutive terms = 27
\( \implies \) \( \frac { a }{ r } \times a \times ar = 27 \)
\( \implies \) \( a^3 = 27 \) We know that \( 27 = 3^3 \).
\( \implies \) \( a^3 = 3^3 \)
\( \implies \) \( a = 3 \) Now, given: Sum of the product of two terms taken at a time is \( \frac { 57 }{ 2 } \)
\( \implies \) \( \left( \frac { a }{ r } \times a \right) + ( a \times ar ) + \left( ar \times \frac { a }{ r } \right) = \frac { 57 }{ 2 } \)
\( \implies \) \( \frac { a^2 }{ r } + a^2r + a^2 = \frac { 57 }{ 2 } \) Substitute \( a = 3 \) into the equation:
\( \implies \) \( \frac { 3^2 }{ r } + 3^2r + 3^2 = \frac { 57 }{ 2 } \)
\( \implies \) \( \frac { 9 }{ r } + 9r + 9 = \frac { 57 }{ 2 } \) Multiply the entire equation by 2r to clear the denominators:
\( \implies \) \( 2r \left( \frac { 9 }{ r } + 9r + 9 \right) = 2r \left( \frac { 57 }{ 2 } \right) \)
\( \implies \) \( 18 + 18r^2 + 18r = 57r \) Rearrange the terms into a quadratic equation:
\( \implies \) \( 18r^2 + 18r - 57r + 18 = 0 \)
\( \implies \) \( 18r^2 - 39r + 18 = 0 \) Divide the equation by 3 to simplify:
\( \implies \) \( 6r^2 - 13r + 6 = 0 \) Factorize the quadratic equation: We need two numbers that multiply to \( 6 \times 6 = 36 \) and add up to -13. These numbers are -9 and -4.
\( \implies \) \( 6r^2 - 9r - 4r + 6 = 0 \) Group terms and factor:
\( \implies \) \( 3r(2r - 3) - 2(2r - 3) = 0 \)
\( \implies \) \( (2r - 3)(3r - 2) = 0 \) This gives two possible values for r:
\( \implies \) \( 2r - 3 = 0 \) or \( 3r - 2 = 0 \)
\( \implies \) \( 2r = 3 \) or \( 3r = 2 \)
\( \implies \) \( r = \frac { 3 }{ 2 } \) or \( r = \frac { 2 }{ 3 } \) Case 1: When \( a = 3 \) and \( r = \frac { 3 }{ 2 } \) The three terms are: \( \frac { a }{ r } = \frac { 3 }{ \frac { 3 }{ 2 } } = 3 \times \frac { 2 }{ 3 } = 2 \) \( a = 3 \) \( ar = 3 \times \frac { 3 }{ 2 } = \frac { 9 }{ 2 } \) The terms are 2, 3, \( \frac { 9 }{ 2 } \). Case 2: When \( a = 3 \) and \( r = \frac { 2 }{ 3 } \) The three terms are: \( \frac { a }{ r } = \frac { 3 }{ \frac { 2 }{ 3 } } = 3 \times \frac { 3 }{ 2 } = \frac { 9 }{ 2 } \) \( a = 3 \) \( ar = 3 \times \frac { 2 }{ 3 } = 2 \) The terms are \( \frac { 9 }{ 2 } \), 3, 2. Both sets of terms are essentially the same sequence, just in reverse order. The three terms of the G.P. are 2, 3, and \( \frac { 9 }{ 2 } \).
In simple words: Start by using \( \frac{a}{r}, a, ar \) for the three terms, as it simplifies the product. Then use the given conditions to find 'a' and 'r'. Once you have these, list out the terms.

🎯 Exam Tip: Choosing the terms of a G.P. as \( \frac{a}{r}, a, ar \) is a standard technique when their product is given, as it helps 'r' cancel out, simplifying the calculation for 'a'.

 

Question 10. A man joined a company as Assistant Manager. The company gave him a starting salary of Rs.60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Answer:
This problem describes a Geometric Progression where the salary increases by a fixed percentage each year. Starting salary (first term) \( a = \text{Rs. } 60000 \) Annual increase = 5% So, the salary each year will be 100% + 5% = 105% of the previous year's salary. The common ratio \( r = 105\% = \frac { 105 }{ 100 } = \frac { 21 }{ 20 } \) We need to find the salary after 5 years. This means we are looking for the salary at the *start* of the 6th year, or the 5th increase applied to the initial salary. If we consider the starting salary as \( t_1 \), then the salary after 1 year is \( t_2 \), and the salary after 5 years is \( t_6 \). Using the formula for the nth term of a G.P., \( t_n = ar^{n-1} \). For salary after 5 years, \( n = 5 + 1 = 6 \). So, \( t_6 = ar^{6-1} = ar^5 \)
\( \implies \) \( t_6 = 60000 \times \left( \frac { 21 }{ 20 } \right)^5 \)
\( \implies \) \( t_6 = 60000 \times \frac { 21^5 }{ 20^5 } \)
\( \implies \) \( t_6 = 60000 \times \frac { 4084101 }{ 3200000 } \)
\( \implies \) \( t_6 = \frac { 60000 \times 4084101 }{ 3200000 } \)
\( \implies \) \( t_6 = \frac { 6 \times 4084101 }{ 320 } \)
\( \implies \) \( t_6 = \frac { 3 \times 4084101 }{ 160 } \)
\( \implies \) \( t_6 = \frac { 12252303 }{ 160 } \)
\( \implies \) \( t_6 = 76576.89375 \) Rounding to the nearest whole rupee, the salary after 5 years will be Rs. 76577. *(The provided solution calculates for \( t_5 \) and then applies 5% increase for the 5th year separately. Let's follow the standard G.P. approach for "after 5 years" which usually means \( t_6 \). However, the original solution calculates \( t_5 \) then adds 5% of \( t_5 \) to it. Let's re-align with the source's logic of calculating each year's increase individually to match its \( t_5 \) equivalent and final answer.)* Let's recalculate based on the source's interpretation for 5 years total earnings (meaning the salary at the end of 5th year, which is \( t_6 \) in standard GP, or \( a(1+r)^5 \) if 'a' is the initial salary). Initial salary (a) = Rs. 60000 Annual increase rate = 5% The salary at the end of each year forms a G.P. where the initial salary is the first term, and the ratio is \( 1 + \text{rate} \). Common ratio \( r = 1 + \frac { 5 }{ 100 } = 1 + 0.05 = 1.05 = \frac { 105 }{ 100 } = \frac { 21 }{ 20 } \) We need the salary *after* 5 years. This means the salary at the start of the 6th year, or after 5 increases have occurred. If the starting salary is \( t_1 \), then the salary after 5 years is \( t_6 \). \( t_6 = a r^{6-1} = a r^5 \) \( t_6 = 60000 \times \left( \frac { 21 }{ 20 } \right)^5 \) \( t_6 = 60000 \times \frac { 21 \times 21 \times 21 \times 21 \times 21 }{ 20 \times 20 \times 20 \times 20 \times 20 } \) \( t_6 = 60000 \times \frac { 4084101 }{ 3200000 } \) \( t_6 = \frac { 60000 \times 4084101 }{ 3200000 } = \frac { 60 \times 4084101 }{ 3200 } = \frac { 6 \times 4084101 }{ 320 } = \frac { 3 \times 4084101 }{ 160 } = \frac { 12252303 }{ 160 } = 76576.89375 \) Rounding to the nearest whole number gives Rs. 76577. The original solution calculates in steps: Starting salary = Rs. 60000 Increase in 1st year = 5% of 60000 = Rs. 3000 Salary at end of 1st year (start of 2nd) = 60000 + 3000 = Rs. 63000 Increase in 2nd year = 5% of 63000 = Rs. 3150 Salary at end of 2nd year (start of 3rd) = 63000 + 3150 = Rs. 66150 This sequence 60000, 63000, 66150... forms a G.P. with a = 60000 and r = 1.05. Salary after 5 years means the 6th term, \( t_6 \). The solution in the OCR calculates \( t_5 \) and then adds 5% of \( t_5 \) to get the final answer. Let's follow that exact calculation from the source, rephrasing as needed for clarity, but ensuring the numbers are consistent with the OCR. Salary at the end of the 1st year (t2) = 60000 * 1.05 = Rs. 63000 Salary at the end of the 2nd year (t3) = 63000 * 1.05 = Rs. 66150 Salary at the end of the 3rd year (t4) = 66150 * 1.05 = Rs. 69457.50 Salary at the end of the 4th year (t5) = 69457.50 * 1.05 = Rs. 72930.375 Salary at the end of the 5th year (t6) = 72930.375 * 1.05 = Rs. 76576.89375 Rounding Rs. 76576.89375 to the nearest whole number gives Rs. 76577. The interpretation "salary after 5 years" typically means the value after 5 annual increases. If the starting salary is \( S_0 \), then after 1 year it's \( S_0(1+r) \), after 2 years \( S_0(1+r)^2 \), and after 5 years it's \( S_0(1+r)^5 \). This is exactly \( t_6 \) if \( t_1 = S_0 \). The source's calculation \( t_5 = (60000) (\frac { 21 }{ 20 } )^4 \) calculates the salary *after 4 years*, then adds 5% of *that* to get the salary after 5 years. This is equivalent to \( t_6 \). Let's present it as a standard G.P. calculation for \( t_6 \). Starting salary (a) = Rs. 60000 Common ratio (r) for annual increase = \( 1 + \frac { 5 }{ 100 } = 1 + 0.05 = 1.05 = \frac { 21 }{ 20 } \) We need to find the salary after 5 years, which is the 6th term of the G.P. (since \( t_1 \) is the starting salary). Using the formula \( t_n = ar^{n-1} \): \( t_6 = a r^{6-1} = ar^5 \)
\( \implies \) \( t_6 = 60000 \times \left( \frac { 21 }{ 20 } \right)^5 \)
\( \implies \) \( t_6 = 60000 \times \frac { 4084101 }{ 3200000 } \)
\( \implies \) \( t_6 = 60000 \times 1.2762815625 \)
\( \implies \) \( t_6 = 76576.89375 \) Rounding to the nearest whole rupee, the salary after 5 years will be Rs. 76577.
In simple words: When a salary increases by the same percentage each year, it forms a Geometric Progression. Calculate the common ratio (1 + percentage increase) and then use the G.P. formula to find the salary after the given number of years.

🎯 Exam Tip: For problems involving percentage growth over time, the formula \( A = P(1+r)^n \) (Compound Interest formula) is directly applicable, where A is the final amount, P is the initial amount, r is the growth rate, and n is the number of periods.

 

Question 11. Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: Rs.20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years. Offer B: Rs.22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years. What is his salary in the 4th year with respect to the offers A and B?
Answer:
We need to calculate the salary in the 4th year for both offers. If the starting salary is \( t_1 \), then the salary in the 4th year means \( t_4 \). **For Offer A:** Starting salary \( a = \text{Rs. } 20000 \) Annual increase = 6% Common ratio \( r = 1 + \frac { 6 }{ 100 } = 1.06 = \frac { 106 }{ 100 } = \frac { 53 }{ 50 } \) We need to find the salary in the 4th year, so \( n = 4 \). Using the formula \( t_n = ar^{n-1} \): \( t_4 = a r^{4-1} = ar^3 \)
\( \implies \) \( t_4 = 20000 \times \left( \frac { 53 }{ 50 } \right)^3 \)
\( \implies \) \( t_4 = 20000 \times \frac { 53 \times 53 \times 53 }{ 50 \times 50 \times 50 } \)
\( \implies \) \( t_4 = 20000 \times \frac { 148877 }{ 125000 } \)
\( \implies \) \( t_4 = \frac { 20000 \times 148877 }{ 125000 } \)
\( \implies \) \( t_4 = \frac { 200 \times 148877 }{ 1250 } \)
\( \implies \) \( t_4 = \frac { 4 \times 148877 }{ 25 } \)
\( \implies \) \( t_4 = \frac { 595508 }{ 25 } \)
\( \implies \) \( t_4 = 23820.32 \) Salary in the 4th year for Offer A is Rs. 23820.32. **For Offer B:** Starting salary \( a = \text{Rs. } 22000 \) Annual increase = 3% Common ratio \( r = 1 + \frac { 3 }{ 100 } = 1.03 = \frac { 103 }{ 100 } \) We need to find the salary in the 4th year, so \( n = 4 \). Using the formula \( t_n = ar^{n-1} \): \( t_4 = a r^{4-1} = ar^3 \)
\( \implies \) \( t_4 = 22000 \times \left( \frac { 103 }{ 100 } \right)^3 \)
\( \implies \) \( t_4 = 22000 \times \frac { 103 \times 103 \times 103 }{ 100 \times 100 \times 100 } \)
\( \implies \) \( t_4 = 22000 \times \frac { 1092727 }{ 1000000 } \)
\( \implies \) \( t_4 = \frac { 22000 \times 1092727 }{ 1000000 } \)
\( \implies \) \( t_4 = \frac { 22 \times 1092727 }{ 1000 } \)
\( \implies \) \( t_4 = \frac { 24039994 }{ 1000 } \)
\( \implies \) \( t_4 = 24039.994 \) Salary in the 4th year for Offer B is Rs. 24039.99. Comparing the salaries: Offer A salary in 4th year = Rs. 23820.32 Offer B salary in 4th year = Rs. 24039.99 Offer B provides a higher salary in the 4th year.
In simple words: For each job offer, use the starting salary and the annual increase rate to calculate the salary for the 4th year. The annual increase means the salary forms a G.P. Compare the calculated amounts to see which offer is better at that point.

🎯 Exam Tip: When comparing offers with percentage increases, calculate the common ratio carefully for each. Pay attention to whether "after X years" refers to \( t_{X+1} \) or \( t_X \) in the G.P. sequence.

 

Question 12. If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that \( x^{b-c} \times y^{c-a} \times z^{a-b} = 1 \).
Answer:
Given that a, b, c are three consecutive terms of an A.P. The property of an A.P. is that the common difference is constant: \( b - a = c - b \)
\( \implies \) \( 2b = a + c \) From this, we can also derive: \( b - c = a - b \) (multiplying by -1) \( c - a = (c - b) + (b - a) = (b-a) + (b-a) = 2(b-a) \) Given that x, y, z are three consecutive terms of a G.P. The property of a G.P. is that the common ratio is constant: \( \frac { y }{ x } = \frac { z }{ y } \)
\( \implies \) \( y^2 = xz \) We can express y and z in terms of x and the common ratio, say 'r'. Let \( x = x \), \( y = xr \), \( z = xr^2 \). Now, let's substitute these into the expression we need to prove: \( x^{b-c} \times y^{c-a} \times z^{a-b} = 1 \) Left Hand Side (L.H.S.) \( = x^{b-c} \times (xr)^{c-a} \times (xr^2)^{a-b} \) Using exponent rules \( (pq)^m = p^m q^m \) and \( (p^q)^r = p^{qr} \):
\( \implies \) L.H.S. \( = x^{b-c} \times x^{c-a} r^{c-a} \times x^{a-b} r^{2(a-b)} \) Group the terms with base 'x' and terms with base 'r':
\( \implies \) L.H.S. \( = x^{(b-c) + (c-a) + (a-b)} \times r^{(c-a) + 2(a-b)} \) Let's simplify the exponent for 'x': \( (b-c) + (c-a) + (a-b) = b - c + c - a + a - b = 0 \) So, the 'x' term becomes \( x^0 = 1 \). Now, simplify the exponent for 'r': \( (c-a) + 2(a-b) \) From A.P. property \( 2b = a + c \), we have: \( c - a = 2b - 2a = 2(b-a) \) And \( a - b = -(b-a) \) Substitute these back into the 'r' exponent: Exponent for 'r' \( = 2(b-a) + 2(-(b-a)) \)
\( \implies \) Exponent for 'r' \( = 2(b-a) - 2(b-a) = 0 \) So, the 'r' term becomes \( r^0 = 1 \). Therefore, L.H.S. \( = 1 \times 1 = 1 \). Since L.H.S. = 1, and the Right Hand Side (R.H.S.) is 1, L.H.S. = R.H.S. Hence, it is proved that \( x^{b-c} \times y^{c-a} \times z^{a-b} = 1 \).
In simple words: This proof uses the special rules of Arithmetic and Geometric Progressions. First, replace the G.P. terms (y, z) with 'x' and 'r'. Then, use the relationship between 'a, b, c' from A.P. to simplify the powers. All the terms will cancel out, leaving 1.

🎯 Exam Tip: When dealing with A.P. and G.P. properties in exponents, always substitute the A.P. relationships (like \( 2b=a+c \)) and G.P. terms (\( x, xr, xr^2 \)) strategically. The exponents often simplify to zero, leading to the result of 1.