Maths Arithmetic and Geometric Progression PDF

# Arithmetic and Geometric Progression ## Introduction - We see many things in nature which follow certain patterns. For example, the holes on a honeycomb, the grains on a corncob and petals of a marigold etc. - Similarly, numbers can be written in a way that follow certain pattern or rule. - Each...

# Arithmetic and Geometric Progression ## Introduction - We see many things in nature which follow certain patterns. For example, the holes on a honeycomb, the grains on a corncob and petals of a marigold etc. - Similarly, numbers can be written in a way that follow certain pattern or rule. - Each number in a list of numbers is called a **term** of the list. - We observe the following about the lists of numbers: - list 1: each succeeding term is obtained by adding 2 to its previous term. - list 2: each succeeding term is obtained by subtracting 3 from its previous term. - list 3: each succeeding term is obtained by multiplying its previous term by 2. - list 4: each succeeding term is obtained by multiplying its previous term by 1/2. - list 5: each succeeding term is obtained by squaring the next natural number. - list 6: each succeeding term is obtained by taking the reciprocal of the next natural number. ## Arithmetic Progression (A.P) - A **list of numbers** in which each term is obtained by adding a **fixed number** to its preceding term (except the first term), is called an **arithmetic progression** (abbreviated A.P.). - In other words, a list of numbers is called an A.P. if and only if the difference of any term from its preceding term is **constant** i.e. a fixed number. - This **constant (a fixed number)** is usually denoted by *d* and is called **common difference**. - The **common difference** *d* may be **positive, zero or negative**. - The terms of an A.P. are respectively denoted by *t1, t2, t3, ...* or *a1, a2, a3, ...* - **a1, a2, a3, a4, ...** is an A.P. if and only if *an+1 - an = d*, a constant (independent of *n*) - The first term of an A.P. is also denoted by *a*. - An A.P. is called **finite** if and only if it contains a **finite** number of terms. - An A.P. is called **infinite** if and only if it contains a **infinite** number of terms. - **a1, a2, a3, ..., an** is an A.P. then *a1 + a2 + a3 + ...* is called an arithmetic series. ### General term of an A.P. - The **general term** *an* of an A.P. is denoted by a and *an = a + (n -1)d* - If a **finite A.P. a, a + d, a + 2d, ...** contains *n* terms and *l* is its last term, then *l = a + (n - 1)d*. ### The *n*th term from the end of a finite A.P. - If *l* is the last term of a ** finite A.P. a, a + d, a + 2d, ...**, then its *n*th term from the end is *l - (n - 1)d*. ### Difference of any two terms of an A.P. - If *a, a + d, a + 2d, ...* is an A.P., then *nth term - mth term = (n - m)d*. ### Middle term(s) of an A.P. - If an A.P. consists of *n* terms, then - It has one **middle term** if *n* is **odd** and this term is *(n + 1)/2* th term, - It has two **middle terms** if *n* is **even** and these terms are *n/2* th term and *(n/2 +1)* th term. ### Numbers in an A.P. - Three numbers *a, b, c* are in A.P. if and only if *2b = a + c*. - If the **sum of numbers** in an A.P. is given, then - **Three** numbers are taken as *a -d, a, a + d* - **Four** numbers are taken as *a - 3d, a - d, a + d, a + 3d* - **Five** numbers are taken as *a -2d, a - d, a, a + d, a + 2d*. ### Sum of first *n* terms of an A.P. - If **a, a + d, a + 2d, ...** is an A.P., then sum of its first *n* term denoted by *Sn* (or *S*) is given by *Sn = (2a + (n - 1)d)*. - If *l* is the last term of the finite A.P. consisting of *n* terms then the sum of all the terms is given by *S = (a + 1)*. - Sum of first *n* natural numbers is *n(n+1)/2*. ## Geometric progression (G.P) - A list of (non-zero) numbers is called a ** geometric progression** (abbreviated G.P.) if and only if the ratio of its any term to its preceding term is **constant** i.e. a fixed number. - This (non-zero) constant (a fixed number) is usually denoted by r and is called **common ratio**. - The **common ratio** *r* may be positive or negative. - The terms of a G.P. are respectively denoted by *a1, a2, a3, ...* or *t1, t2, t3*.. - **a1, a2, a3, ...** is a G.P. if and only if *an+1/an = r*, a constant (independent of n) i.e. an+1 = an*r* (except the first term). - The **first term** of a G.P. is also denoted by *a*. - A G.P. is called **finite** if and only if it contains a **finite** number of terms. - If **a1, a2, a3, ..., an** is a G.P., then *a1 + a2 + a3 + ...* is called a geometric series. ### General term of a G.P. - The **general term** *an* of an A.P. is denoted by a and *an = ar^(n-1)*. - If a **finite G.P. a, ar, ar^2, ...** contains *n* term and its last term is *l*, then *l = ar^(n-1)*. ### The nth term from the end of a finite G.P. - If *a, ar, ar^2, ...* is a finite G.P. with last term *l*, then *nth term from end = l/r^(n-1)*. ### Numbers in G.P. - The numbers *a, b, c* are in G.P. if and only if *b/a = c/b*, i.e. iff *b^2 = ac*. - If the product of numbers in G.P. is given, then - Three numbers are taken as *a/r, a, ar*. - Four numbers are taken as *a/r^3, a/r, ar, ar^3*. - Five numbers are taken as *a/r^2, a/r, a, ar, ar^2*. ### Sum of first *n* terms of a G.P. - If *a* is the first term and *r* is the common ratio of a G.P. and the sum of its first *n* terms is denoted by *Sn*, then *Sn = a(1 - r^n)/(1-r)* or *Sn = a(r^n - 1)/(r - 1)*, *r* ≠ 1. ## Chapter Test ### 1 Write the first four terms of the A.P. when its first term is -5 and common difference is -3. - Answer: -5, -8, -11, -14 ### 2 Verify that each of the following lists of numbers is an A.P., and then write its next three terms: - (i) 0, 1/3, 2/3, ... - Answer: Yes, it's an A.P. with a common difference of 1/3. The next three terms are 1, 4/3, 5/3. - (ii) 5, 14/3, 13/3, 4. ... - Answer: Yes, it's an A.P. with a common difference of -1/3. The next three terms are 11/3, 10/3, 3. ### 3 The nth term of an A.P. is 6n + 2. Find the common difference. - Answer: The common difference is 6. ### 4 Show that the list of numbers 9, 12, 15, 18, ... form an A.P. Find its 16th term and the 6th term from the end of the A.P. 17, 14, 11, ... -40. - Answer: - To show that 9, 12, 15, 18, ... is an A.P., check if the difference between consecutive terms is constant. In this case, the difference is 3. Therefore, it's an A.P. - The 16th term of this A.P. is 9 + (16 - 1) * 3 = 54. - The 6th term from the end of the A.P. 17, 14, 11, ..., −40 can be calculated by finding the 6th term from the beginning of the reversed A.P. The reversed A.P. is -40, -37, -34,... The 6th term of this A.P. is -40 + (6 - 1) * 3 = -25. ### 5 Find the 8th term of an A.P. is 31 and the 15th term is 16 more than its 11th term, then find the A.P. - Answer: - Let the first term be a and the common difference be d. - We know that a8 = 31 and a15 = a11 + 16 - Using the formula an = a + (n-1)d: - a8 = a + 7d = 31 ... (1) - a15 = a + 14d ... (2) - a11 = a + 10d ... (3) - Substitute (3) into (2) to get: a + 14d = (a + 10d) + 16 - Solving this equation we get d = 4 - Substitute d in (1) to get: a + 7 * 4 = 31 => a = 3 - Therefore, the A.P is 3, 7, 11, 15,... ### 6 The 17th term of an A.P is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the *n*th term. - Answer: - Let the first term be a, and the common difference be d. - We know that a17 = 2a8 + 5 and a11 = 43 - Using the formula an = a + (n-1)d: - a17 = a + 16d ... (1) - a8 = a + 7d ... (2) - a11 = a + 10d = 43 ... (3) - Substitute (2) into (1) to get: a + 16d = 2(a + 7d) + 5 - Solve this equation we get: 2d - a = 5... (4) - Solving (3) and (4) for a and d, we get a = 23 and d = 14. - The *n*th term of the A.P. is therefore: a + (n-1)d = 23 + (n-1)*14 = 14n + 9. ### 7 The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19. Find the A.P. - Answer: - Let the first term be a and the common difference be d. - We know that a19 = 3a6 and a9 = 19. - Using the formula an = a + (n-1)d: - a19 = a + 18d ... (1) - a6 = a + 5d ... (2) - a9 = a + 8d = 19 ... (3) - Substitute (2) into (1) to get: a + 18d = 3(a + 5d) = 3a + 15d - Solving this equation we get: 2a - 3d = 0 ... (4) - Solving (3) and (4) for a and d, we get: a = 9/2 and d = 3. - The A.P. is therefore: 9/2, 21/2, 33/2, 45/2, ... ### 8 If the 3rd and the 9th terms of an A.P. are 4 and -8 respectively, then which term of this A.P. is zero? - Answer: - Let the first term be a and the common difference be d. - We know that a3 = 4 and a9 = -8. - Using the formula an = a + (n-1)d: - a3 = a + 2d = 4 ... (1) - a9 = a + 8d = -8 ... (2) - Solving (1) and (2) for a and d, we get: a = 8 and d = -3. - Let the *n*th term be 0. So, a + (n - 1)d = 0. - Substituting the values for a and d we get: 8 + (n - 1)(-3) = 0 - Solving this equation we get n = 11/3 which is not a whole number. - Therefore, no term of this A.P. is 0. ### 9 Which term of the list of numbers 5, 2, -1, -4, ... is -55? - Answer: - The list of numbers is an A.P. with a first term of 5 and a common difference of -3. - Let the *n*th term be -55. - Using the formula *an = a + (n-1)d* we get: 5 + (n - 1)(-3) = -55. - Solving this equation we get: n = 21. - The 21st term of the arithmetic progression is -55. ### 10 The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term. - Answer: - Let the first term be a and the common difference be d. - We know that a24 = 2a10. - Using the formula an = a + (n-1)d: - a24 = a + 23d ... (1) - a10 = a + 9d ... (2) - Substitute (2) into (1): a + 23d = 2(a + 9d) - Solving this equation we get: a - 5d = 0 ... (3) - Using the formula an = a + (n-1)d: - a72 = a + 71d ... (4) - a15 = a + 14d ... (5) - Substitute (3) into (4): a72 = 5d + 71d = 76d - Substitute (3) into (5): a15 = 5d + 14d = 19d - Therefore, a72 = 4a15. The 72nd term is four times the 15th term. ### 11 How many three-digit numbers are divisible by 7? - Answer: - The sequence of three-digit numbers divisible by 7 is 105, 112, 119, ... 994. - This is an A.P. with a = 105 and d = 7. - Let the *n*th term be 994. - Using the formula *an = a + (n-1)d* we get: 105 + (n-1)7 = 994 - Solving this equation we get: n = 128. - Therefore, there are 128 three-digit numbers divisible by 7. ### 12 Which term of the list of numbers 20, 19 1/2, 18, 17 1/2, ... is the first negative term? - Answer: - This is an A.P. with a = 20 and d = -1/2. - We want to find the *n*th term that is less than 0 (the first negative term). - Again, using an = a + (n-1)d: - 20 + (n - 1)(-1/2) < 0. - Solving for n, we get: n > 41. - Therefore, the 42nd term is the first negative term in the sequence. ### 13 The angles of a quadrilateral are in A.P. If the greatest angle is double the smallest angle, find all the four angles. - Answer: - Let the angles of the quadrilateral be a - 3d, a - d, a + d, a + 3d (since they are in A.P.). - The largest angle, a + 3d, is twice the smallest angle, a - 3d: a + 3d = 2(a - 3d) - Simplifying this equation we get: a = 9d. - The sum of angles in a quadrilateral is 360 degrees: (a - 3d) + (a - d) + (a + d) + (a + 3d) = 360 - Simplifying this equation we get: 4a = 360, and so a = 90. - Substituting a = 90 into a = 9d we get d = 10. - Therefore, the four angles are: 60 degrees, 80 degrees, 100 degrees and 120 degrees. ### 14 The sum of three numbers in A.P. is -3 and the product is 8. Find the numbers. - Answer: - Let the three numbers be a - d, a, a + d. - Their sum is -3: (a - d) + a + (a + d) = -3 - Simplifying this equation we get: a = -1 - The product of the numbers is 8: (a - d) * a * (a + d) = 8. - Substituting a = -1 in this equation: (-1-d)*(-1)*(-1+d) = 8 - Simplifying this equation we get: d^2 - 1 = 8 => d^2 = 9 => d = 3, -3. - The three numbers are therefore: -4, -1, 2 or 2, -1, -4. ### 15 The sum of first 20 terms of an A.P. whose nth term is **15 - 4n**. - Answer: - *a1 = 15 - 4 \* 1 = 11*. - Therefore, *a = 11* and *d = -4* - *S20 = (2a + (20 - 1)d) = (2 \* 11+ (20-1)(-4)) = (-50)* ### 16 Find the sum: 18 + 15 1/2 + 13 + ... + (-49). - Answer: - *a =18, d = -2.5* - We need to find the value of n in *an = -49* to get the sum. - Applying the formula * an = a + (n-1)d* and solving for *n* we get *n = 28*. - *S28 = 28/2 * (2 \* 18 + (28-1) \* -2.5) = -637* ### 17 (i) How many terms of the A.P. -6, - 1/2, 5, ... make the sum - 25? (ii) Solve the equation **2 + 5 + 8 + ... + x = 155**. - Answer: - (i) *a = -6, d = -1/2 +6 = 11/2* - *Sn = -25 = n/2 * (2* -6 + (n - 1)*11/2 )* - Solving for *n* we get *n = -11/3* and * n = 10*. Since * n* must be positive, we disregard the negative answer. Therefore, *n = 10*. - (ii) This is an A.P. with *a = 2, d = 3*. We need to find *n*. -*S = 155 = n/2 * (2 \* 2 + (n - 1) \* 3)* - Solving for *n*, we get: * n = 10*. Therefore, *x = 2 + 9 \* d = 29*. ### 18 If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is 7:13, then find the sum of first 20 terms of this A.P. - Answer: - Let the first term be a, and the common difference be d. - We know that *a3 = 5* - Using the formula *an = a + (n-1)d*: - *a3 = a + 2d = 5 ... (1)* - *a6 = a + 5d ... (2)* - *a10 = a + 9d ... (3)* - Given *a6/a10 = 7/13*, substitute (2) and (3) to get: *(a + 5d)/(a + 9d) = 7/13*. - Solve this equation to get *d = 2*. - Substitute d in (1) to get: **a = 1**. - *S20 = (2a + (20 - 1)d) = 20/2 * (2 \* 1 + (20 - 1) \* 2) = 400*. ### 19 The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term. - Answer: - We know that *S14 = 1505, a = 10*. - Using the formula *Sn = n/2 * (2a + (n -1)d)* we get: *1505 = 14/2 * (2 \* 10 + (14- 1)d)*. - Solving this equation we get: *d = 20*. - Applying the formula *an = a + (n - 1)d* we get: *a25 = 10 + (25 -1) * 20 = 490*. ### 20 Find the geometric progression whose 4th term is 54 and 7th term is 1458. - Answer: - Let the first term be *a* and the common ratio be *r*. - We know that *a4 = 54, a7 = 1458* - Applying the formula *an = a + (n - 1)d* we get: - *a4 = ar^3 = 54 ... (1)* - *a7 = ar^6 = 1458 ... (2)* - Divide (2) by (1): *r^3 = 1458/54 = 27*. *r = 3*. - Substitute this value of *r* in (1) to get: *a * 27 = 54*. Solving this equation we get: *a = 2*. - Therefore, the geometric progression is: **2, 6, 18, 54,...** ### 21 The fourth term of a G.P. is the square of its second term and the first term is -3. Find its 7th term. - Answer: - Let the first term be *a* and the common ratio be *r*. - We know that *a = -3, a4 = (a2)^2*. - Applying the formula *an = a + (n - 1)d* we get: *a4 = ar^3, a2 = ar*. - Substituting this in the given equation, we get: *ar^3 = (ar)^2* => *r = a = -3*. - *a7 = ar^6 = (-3)(-3)^6 = -2187*. ### 22 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: - Let *a* be the first term and *r* be the common ratio of the given G.P. - We know that *a4 = x, a10 = y, a16 = z* - Applying the formula *an = a + (n - 1)d* we get: - *a4 = ar^3 = x* - *a10 = ar^9 = y* - *a16 = ar^15 = z*. - Therefore, multiplying *a4* and *a16*, we get: *(ar^3)*(ar^15) = x * z = a^2 * r^18 = (ar^9)^2 = y^2*. - This proves that *y^2 = x*z, which confirms *x, y, z* form a G.P. ### 23 How many terms of the G.P. 3, 3/2, 3/4, ... are needed to give the sum 3069/512? - Answer: - *a = 3, r = 1/2*. We need to find *n*. - Using the formula *Sn = a(1 - r^n)/(1 -r)* we get: *3069/512 = 3 ( 1 - (1/2)^n) / (1 - 1/2)* - Solving for *n* we get: *n = 10*. ## Exercise 9.5 ### 1 Find the sum of: (i) 20 terms of the series 2 + 6 + 18 + ... (ii) 10 terms of series 1 + √3 + 3 + ... (iii) 6 terms of the G.P. 1, 3/2, 9/4, ... (iv) 5 terms and *n* terms of the series 1 + 2/3 + 4/9 + ... - Answer: - (i) * a = 2, r = 3*. *S20 = 2(3^20 - 1) / (3 - 1) = 3^20 - 1* - (ii) *a = 1, r = √3*. *S10 = 1(√3^10 - 1) / (√3 - 1) = (9√3 + 1) / 2* - (iii) *a = 1 , r = 3/2*. *S6 = 1 ( (3/2)^6 -1) / (3/2 - 1) = 729 / 16 -1 =(713/16)* - (iv) *a = 1, r = 2/3*. *S5 = 1( (2/3)^5 -1) / (2/3 -1) = 32/243 -1 = 79/243*. - *Sn = 1( (2/3)^n - 1) / (2/3 -1) = 3((2/3)^n -1) / (2 - 3) = 3(1 - (2/3)^n) / (3 - 2) = 3 - 3(2/3)^n* ### 2 Find the sum of the series 81 - 27 + 9 - ... - Answer: - *a = 81, r = -1/3*. - *Sn = a(1 - r^n)/(1-r)*. ### 3 The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term. - Answer: - *an = ar^(n-1) = 128, r = 2*. Therefore, *a = 128 / 2^(n-1)* - *Sn = a(1 - r^n) / (1 -r)* - Substitute to get *255 = 128/2(n-1) * (1 - 2^n) / (1-2)* - Solving for *a*, we get: *a = 1*. ### 4 (i) How many terms of the G.P. 3, 3^2, 3^3, ... are needed to give the sum 120? (ii) How many terms of the G.P. 1, 4, 16, ... must be taken to have their sum equal to 341? - Answer: - (i) *a = 3, r = 3*. - *Sn = 120 = 3(1 - 3^n) / (1 - 3)* - Solving for *n*, we get: *n = 4*. - (ii) *a = 1, r = 4*. - *Sn = 341 = 1(1 - 4^n) / (1 - 4)* - Solving for * n*, we get: *n = 4*. ### 5 How many terms of the series 1/2 + 1/4 + 1/8 + ... will make the sum 55/72? - Answer: - *a = 1/2, r = 1/2*. - Apply the formula *Sn = a(1 - r^n) / (1-r)*: - *55/72 = (1/2)(1 - (1/2)^n)/ (1 - 1/2)* - Solving for *n* we get: *n = 6*. ### 6 The 2nd and 5th terms of a geometric series are 1/2 and 1/16, respectively. Find the sum of the series upto 8 terms. - Answer: - Let the first term be a, and the common ratio be r. - a2 = ar = 1/2, a5 = ar^4 = 1/16. - Divide the second equation by the first equation: *r^3 = 1/16 / 1/2 = 1/8*. Therefore *r = 1/2*. - Substitute the value of r in *ar = 1/2* and get *a = 1*. - *S8 = 1[(1 - (1/2) ^ 8) / (1 - 1/2) = 255/128*. ### 7 The first term of a G.P. is 27 and its 8th term is 1. Find the sum of its first 10 terms. - Answer: - *a = 27* and the 8th term is 1. - We know that *a8 = ar^7 = 1*. - Therefore, *r = (1/27)^(1/7)* - *S10 = 27(1 - (1/27)^(10/7)) / (1 - (1/27)^(1/7)) = (27^10 -1) / (27^(7/7) - 1)* ### 8 Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728. - Answer: - Let *a* be the first term. - *l = ar^(n-1) = 486, r = 3*. Therefore, *a = 486/3^(n-1)*. - We are given *Sn = 728 = a(1 - r^n)/ (1 - r)* - Substitute the values of *a* and *r* into the equation, and solve for *n* to get