Answers: (i) A be the Arithmetic mean 7, A, 13 are in AP A-7 = 13-A 2A = 13 + 7 A = 10 (ii) A be the Arithmetic mean 12, A, – 8 are in AP A – 12 = – 8 – A 2A = 12 + 8 A =...
Find the A.M. between: (i) 7 and 13 (ii) 12 and – 8
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Arithmetic Progressions
Find the A.M. between: (x – y) and (x + y)
Answer: A be the Arithmetic mean x – y, A, x + y are in AP A – (x – y) = (x + y) – A 2A = x + y + x – y A = x
Insert 4 A.M.s between 4 and 19.
Answer: A1, A2, A3, A4 - 4 AM Between 4 and 19 4, A1, A2, A3, A4, 19 are in AP. By using the formula, d = (b-a) / (n+1) d = (19 – 4) / (4 + 1) d = 15/5 d = 3 A1 = a + d = 4 + 3 = 7 A2 = A1 + d = 7 +...
Insert 7 A.M.s between 2 and 17.
Answer: A1, A2, A3, A4, A5, A6, A7 - 7 AMs between 2 and 17 2, A1, A2, A3, A4, A5, A6, A7, 17 are in AP By using the formula, an = a + (n – 1)d an = 17, a = 2, n = 9 17 = 2 + (9 – 1)d 17 = 2 + 9d –...
Insert six A.M.s between 15 and – 13.
Answer: A1, A2, A3, A4, A5, A6 - 7 AM between 15 and – 13 15, A1, A2, A3, A4, A5, A6, – 13 are in AP an = a + (n – 1)d an = -13, a = 15, n = 8 -13 = 15 + (8 – 1)d -13 = 15 + 7d 7d = -13 – 15 7d =...
There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n
Answer: Let the series be 3, A1, A2, A3, …….., An, 17 Given, an/a1 = 3/1 Total terms in AP are n + 2 17 is the (n + 2)th term By using the formula, An = a + (n – 1)d An = 17, a = 3 So, 17 = 3 + (n +...
Insert A.M.s between 7 and 71 in such a way that the 5th A.M. is 27. Find the number of A.M.s.
Answer: Let the series be 7, A1, A2, A3, …….., An, 71 Total terms in AP are n + 2 71 is the (n + 2)th term By using the formula, An = a + (n – 1)d An = 71, n = 6 A6 = a + (6 – 1)d a + 5d = 27...
If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.
Answer: Let a and b be the first and last terms The series be a, A1, A2, A3, …….., An, b Mean = (a+b)/2 Mean of A1 and An = (A1 + An)/2 A1 = a+d An = a – d AM = (a+d+b-d)/2 AM = (a+b)/2 AM between...
If x, y, z are in A.P. and A1is the A.M. of x and y, and A2 is the A.M. of y and z, then prove that the A.M. of A1 and A2 is y.
Answer: Given, A1 = AM of x and y A2 = AM of y and z A1 = (x+y)/2 A2 = (y+x)/2 AM of A1 and A2 = (A1 + A2)/2 => [(x+y)/2 + (y+z)/2]/2 => [x+y+y+z]/2 => [x+2y+z]/2 x, y, z are in AP, y =...
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P
Answer: A1, A2, A3, A4, A5 - 5 numbers between 8 and 26 8, A1, A2, A3, A4, A5, 26 are in AP By using the formula, An = a + (n – 1)d An = 26, a = 8, n = 7 26 = 8 + (7 – 1)d 26 = 8 + 6d 6d = 26 – 8 6d...
If ${a^2},{b^2},{c^2}$ are in AP., prove that a/(b+c), b/(c+a), c/(a+b) are in AP.
Answer: If a2, b2, c2 are in AP then, b2 – a2 = c2 – b2 If a/(b+c), b/(c+a), c/(a+b) are in AP then, b/(c+a) – a/(b+c) = c/(a+b) – b/(c+a) Let us take LCM on both the sides, b2 – a2 = c2 – b2...
If a, b, c are in A.P., then show that: $(i){a^2}(b + c),{b^2}(c + a),{c^2}(a + b)$ are also in A.P. (ii) b + c – a, c + a – b, a + b – c are in A.P.
Answers: (i) If b2(c + a) – a2(b + c) = c2(a + b) – b2(c + a) b2c + b2a – a2b – a2c = c2a + c2b – b2a – b2c Given, b – a = c – b a, b, c are in AP, c(b2 – a2 ) + ab(b – a) = a(c2 – b2 ) + bc(c – b)...
If a, b, c are in A.P., then show that: $bc – {a^2},ca – {b^2},ab – {c^2}$
Answer: If (ca – b2) – (bc – a2) = (ab – c2) – (ca – b2) bc – a2, ca – b2, ab – c2 are in A.P. Consider LHS and RHS, (ca – b2) – (bc – a2) = (ab – c2) – (ca – b2) (a – b2 – bc + a2) = (ab – c2 – ca...
If (b+c)/a, (c+a)/b, (a+b)/c are in AP., prove that: (i) 1/a, 1/b, 1/c are in AP (ii) bc, ca, ab are in AP
Answer: (i) If 1/a, 1/b, 1/c are in AP 1/b – 1/a = 1/c – 1/b Consider LHS, 1/b – 1/a = (a-b)/ab = c(a-b)/abc Consider RHS, 1/c – 1/b = (b-c)/bc = a(b-c)/bc [by multiplying with ‘a’ on both the...
If a, b, c are in A.P., prove that: $\begin{array}{l} (i){(a – c)^2} = 4(a – b)(b – c)\\ (ii){a^2} + {c^2} + 4ac = 2(ab + bc + ca) \end{array}$
Answer: (i) Expanding, a2 + c2 – 2ac = 4(ab – ac – b2 + bc) a2 + 4c2b2 + 2ac – 4ab – 4bc = 0 (a + c – 2b)2 = 0 a + c – 2b = 0 a, b, c are in AP b – a = c – b a + c – 2b = 0 a + c = 2b (a – c)2 = 4...
If a, b, c are in A.P., prove that: ${a^3} + {c^3} + 6abc = 8{b^3}$
Answer: Expanding, a3 + c3 + 6abc = 8b3 a3 + c3 – (2b)3 + 6abc = 0 a3 + (-2b)3 + c3 + 3a(-2b)c = 0 if a + b + c = 0, a3 + b3 + c3 = 3abc (a – 2b + c)3 = 0 a – 2b + c = 0 a + c = 2b b = (a+c)/2 a, b,...
If a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP., prove that a, b, c are in AP.
Answer: Given, a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP Also, a(1/b + 1/c) + 1, b(1/c + 1/a) + 1, c(1/a + 1/b) + 1 are in AP Let us take LCM for each expression, (ac+ab+bc)/bc ,...
Show that ${x^2} + xy + {y^2},{z^2} + zx + {x^2}and{y^2} + yz + {z^2}$ are in consecutive terms of an A.P., if x, y and z are in A.P.
Answer: Given, x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are in AP (z2 + zx + x2) – (x2 + xy + y2) = (y2 + yz + z2) – (z2 + zx + x2) d = common difference, Y = x + d and x = x + 2d Consider the...
If 1/a, 1/b, 1/c are in A.P., prove that: (i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P. (ii) a(b + c), b(c + a), c(a + b) are in A.P.
Answer: (i) If a, b, c are in AP b – a = c – b 1/a, 1/b, 1/c are in AP 1/b – 1/a = 1/c – 1/b If (b+c)/a, (c+a)/b, (a+b)/c are in AP (c+a)/b – (b+c)/a = (a+b)/c – (c+a)/b Let us take LCM, 1/a, 1/b,...
Find the sum of the following arithmetic progressions: (i) 50, 46, 42, …. to 10 terms (ii) 1, 3, 5, 7, … to 12 terms
Answers: (i) n = 10 First term, a = a1 = 50 Common difference, d = a2 – a1 = 46 – 50 = -4 By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values of ‘a’ and ‘d’, we get S = 10/2 (100 +...
Find the sum of the following arithmetic progressions: (i) 3, 9/2, 6, 15/2, … to 25 terms (ii) 41, 36, 31, … to 12 terms
Answers: (i) n = 25 First term, a = a1 = 3 Common difference, d = a2 – a1 = 9/2 – 3 = (9 – 6)/2 = 3/2 By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values of ‘a’ and ‘d’, we get S =...
Find the sum of the following arithmetic progressions: (i) a+b, a-b, a-3b, … to 22 terms $(ii){(x – y)^2},({x^2} + {y^2}),{(x + y)^2},…$ to n terms
Answers: (i) n = 22 First term, a = a1 = a+b Common difference, d = a2 – a1 = (a-b) – (a+b) = a-b-a-b = -2b By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values of ‘a’ and ‘d’, we...
Find the sum of the following arithmetic progression: (x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), … to n terms
Answer: n = n First term, a = a1 = (x-y)/(x+y) Common difference, d = a2 – a1 = (3x – 2y)/(x + y) – (x-y)/(x+y) = (2x – y)/(x+y) By using the formula, S = n/2 (2a + (n – 1) d) Substitute the values...
Find the sum of the following series: (i) 2 + 5 + 8 + … + 182 (ii) 101 + 99 + 97 + … + 47
Answers: (i) First term, a = a1 = 2 Common difference, d = a2 – a1 = 5 – 2 = 3 an term of given AP is 182 an = a + (n-1) d 182 = 2 + (n-1) 3 182 = 2 + 3n – 3 182 = 3n – 1 3n = 182 + 1 n = 183/3 n =...
Find the sum of the following series: ${(a – b)^2} + ({a^2} + {b^2}) + {(a + b)^2} + s…. + [{(a + b)^2} + 6ab]$
Answer: First term, a = a1 = (a-b)2 Common difference, d = a2 – a1 = (a2 + b2) – (a – b)2 = 2ab an term of given AP is [(a + b)2 + 6ab] an = a + (n-1) d [(a + b)2 + 6ab] = (a-b)2 + (n-1)2ab ...
Find the sum of first n natural numbers.
Answer: First term, a = a1 = 1 Common difference, d = a2 – a1 = 2 – 1 = 1 l = n Sum of n terms = S S = n/2 [2a + (n-1) d] S = n/2 [2(1) + (n-1) 1] S = n/2 [2 + n – 1] S = n/2 [n – 1] ∴ The sum of...
Find the sum of all – natural numbers between 1 and 100, which are divisible by 2 or 5
Answer: The natural numbers which are divisible by 2 or 5 are: 2 + 4 + 5 + 6 + 8 + 10 + … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) are AP with...
Find the sum of first n odd natural numbers.
Answer: Given, AP of first n odd natural numbers whose first term a is 1 Common difference d = 3 The sequence is 1, 3, 5, 7……n a = 1, d = 3-1 = 2, n = n By using the formula, S = n/2 [2a + (n-1)d] S...
Find the sum of all odd numbers between 100 and 200
Answer: The series is 101, 103, 105, …, 199 Number of terms be n a = 101, d = 103 – 101 = 2, an = 199 an = a + (n-1)d 199 = 101 + (n-1)2 199 = 101 + 2n – 2 2n = 199 – 101 + 2 2n = 100 n = 100/2 n =...
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667
Answer: The odd numbers between 1 and 1000 divisible by 3 are 3, 9, 15,…,999 Number of terms be ‘n’, so the nth term is 999 a = 3, d = 9-3 = 6, an = 999 an = a + (n-1)d 999 = 3 + (n-1)6 999 = 3 + 6n...
Find the sum of all integers between 84 and 719, which are multiples of 5
Answer: The series is 85, 90, 95, …, 715 ‘n’ terms in the AP a = 85, d = 90-85 = 5, an = 715 an = a + (n-1)d 715 = 85 + (n-1)5 715 = 85 + 5n – 5 5n = 715 – 85 + 5 5n = 635 n = 635/5 n = 127 By using...
Find the sum of all integers between 50 and 500 which are divisible by 7
Answer: The series of integers divisible by 7 between 50 and 500 are 56, 63, 70, …, 497 Number of terms be ‘n’ a = 56, d = 63-56 = 7, an = 497 an = a + (n-1)d 497 = 56 + (n-1)7 497 = 56 + 7n – 7 7n...
Find the sum of all even integers between 101 and 999
Answer: All even integers will have a common difference of 2. AP is 102, 104, 106, …, 998 a = 102, d = 104 – 102 = 2, an = 998 By using the formula, an = a + (n-1)d 998 = 102 + (n-1)2 998 = 102 + 2n...
The Sum of the three terms of an A.P. is 21 and the product of the first, and the third terms exceed the second term by 6, find three terms.
Answer: Given. The sum of first three terms is 21 Assume, the first three terms as a – d, a, a + d [where a is the first term and d is the common difference] Sum of first three terms is a – d + a +...
Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers
Answer: Given, Sum of first three terms is 27 Assume, the first three terms as a – d, a, a + d [where a is the first term and d is the common difference] Sum of first three terms is a – d + a + a +...
Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Answer: Given, Sum of four terms is 50. Assume these four terms as a – 3d, a – d, a + d, a + 3d Sum of these terms - 4a = 50 a = 50/4 = 25/2 … (i) It is also given that the greatest number is 4 time...
The sum of three numbers in A.P. is 12, and the sum of their cubes is 288. Find the numbers.
Answer: Given, Sum of three numbers = 12 Assume the numbers in AP are a – d, a, a + d 3a = 12 a = 4 It is also given that the sum of their cube is 288 (a – d)3 + a3 + (a + d)3 = 288 a3 – d3 – 3ad(a...
If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.
Answer: Given, Sum of first three terms is 24 Assume the first three terms are a – d, a, a + d [where a is the first term and d is the common difference] Sum of first three terms is a – d + a + a +...
The angles of a quadrilateral are in A.P. whose common difference is 10. Find the angles
Answer: Given, d = 10 Sum of all angles in a quadrilateral = 360 Assume the angles are a – 3d, a – d, a + d, a + 3d a – 2d + a – d + a + d + a + 2d = 360 4a = 360 a = 90… (i) (a – d) – (a –...
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Answer: Given, 24th term is twice the 10th term a24 = 2a10 an = a + (n – 1) d When n = 10, a10 = a + (10 – 1)d = a + 9d When n = 24, a24 = a + (24 – 1)d = a + 23d When n = 34, a34 = a + (34 – 1)d =...
The 10th and 18th term of an A.P. are 41 and 73 respectively, find 26th term.
Answer: Given, 10th term of an A.P is 41, and 18th terms of an A.P. is 73 a10 = 41 a18 = 73 an = a + (n – 1) d When n = 10, a10 = a + (10 – 1)d = a + 9d When n = 18, a18 = a + (18 – 1)d = a + 17d...
If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that the 25th term of the A.P. is Zero.
Answer: Given, 10 times the 10th term of an A.P. is equal to 15 times the 15th term 10a10 = 15a15 an = a + (n – 1) d When n = 10, a10 = a + (10 – 1)d = a + 9d When n = 15, a15 = a + (15 – 1)d = a +...
If 9th term of an A.P. is Zero, prove that its 29th term is double the 19th term.
Answer: Given, 9th term of an A.P is 0 a9 = 0 an = a + (n – 1) d [w When n = 9, a9 = a + (9 – 1)d = a + 8d a9 = 0 a + 8d = 0 a = -8d When n = 19, a19 = a + (19 – 1)d = a + 18d = -8d + 18d =...
The 6th and 17th terms of an A.P. are 19 and 41 respectively. Find the 40th term.
Answer: Given, 6th term of an A.P is 19 and 17th terms of an A.P. is 41 a6 = 19 a17 = 41 an = a + (n – 1) d When n = 6, a6 = a + (6 – 1) d = a + 5d When n = 17, a17 = a + (17 – 1)d = a + 16d a6 = 19...
The first term of an A.P. is 5, the common difference is 3, and the last term is 80; find the number of terms.
Answer: Given, First term, a = 5; last term, l = an = 80 Common difference, d = 3 an = a + (n – 1) d an = 5 + (n – 1)3 = 5 + 3n – 3 = 3n + 2 Put an = 80 as 80 is last term of A.P. 3n + 2 = 80 3n =...
How many terms are there in the A.P. -1, -5/6, -2/3, -1/2, …, 10/3 ?
Answer: Given, AP of -1, -5/6, -2/3, -1/2, … a1 = a = -1 a2 = -5/6 Common difference, d = a2 – a1 = -5/6 – (-1) = -5/6 + 1 = (-5+6)/6 = 1/6 an = a + (n – 1) d an = -1 + (n – 1) 1/6 = -1 + 1/6n – 1/6...
How many terms are in A.P. 7, 10, 13,…43?
Answer: Given, AP of 7, 10, 13,… a1 = a = 7 a2 = 10 Common difference, d = a2 – a1 = 10 – 7 = 3 an = a + (n – 1) d an = 7 + (n – 1)3 = 7 + 3n – 3 = 3n + 4 Put, an = 43 3n + 4 = 43 3n = 43 – 4 3n =...
Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, … is (a) purely real (b) purely imaginary ?
Answer: Given, AP of 12 + 8i, 11 + 6i, 10 + 4i, … a1 = a = 12 + 8i a2 = 11 + 6i Common difference, d = a2 – a1 = 11 + 6i – (12 + 8i) = 11 – 12 + 6i – 8i = -1 – 2i an = a + (n – 1) d an = 12 + 8i +...
Which term of the sequence 24, 23 ¼, 22 ½, 21 ¾ is the first negative term?
Answer: Given, AP of 24, 23 ¼, 22 ½, 21 ¾, … = 24, 93/4, 45/2, 87/4, … a1 = a = 24 a2 = 93/4 Common difference, d = a2 – a1 = 93/4 – 24 = (93 – 96)/4 = – 3/4 an = a + (n – 1) d an = a + (n – 1) d...
(i) Is 68 a term of the A.P. 7, 10, 13,…? (ii) Is 302 a term of the A.P. 3, 8, 13,…?
Answers: (i) Given, A.P is 7, 10, 13,… a1 = a = 7 a2 = 10 Common difference, d = a2 – a1 = 10 – 7 = 3 an = a + (n – 1)d an = 7 + (n – 1)3 = 7 + 3n – 3 = 3n + 4 Put, an = 68 3n + 4 = 68 3n = 68 – 4...
Which term of the A.P. 4, 9, 14,… is 254 ?
Answer: Given, A.P is 4, 9, 14,… a1 = a = 4 a2 = 9 Common difference, d = a2 – a1 = 9 – 4 = 5 an = a + (n – 1)d an = 4 + (n – 1)5 = 4 + 5n – 5 = 5n – 1 Put, an = 254 5n – 1 = 254 5n = 254 + 1 5n =...
(i) Which term of the A.P. 3, 8, 13,… is 248 ? (ii) Which term of the A.P. 84, 80, 76,… is 0 ?
Answer: (i) Given, A.P is 3, 8, 13,… a1 = a = 3 a2 = 8 Common difference, d = a2 – a1 = 8 – 3 = 5 an = a + (n – 1)d an = 3 + (n – 1)5 = 3 + 5n – 5 = 5n – 2 (Put, an = 248) ∴ 5n – 2 = 248 = 248 + 2...
In an A.P., show that am+n + am–n = 2am.
Answer: Using the formula, an = a + (n – 1)d LHS: am+n + am-n am+n + am-n = a + (m + n – 1)d + a + (m – n – 1)d = a + md + nd – d + a + md – nd – d = 2a + 2md – 2d = 2(a + md – d) = 2[a + d(m – 1)]...
Find: nth term of the A.P 13, 8, 3, -2, ….
Answer: nth term of the A.P 13, 8, 3, -2, …. Arithmetic Progression (AP) whose common difference is = an – an-1 where n > 0 Consider, a = a1 = 13, a2 = 8 … Common difference, d = a2 – a1 = 8 – 13...
Find: (i) 10th term of the A.P. 1, 4, 7, 10, ….. (ii) 18th term of the A.P. √2, 3√2, 5√2, …
Answer: (i) Arithmetic Progression (AP) whose common difference is = an – an-1 where n > 0 Consider, a = a1 = 1, a2 = 4 … Common difference, d = a2 – a1 = 4 – 1 = 3 Finding an an = a + (n-1) d =...
If the nth term of a sequence is given by an = n2 – n+1, write down its first five terms.
Answer: Using n = 1, 2, 3, 4, 5, the first five terms can be calculated. If n = 1, a1 = (1)2 – 1 + 1 a1 = 1 – 1 + 1 a1 = 1 If n = 2, a2 = (2)2 – 2 + 1 a2 = 4 – 2 + 1 a2 = 3 If n = 3, a3 = (3)2 – 3 +...
A sequence is defined by an = n3 – 6n2 + 11n – 6, n ∈ N. Show that the first three terms of the sequence are zero and all other terms are positive.
Answer: Using n = 1, 2, 3, the first three terms can be calculated. If n = 1, a1 = (1)3 – 6(1)2 + 11(1) – 6 a1 = 1 – 6 + 11 – 6 a1 = 12 – 12 a1 = 0 If n = 2, a2 = (2)3 – 6(2)2 + 11(2) – 6 a2 = 8 –...
Find the first four terms of the sequence defined by a1 = 3 and an = 3an–1 + 2, for all n > 1.
Answer: Using n = 1, 2, 3, 4, the first four terms can be calculated. If n = 1, a1 = 3 If n = 2, a2 = 3a2–1 + 2 a2 = 3a1 + 2 a2 = 3(3) + 2 a2 = 9 + 2 a2 = 11 If n = 3, a3 = 3a3–1 + 2 a3 = 3a2 + 2...
Write the first five terms in each of the following sequences: (i) a1 = 1, an = an–1 + 2, n > 1 (ii) a1 = 1 = a2, an = an–1 + an–2, n > 2
Answer: (i) Using n = 1, 2, 3, 4, 5, the first five terms can be calculated. If n = 1, a1 = 1 If n = 2, a2 = a2–1 + 2 a2 = a1 + 2 a2 = 1 + 2 a2 = 3 If n = 3, a3 = a3–1 + 2 a3 = a2 + 2 a3 = 3 + 2...
Write the first five terms in each of the following sequence: a1 = a2 =2, an = an–1 – 1, n > 2
Answer: Using n = 1, 2, 3, 4, 5, the first five terms can be calculated. If n = 1, a1 = 2 If n = 2, a2 = 2 If n = 3, a3 = a3–1 – 1 = a2 – 1 = 2 – 1 = 1 If n = 4, a4 = a4–1 – 1 = a3 – 1 = 1 –...
The Fibonacci sequence is defined by a1 = 1 = a2, an = an–1 + an–2 for n > 2. Find (an+1)/an for n = 1, 2, 3, 4, 5.
3Answer: an = an–1 + an–2 If n = 1, (an+1)/an = (a1+1)/a1 = a2/a1 = 1/1 = 1 a3 = a3–1 + a3–2 = a2 + a1 = 1 + 1 = 2 If n = 2, (an+1)/an = (a2+1)/a2 = a3/a2 = 2/1 = 2 a4 = a4–1 + a4–2 = a3 + a2 = 2 +...
Show that the sequence defined by $a_{n}=2 / 3^{n}, n \in N$ is a G.P.
Solution: Given that, $a_{n}=2 / 3^{n}$ Consider $\mathrm{n}=1,2,3,4, \ldots$ since $\mathrm{n}$ is a natural number. Therefore, $\begin{array}{l} a_{1}=2 / 3 \\ a_{2}=2 / 3^{2}=2 / 9 \\ a_{3}=2 /...
A farmer buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount. How much the tractor cost him?
Solution: We have to find the total cost of the tractor if he buys it in installments. Total price $=$ ₹ 12000 Paid amount $=$ ₹ 6000 Unpaid amount $=$ ₹ $12000-6000=$ ₹ 6000 He pays remaining ₹...
A piece of equipment cost a certain factory 600,000 . If it depreciates in value $15 \%$ the first, $13.5 \%$ the next year, $12 \%$ the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?
Solution: Given that a piece of equipment cost a certain factory is ₹ 600,000 We have to find the value of the equipment at the end of 10 years. The price of equipment depreciates $15 \%, 13.5 \%,...
A man is employed to count ₹ 10710 . He counts at the rate of 180 per minute for half an hour. After this he counts at the rate of ₹3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.
Solution: Given that the amount to be counted is ₹ 10710 We have to find the time taken by man to count the entire amount. He counts the amount at the rate of ₹ 180 per minute for 30 minutes. Amount...
There are 25 trees at equal distances of 5 meters in a line with a well, the distance of well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.
Solution: It is given that total number of trees are 25 and the distance between two adjacent trees are 5 meters To find the total distance the gardener will cover. As given the gardener is coming...
A manufacturer of the radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find
(і) the product in the 10th year.
Solution: Given that, In the third and seventh year 600 and 700 radio sets units are produced, respectively. $a_3 = 600$ and $a_7 = 700$ (i) The product in the $10^{\text {th }}$ year. Find the...
A manufacturer of the radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find
(і) the production in the first year
(іі) the total product in the 7 years and
Solution: Given that, In the third and seventh year 600 and 700 radio sets units are produced, respectively. $a_3 = 600$ and $a_7 = 700$ (і) The production in the first year Find the production in...
A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the instalment.
Solution: As per the question: There are 40 annual instalments that form an arithmetic series. Let '$a$' be the first instalment $S_{40}=3600, n=40$ Using the formula, $\begin{array}{l} S_{n}=n /...
A man saves ₹ 32 during the first year, ₹ 36 in the second year and in this way he increases his savings by ₹ 4 every year. Find in what time his saving will be ₹ 200.
Solution: As per the question: Savings for the first year is ₹ 32 Savings for the second year is ₹ 36 Every year he increases his savings by ₹ 4. Therefore, A.P. will be $32,36,40, \ldots \ldots...
A man saved ₹ 16500 in ten years. In each year after the first he saved ₹ 100/- more than he did in the preceding year. How much did he saved in the first year?
Solution: As per the question: A man saved ₹$16500$ in ten years Let be his savings in the first year be ₹ $x$ Every year his savings increased by ₹ 100. Therefore, A.P will be $x$, $100 + x$, $200...
1. Write the first terms of each of the following sequences whose ${{n}^{th}}$ term are: (iii) ${{a}_{n}}={{3}_{n}}$ (iv) ${{a}_{n}}=\left( 3n-2 \right)/5$
An arithmetic progressions or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant. Solutions: (iii) ${{a}_{n}}={{3}^{n}}$ Given sequence...
Write the first five terms of each of the following sequences whose ${{n}^{th}}$ term are: (i) ${{a}_{n}}=3n+2$ (ii) ${{a}_{n}}=\left( n-2 \right)/3$ An arithmetic progressions or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.
Solutions: (i) ${{a}_{n}}=3n+2$ Given sequence whose ${{a}_{n}}=3n+2$ To get the first five terms of given sequence, put $n=1,2,3,4,5$ and we get ${{a}_{1}}=\left( 3\times 1 \right)+2=3+2=5$...