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CLASS 10MATHEMATICS( )PART - 1
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PREFACE
CHAPTER
1 ARITHMETIC SEQUENCES
OUTLINE ¾ Numbers in order ¾ Position and term ¾ Algebra of sequences ¾ Sum of arithmetic sequences ¾ Arithmetic progression
TOPIC-1Sequence of Numbers
QUICK REVIEW ¾ A collection of numbers written in order as the first, second,
third and so on acccording to a definite rule is called a number sequence.
¾ A sequence which starts with any number and proceeds by the addition of one number again and again is called an arithmetic sequence or an arithmetic progression.
KNOW THE TERMS
¾ Sequence : An ordered list of numbers. ¾ Arithmetic progression : A sequence of numbers such that the difference between the consecutive terms
is constant.
GREENBOARD ?How it is done on
Q. The 7th term of an arithmetic sequence is 19 and its 11th term is 31. What is the algebraic expression for this sequence.
Sol. Step 1 : Nth term of arithmetic sequence is a + (n – 1) d
\ Tn = a + (n – 1) d a is first number, d difference between two
consecutive numbers and n number of terms 19 = a + 6d ...(i) 31 = a + 10d ...(ii)Step 2 : Subtract equation (i) from equation (ii), a + 10d = 31 – a – 6d = – 19 4d = 12 Þ d = 3
Step 3 : Putting the value of d in equation (i), 19 = a + 6 × 3 Þ 19 – 18 = a Þ a = 1 Step 4 : We have first number is one and its common
difference 3.Step 5 : Putting the values of a and d in general term Tn = a + (n – 1) d Tn = 1 + (n – 1) 3 = 1 + 3n – 3 Þ Tn = 3n – 2 \ Required algebraic expression (3n – 2).
TOPIC - 1Sequence of Numbers .... P. 05
TOPIC - 2Addition, Subtraction, Position and Terms .... P. 07
TOPIC - 3Algebra of Arithmetic Sequences .... P. 12
6 ] Oswaal Kerala SSLC Question Bank, MATHEMATICS, Class – X
Very Short Answer Type Questions (2 marks each)
Q.1. Write an arithmetic sequence with common difference 3. Find its 11th term. [March 2016]
Sol. Let first term = a and given common difference be 3.
\ arithmetic sequence is a, a + 3, a + 6, a + 9 ....... Hence, its 11th term = a + 3 (11 – 1) = a + 30. Q.2. 6, 11, 16, .... is an arithmetic sequence. What is its
next term ? What is the least three digit number that comes as a term of this sequence ? [March 2014]
Sol. The arithmetic sequence is 6, 11, 16... a1 = a = 6; a2 = 11; a3 = 16 Common difference = d = a2 – a1 = 11 – 6 = 5 Hence, the next term a4 = a3 + d = 16 + 5 = 21 Again tn ³100 \a + (n – 1) × d ³ 100 Þ6 + (n – 1) × 5 ³ 100 Þ (n – 1) × 5 ³ 94 Þ (n – 1) ³ 18.8 n ³ 19.8 = 20th term Hence, t20 = 6 + (19) × (5) = 6 + 95 = 101. Q.3. The first term of an arithmetic sequence is 10 and
its common difference 3. Write the first three terms of the sequence. Verify whether 100 is a term of this sequence. [March 2014]
Sol. The first three terms are 10, 13, 16. So, first term, a = 10 and common difference, d =
13 – 10 = 3 nth term, tn = a + (n – 1) d tn = 10 + (n – 1) 3 tn = 10 + 3n – 3 = 7 + 3n
To show, 100 is a term of this sequence. If 7 is subtracted from any term of this sequence, the
resulting number is a multiple of 3. Now, 100 – 7 = 93 = 31 × 3, which is a multiple of
3. Hence, 100 is a term of this sequence. Q.4. The second and fourth terms of the following
arithmetic sequence are missing. Find the numbers at these positions : 11, ...., 19, ..... [March 2013]
Sol. The first term of the arithmetic sequence = 11 The third term = 19 Let the second term be x and the common difference
be d. We have now d = x – 11 = 19 – x Þ 2x = 30 \ x = 15 So, d = 15 – 11 = 4 \The second term of the sequence = 11 + 4 = 15. and fourth term = 19 + 4 + 23. Q.5. Arithmetic sequence starts as 5, 9, 13. What is
the next term ? Is 2012 a term of this sequence ? Why ? [March 2012]
Sol. Given sequence is 5, 9, 13, ... Common difference, d = 9 – 5 = 4 The next term = 13 + 4 = 17 tn = a + (n – 1 ) × d = 5 + (n – 1 ) × 4 = 5 + 4n – 4 = 1 + 4n If 1 is subtracted from any term of this sequence,
the resulting number is a multiple of common difference 4.
2012 – 1 = 2011 but 2011 is not a multiple of 4. Hence, 2012 is not a term of this sequence.
Short Answer Type Questions-I (3 marks each)
Q.1. Look at these triangles made with dots. How many dots are there in each ?
Compute the number of dots needed to make the next two triangles. (TBQ)
Sol. Number of dotes in firstD = 3 No. of dots in second D= 3 + 3 = 6 No. of dots in third D= 6 + 4 = 10 In the same way no. of dotes in next two triangles
are 10 + 5 = 15 and 15 + 6 = 21. Q.2. Make the following number sequences, from the
sequence of equilateral triangles, squares, regular pentagons and so on, of regular polygons :
Number of sides 3, 4, 5, .... (i) Sum of interior angles (ii) Sum of exterior angles (iii) One interior angle (iv) One exterior angle (TBQ) Sol. (i) Sum of interior angles of regular polygon of n
sides
= n × ( )n
n− °2 180
= (n – 2) 180°
\ Sum of interior angles of equilateral triangle (n = 3), square (n = 4), regular pentagon (n = 5) etc. are
(3 – 2) 180°, (4 – 2) 180°, (5 – 2)180°, ......... i.e., 180°, 360°, 540°, ......... (ii) Sum of exterior angles of regular polygon of n
sides = n × 180° – (n – 2)180° = 360°, which is free from n.
ARITHMETIC SEQUENCES [ 7
(iii) One interior angle of triangle, square, pentagon etc. are
1803
° , 3604
° , 5405
° , ...........
i.e., 60°, 90°, 108°, ........ (iv) One exterior angle of triangle, square, pentagon
etc. are
3603
° , 3604
° , 3605
° , ...........
i.e., 120°, 90°, 72°, ...... Q.3. Write down the sequence of natural numbers
leaving remainder 1 on division by 3 and the sequence of natural numbers leaving remainder 2 on division by 3. (TBQ)
Sol. (i) Sequence of natural number leaving remainder 1 on division by 3 are 3 × 1 + 1, 3 × 2 + 1, 3 × 3 + 1, 3 × 4 + 1, 3 × 5 + 1, .......
i.e. 4, 7, 10, 13, 16, ...... (ii) Sequence of natural number leaving remainder
2 on division by 3 are 3 × 1 + 2, 3 × 2 + 2, 3 × 3 + 2, 3 × 4 + 2, 3 × 5 + 2, .......
i.e., 5, 8, 11, 14, 17, ....... Q.4. Write down the sequence of natural numbers
ending in 1 or 6 and describe it in two other ways. (TBQ)
Sol. (i) Sequence of numbers that end in 1 or 6 is : 1, 6, 11, 16, 21, 26, 31, 36, ..... (ii) Sequence of numbers other way a, a + d,
a + 2d, a+ 3d, ....., where a = 1 and d = 5. Q.5. One cubic centimetre of iron weighs 7.8 gram.
Write as sequences, the volumes of weights of iron cubes of sides 1 centimetre, 2 centimetre and so on . (TBQ)
Sol. Given, 1 cm3 = 7·8 gram The volumes of iron cubes of sides 1 cm, 2 cm, 3 cm
and so on are 13 cm3, 23 cm3, 33 cm3, ......... So, their corresponding weights (in gram) are 7.8 × 13, 7.8 × 23, 7.8 × 33, ............ , which is the required sequence whose nth term is
tn = 7.8 n3, where n ÎN.
Q.6. Write four arithmetic sequences with 100 as the sum of the first four terms. (TBQ)
Sol. Let first term of arithmetic sequence be ‘a’ and its common difference ‘d’
\Sequence is a + a + d, a+ 2d, a+ 3d, ........ Given, sum of first four terms = 100 \ 4a + 6d = 100 Þ 2a + 3d = 50 (i) If a = 1, 3d = 50 – 2 × 1 = 48
or d = 483
= 16
\required sequence is 1, 17, 33, 49, ....... (ii) If a = 4, 3d = 50 – 2 × 4 = 42
or d = 423
= 14
\sequence is 4, 18, 32, 46, ....... (iii) If a = 7, 3d = 50 – 2 × 7 = 36
or d = 363
= 12
\ sequence is 7, 19, 31, 43, ....... (iv) If a = 10, 3d = 50 – 2 × 10 = 30
or d = 303
= 10
\ sequence is 10, 20, 30, 40. Q.7. The 8th term of an arithmetic sequence is 12 and its
12th term is 8. What is the algebraic expression for this sequence? (TBQ)
Sol. Given, T8 = 12 and T12 = 8 Þ a + 7d = 12 .....(i) and a + 11d = 8 .....(ii) \(i) – (ii), – 4d = 4 Þ d = –1 and a = 12 – 7 (–1) = 19 Now, Tn = 19 + (n – 1) × –1 = 20 – n Algebraic expression is (20 – n).
KNOW THE LINKS
¾ http://www.mathsisfun.com/definitions/arithmetic-sequence.html ¾ http://www.math10.com/en/algebra/arithmetic-progression.html
TOPIC-2Addition, Subtraction, Position and Terms
QUICK REVIEW ¾ In an arithmetic sequence, if we subtract any number from the number just after it, we get the same number and
this number is called the common difference of this arithmetic sequence.
8 ] Oswaal Kerala SSLC Question Bank, MATHEMATICS, Class – X
¾ The numbers ordered as first or second are called as terms of the sequence. ¾ The sum of three consecutive numbers a – d, a, a + d of an arithmetic sequence is three times the middle number. ¾ The sum of the five consecutive numbers a – 2d, a – d, a, a + d, a + 2d of an arithmetic sequence is five times the
middle number. ¾ If even number of terms are taken, the common difference being d. If four consecutive terms are taken and middle
pair x, y, then the sum of four terms x – d, x, y, y + d = 2 × (x+ y)
= 4 ×
12
× (x+ y)
= 4 × average of the middle pair. ¾ The difference between any two terms is got by multiplying the difference of their positions by the common
difference. ¾ In any arithmetic sequence, the term difference is proportional to the position difference, and the constant of
proportionality is the common difference. ¾ Any arithmetic sequence of natural numbers consists of numbers leaving the same remainder on division by a
specific number. This divisor is the common difference. ¾ In an arithmetic sequence, nth term = a + (n – 1)d, where a is the first term and d is the common difference.
KNOW THE TERMS
¾ Term : The number which occupies a given position in an arithmetic sequence. ¾ Common difference : The number (repeatedly added to get the numbers) in an arithmetic sequence.
GREENBOARD ?How it is done on
Q. Find the sum of the first 20 terms of the arithmetic sequence :
–11, –9, –7, ......
Sol. Step 1 : Sequence is –11, –9, –7 in which first term is –11 and difference –9 – (–11) = –9 + 11 = +2
Now a = –11, d = 2 and n = 20
Step 2 : Sum of nth term of AP
Sn = n
2[2a + (n – 1) d]
= 20
2[2 × –11 + (20 – 1) 2]
= 10 [–22 + (19) 2]
Step 3 : Sum of 20th term
S20 = 10 (–22 + 38)
S20 = 10(16) = 160
Step 4 : Required sum of 20th term of given term is 160.
Very Short Answer Type Questions (2 marks each)
Q.1. Find the sum of the first 25 terms of each of the arithmetic sequence below :
(i) 11, 22, 23, ... (ii) 12, 23, 34, ..... (iii) 21, 32, 43, ..... (iv) 19, 28, 37 ...... (v) 1, 6, 11 ..... (TBQ) Sol. (i) Given sequence is 11, 22, 33. Here, a = 11, d = 22 – 11 = 11
\ Sn = n2
[2a + (n – 1)d]
= 252
[2 × 11 + (25 – 1)11]
= 252
[22 + 24 × 11]
= 252
[22 + 264] = 252
[286]
= 25 × 143 = 3575 (ii) Given sequence is 12, 23, 34. Here, a = 12, d = 23 – 12 = 11
\ S25 = 252
[2 × 12 + (25 – 1) × 11]
= 252
[24 + 24 × 11]
= 252
[24 + 264] = 252
[288]
= 25 × 144 = 3600.
ARITHMETIC SEQUENCES [ 9
(iii) Given sequence is 21, 32, 43 Here, a = 21, d = 32 – 21 = 11
\ S25 = 252
[2 × 21 + (25 – 1)11]
= 252
[42 + 24 × 11]
= 252
[42 + 264] = 252
[306]
= 25 × 153 = 3825. (iv) Given sequence is 19, 28, 37 Here, a = 19, d = 28 – 19 = 9
\ S25 = 252
[2 × 19 + (25 – 1)9]
= 252
[38 + 24 × 9]
= 252
[38 + 216] = 252
[254]
= 25 × 127 = 3175. (v) Given sequence is 1, 6, 11 Here, a = 11, d = 6 – 1 = 5
\ S25 = 252
[2 × 1 + (25 – 1)5]
= 252
[2 + 24 × 5] = 252
[2 + 120]
= 252
[122] = 25 × 61 = 1525.
Q.2. The expression for the sum of n terms of some arithmetic sequences are given below. Find the expression for the nth term of each : [TBQ]
(i) n2 + 2n (ii) 2n2 + n (iii) n2 – 2n (iv) 2n2 – n (v) n2 – n. Sol. (i) Given, Sn = n2 + 2n, then Sn–1 = (n – 1)2 + 2 (n – 1) = n2 – 2n + 1 + 2n – 2 = n2 – 1 \ tn = Sn – Sn–1
= (n2 + 2n) – (n2 – 1) = 2n + 1 (ii) Given, Sn = 2n2 + n, then Sn–1 = 2(n – 1)2 + n – 1 = 2(n2 – 2n + 1) + n – 1 = 2n2 – 3n + 1 \ tn = Sn – Sn–1
= (2n2 + n) – (2n2 – 3n + 1) = 4n – 1 (iii) Given, Sn = n2 – 2n, then Sn–1 = (n – 1)2 – 2 (n – 1) = n2 – 2n + 1 – 2n + 2 = n2 – 4n + 3 \ tn = Sn – Sn–1
= (n2 – 2n) – (n2 – 4n + 3) = 2n – 3 (iv) Given, Sn = 2n2 – n, then Sn–1 = 2 (n – 1)2 – (n – 1) = 2 (n2 – 2n + 1) – n + 1 = 2n2 – 5n + 3
\ tn = Sn – Sn–1
= (2n2 – n) – (2n2 – 5n + 3) = 4n – 3. (v) Given, Sn = n2 – n, then Sn–1 = (n – 1)2 – (n – 1) = n2 – 2n + 1 – n + 1 = 2n2 – 3n + 2 \ tn = Sn – Sn–1
= (n2 – n) – (n2 – 3n + 2) = 2n – 2. Q.3. In each of the arithmetic sequences below, some
terms are missing and their positions are marked
with . Find them.
(i) 24, 42, , , ..... (ii) , 24, 42, , ....
(iii) , , 24, 42, ..... (iv) 24, , 42, , .....
(v) , 24, , 42, ..... (vi) 24, , , 42, ..... (TBQ)
Sol. (i) 24, 42, , , .... (ii) , 24, 42, , .....
(iii) , , 24, 42, ..... (iv) 24, , 42, , .....
(v) , 24, , 42, ..... (vi) 24, , , 42, ..... Explanation : (i) Common diff. = 42 – 24 = 18 (ii) Common diff. = 42 – 24 = 18 (iii) Common diff. = 42 – 24 = 18
(iv) Common diff. = 1
2(42 – 24) = 9
(v) Common diff. = 12
(42 – 24) = 9
(vi) Common diff. = 13
(42 – 24) = 6
Q.4. Calculate in head, the sums of the following arithmetic sequences :
(i) 51 + 52 + 53 + .... + 70
(ii) 112
+ 212
+ ..... + 1212
(iii) 12
+ 1 + 112
+ .... + 1212
. (TBQ)
Sol. (i) Given sequence is 51 + 52 + 53 + ....... + 70 Here, a = 51, d = 52 – 51 Tn = a + (n – 1) d \ 70 = 51 + (n – 1)1 Þ n – 1 = 70 – 51 = 29 Þ n = 29 + 1 = 30
Sn = n2
[a + l]
= 302
[51 + 70]
= 15 (121) = 1815.
(ii) Given sequence is 112
+ 212
+ ..... + 1212
i.e., 32
, 52
+ ..... + 252
Here, a = 32
, d = 52
– 32
= 22
= 1
Oswaal Kerala SSLC Question Bank ForClass 10 Mathematics (Part-1) WithComplete Solutions For 2017 Exam
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