Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 3 Arithmetic Sequences and Algebra Questions and Answers Notes Pdf to clear their doubts.
SSLC Maths Chapter 3 Arithmetic Sequences and Algebra Questions and Answers
Arithmetic Sequences and Algebra Class 10 Questions and Answers Kerala State Syllabus
SCERT Class 10 Maths Chapter 3 Arithmetic Sequences and Algebra Solutions
Class 10 Maths Chapter 3 Kerala Syllabus – Algebraic Form
Textbook Page No. 63
Question 1.
Find the algebraic form of the arithmetic sequences given below:
a) 1,6, 11, 16
b) 2, 7, 12, 17….
c) 21, 32, 43, 54
d) 19, 28, 37
e) 1, 1\(\frac{1}{2}\), 2, 2\(\frac{1}{2}\)
f) \(\frac{1}{6}, \frac{1}{3}, \frac{1}{2}\), …………….
Answer:
a) This is the sequence of numbers 4 less than the multiples of 5. So we can write xn = 5n – 4 (Or) xn = dn + (f – d) = 5n – 4
b) This is the sequence of numbers 3 less than the multiples of 5.
So its n* term will be 5n – 3
(Or ) xn = dn + (f – d) = 5n + (2 – 5) = 5n – 3
c) d = 11, xn = 11n + 10
d) d = 9, xn = 9n + 10
e) d = \(\frac{1}{2}\), x = \(\frac{n}{2}+\frac{1}{2}=\frac{1}{2}\)(n + 1)
Question 2.
The terms of some arithmetic seefuences in two specified positions are given below. Find the alge-braic form of each:
(i) 1st term5, 10th term 23
(ii) ls’term5, 7th term23
(iii) 5th term 10, 10th term 5
(iv) 8th term 2, 12th term 8
Answer:
a) 9d = 18 → d = 2. xn = 2n + 3
b) 6d = 23 – 5 = 18, d = 3, xn = 3n + 2
c) 5d = 5 – 10 = -5, d = -1, f = 14, xn = -n + 15
d) 4d = 8 – 2 = 6, d = \(\frac{6}{4}=\frac{3}{2}\),
x1 = \(\frac{17}{2}\), xn = \(\frac{3}{2}\)n – 10
Question 3.
Prove that the arithmetic sequence with first term \(\frac{1}{3}\) and common difference \(\frac{1}{6}\) contains all natural numbers.
Answer:
Algebraically the sequence can be written as
xn = \(\frac{1}{6}\)n + \(\left(\frac{1}{3}-\frac{1}{6}\right)=\frac{n+1}{6}\)
For n = 5, 11, 17, we get the terms
x5 = \(\frac{5+1}{6}\) = 1
x11 = \(\frac{11+1}{6}\) = 2
x17 = \(\frac{17+1}{6}\) = 3
The natural numbers 1, 2, 3 in the order can be obtained by giving the terms of 5, 11, 17….
That is, the sequence having first term \(\frac{1}{6}\) and common difference \(\frac{2}{3}\) contains all natural numbers.
Question 4.
Prove that the arithmetic sequence with first term \(\frac{1}{3}\) and common difference \(\frac{2}{3}\) contains all odd numbers, but no even numbers.
Answer:
Algebraic form of the sequence is \(\frac{2}{3}\) n + \(\left(\frac{1}{3}-\frac{2}{3}\right)\)
xn = \(\frac{2 n-1}{3}\).
x2 = 1, x5 = 3, x8 = 5
Numerator of n’A term is an odd number. When an odd number is divided by 3 we get the quotient odd, no even number.
No even number is a term of the sequence.
Question 5.
Prove that in the arithmetic sequence 4, 7, 10, … the squares of all terms are also terms of the sequence.
Answer:
Algebraic form of an arithmetic sequence is 3n + 1 Any number in the form 3n + 1 is a term of this sequence for a positive integer n Square of the term of this sequence is (3n + 1)2 = 9n2 + 6n + 1 = 3(3n2 + 2n) + 1
3n2 +2n is a positive integer.
So square of the term is also a term of the sequence.
Question 6.
Prove that the arithmetic sequence 5,8,11,…does not contain any perfect square.
Answer:
The common difference of this sequence is 3
In an arithmetic sequence of positive integer terms, the same remainder will be obtained on dividing the terms by the common difference. Here the remainder is 2
The remainder on dividing a perfect square by 3 is 0 or 1, not 2
Refer : Extended Activity Page for a proof of the statement : The remainder on dividing a perfect square by 3 is 0 or 1
So no perfect square is a term of the sequence .
Class 10 Maths Kerala Syllabus Chapter 3 Solutions – Sums
Textbook Page No. 71
Question 1.
Calculate in head the sum of the arithmetic sequences below
(a) 51 + 52 + 53 + …………. + 70
(b) 1\(\frac{1}{2}\) + 2\(\frac{1}{2}\) + ….. + 12 \(\frac{1}{2}\)
(c) \(\frac{1}{2}\) + 1 + 1\(\frac{1}{2}\) + 2 + ………… + 12\(\frac{1}{2}\)
(d) \(\frac{1}{101}+\frac{3}{101}+\cdots+\frac{201}{101}\)
Answer:
(a) 1210
50 × 20 + (1 + 2 + 3 + …………… +20) = 1000 + 210 = 1210
Or by pairing of 20 numbers with pair sum 121, the sum is 121 × 10 = 1210
(b) 84
(1 + 2 + 3 + …… + 12) + 12 × \(\frac{1}{2}\) = 78 + 6 = 84
By pairing the terms equidistant from both ends, sum is 14 × 6 = 84
(c) 162.5
2(1 + 2 + 3 + ………….. + 12) + 13 × \(\frac{1}{2}\)
= 156 + \(\frac{1}{2}\) × 13
= 162.5
(d) \(\frac{101^2}{101}\) = 101
Question 2.
Calculate the sum of the first 25 terms of the arithmetic sequences below:
(a) 11, 22, 33 ……….
(b) 12, 23, 34 ………….
(c) 21, 32, 43 ……………
(d) 19, 28, 37 ……………
Answer:
(a) This is the sequence of multiples of 11.
Its algebraic form is xn = 11 n
Sum of first 25 terms
= 11 × 1 + 11 × 2 + 11 × 3 – 1 – 11 × 25
Let S be the sum of first n terms.
So S25 = 11(1 + 2 + 3 + …. + 25)
= 11 × \(\frac{25(25+1)}{2}\)
= 11 × 25 × 13
= 3575
(b) This is the sequence having algebraic form 11 n +1 Sum of the first 25 terms
= 11(1 + 2 + 3 …….. + 25) + 25
= 3575 + 25
= 3600
(c) This sequence having algebraic form 11n + 10
Sn = 11(1 + 2 + 3 … + 25) + 10 × 25
= 3575 + 250
= 3825
(d) nth term of the sequence is 9n + 10
Sum of the first 25 terms
= 9(1 + 2 + 3 + ………… + 25) + 250
= 3175
Question 3.
Find the sum of all multiples of 9 among three digit numbers
Answer:
108, 117, 126 999 is the sequence of three digit multiples of 9
Algebraically the sequence is 9n + 99.
9n + 99 = 999 ⇒ 9n = 900, n = 100
Sum of the three digit multiples of 9 is
(108 + 999) × \(\frac{100}{2}\) = 55350
Question 4.
The nth term of some arithmetic sequences are given below. Find the sum of first n terms of each:
(a) 2n + 3
(b) 3n + 2
(c) 2n – 3
(d) 3n – 2
Answer:
(a) xn = 2n + 3, x1 = 2 × 1 + 3 = 5
Sum =(x1 + xn) × \(\frac{n}{2}\) → (5 + 2n + 3) × \(\frac{n}{2}\)
Sum = (8 + 2n) × \(\frac{n}{2}\) = n2 + 4n
(b) xn = 3n + 2, x1 = 3 × 1 + 2 = 5
Sum =(x1 + xn) x \(\frac{n}{2}\) = (5 + 3n + 2) x \(\frac{n}{2}\) = \(\frac{7}{2}\)n + \(\frac{3}{2}\)n2
(c) xn = 2n – 3, x1 = -1, Sum = n2 – 2n
(d) xn = 3n – 2, x1 =1, Sum = \(\frac{-n}{2}+\frac{3}{2}\)n2
Question 5.
The sum of the first n terms of some arithmetic sequence are given below. Find the nth term of each.
(a) n2 + 2n
(b) 2n2 + n
(c) n2 – 2n
(d) 2n2 – n
(c) n2 – n
Answer:
(a) Sum = n2 + 2n
x1 = 12 + 2 × 1 = 3, x1 + x2 = 22 + 2 × 2 = 8
x2 = 8 – 3 = 5, common difference = 5 – 3 = 2.
xn = dn + (f – d) = 2n + 3 – 2 = 2n + 1
Note that x and f stands for the first term.
(b) Sum = 2n3 + n,x1 = 2 × 12 + 1 = 3
x1 + x2 = 2 × 22 + 2 = 10
x2 = 10 – 3 = 7, d = 7 – 3 = 4
xn = 4n + (3 – 4) = 4n -1
(c) Sum = n2 – 2n
x1 = 12 – 2 × 1 = -1, x1 + x2 2 = 22 – 2 × 2 = 0
x2 = 0 – (-1) = 1.
Common difference d = 1 – (-1) = 2
xn = dn + (f – d) = 2n + (-1 – 2) = 2n – 3
(d) Sum = 2n2 -n
x1 = 2 × 12 – 1 = 1, x1 + x2 = 2 × 22 – 2 = 6
x2 =6 – 1 = 5, d = 5 – 1 = 4
xn = 4n – 3
(e) Sum = n2 – n
There is a short cut for getting «th term from its algebraic form
pn2 + qn is the general form of the sum. Two times the coefficient of n2 is the common difference.
Here d = 2p
d = 2 × 1 = 2, f = 12 —1 = 0,
xn = 2n + (0 – 2) = 2n – 2
Short cut is useful for answering objective type questions.
Question 6.
(i) Calculate the sum of the first 20 natural numbers
(ii) Calculate the sum of the first 20 numbers got by multiplying the natural numbers by 5 and adding 1. Calculate also the sum of the first n terms.
Answer:
(a) Sum of the first 20 terms = \(\frac{20}{2}\)(20 + 1)
210
(b) 1 × 5 + 1, 2 × 5 + 1, 3 × 5 + 1, ….. is the sequence Its algebra is xn
= 5n +1
Sum of the first 20 terms
y20 = 5(1 + 2 + 3 ……….. + 20) + 20
= 5 × 210 + 20
= 1070
(c) Sum = 5 × \(\frac{1}{2}\)n(n + 1) + n
= \(\frac{5}{2}\) n2 + \(\frac{7}{2}\)n
Question 7.
How much more is the sum of first 25 terms of the arithmetic sequence 15, 21, 27, than the sum of the first 25 terms of the arithmetic sequence 7, 13, 19….?
Answer:
Both sequence have common difference 6 .
So the difference between the terms in the same position are equal. It is 8
Difference of sums = 25 × 8 = 200
Question 8.
10th term of an arithmetic sequence is 50 and 21st term is 75. Calculate the sum of first 30 terms of this sequence.
Answer:
There are 12 terms from 10th to 21st
x10 + x21 = 125.Therefore x1 + x30 = 125
Sum of first 30 terms = (x1 + x30) ×x \(\frac{30}{2}\)
= 125 × 15
= 1875
Arithmetic Sequences and Algebra Class 10 Notes Pdf
Class 10 Maths Chapter 3 Arithmetic Sequences and Algebra Notes Kerala Syllabus
Introduction
This is the continuation of the first unit where we defined arithmetic sequence and its general numerical properties. To understand sequence for higher studies and theoretical analysis the rule behind the sequence should be discussed algebraically.
Algebraic form of the arithmetic sequence:
Algebraic form of an arithmetic sequence is the relation between terms of the sequence and natural numbers. For an arithmetic sequence the nth term will be in the form xn = an + b . If f is the first term and d is the common difference, xn = dn + (f -d) or xn = f + (n – 1)d using nth term we can write the numerical form of the sequence by giving n = 1, 2, 3 ….
Summation of the terms of an arithmetic sequence
In the first unit we add the terms of the sequence by pairing of terms equidistant from both ends. Here we establish the sum as (x1 + xn) × \(\frac{n}{2}\).
The sum of first n natural numbers, sum of first n odd numbers and sum of first n even numbers also discuss on the basis of the algebraic form of the sequence.
→ Algebraic form of an arithmetic sequence is the relation between terms of the sequence and natural numbers in the order written in algebraic language.
→ Algebraic form can be used for problem solving as well as theoretical purposes.
→ If/is the first term and d is the common difference of an arithmetic sequence then n* term or algebraic form can be obtained by adding (« – 1) times common difference to the first term. It is
xn = f + (n – 1)d.
This can be written as xn = dn + (f – d).
nth term of an arithmetic sequence is always in the form an + b. In this form a is the common difference and a + b is the first term.
→ Sum of the first n natural numbers can be calculated by pairing of numbers equidistant from both ends. Sum of first n natural numbers = \(\frac{n}{2}\)(n + 1)
→ Sum of the first n odd numbers = n2. Sum of the first n even numbers – n(n + 1)
→ The sum of the specific number of terms can be calculated using its algebaric form.
If an + b is the algebraic form of an arithmetic sequence then sum of the first n terms ]
= a(1 + 2 + 3 + …….. + n) + nb = a \(\frac{n}{2}\) (n + 1) + nb
The sum of the first n terms can be calculated as = (x1 + xn)x-
→ The sum of the first n terms of an arithmetic sequence is in the form pn2 + qn.
For n = 1 we get p + q as the first term. For n = 2 we get the sum of first two terms.
By subtracting: sum of first two terms – sum of first term gives second term x2
x2 – x1 = d, the common difference. Using first term and common difference we can calculate nth term or algebraic form of the sequence.
→ Note that, in the sum of first n terms, twice the coefficient of n2 will be the common difference.
→ The algebraic form of any arithmetic sequence is .
xn = an + b .
where a and b are specific numbers.
→ The sum of any number of consecutive natural numbers starting from one is the half the product of the last number and the next.
For the arithmetic sequence given by xn = an + b the sum of the first n terms is,
x1 + x2 + ……….. + xn = \(\frac{1}{2}\)an(n + 1) + nb
→ The sum of consecutive terms of an arithmetic sequence is half the product of the sum of the first and last terms by the number of terms.
→ The algebraic form of the sum of the first« terms of an arithmetic sequence is pn2 + qn where p and q are specific numbers.