RD Chapter 19 Arithmetic Progressions Ex 19.1 Solutions
Question - 1 : - If the nth┬аterm of a sequence is given by an┬а=n2┬атАУ n+1, write down its first five terms.
Answer - 1 : -
Given:
an┬а= n2┬атАУn+1
By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.
When n = 1:
a1┬а=(1)2┬атАУ 1 + 1
= 1 тАУ 1 + 1
= 1
When n = 2:
a2┬а=(2)2┬атАУ 2 + 1
= 4 тАУ 2 + 1
= 3
When n = 3:
a3┬а=(3)2┬атАУ 3 + 1
= 9 тАУ 3 + 1
= 7
When n = 4:
a4┬а=(4)2┬атАУ 4 + 1
= 16 тАУ 4 + 1
= 13
When n = 5:
a5┬а=(5)2┬атАУ 5 + 1
= 25 тАУ 5 + 1
= 21
тИ┤┬аFirst five termsof the sequence are 1, 3, 7, 13, 21.
Question - 2 : - A sequence is defined by an┬а= n3┬атАУ 6n2┬а+11n тАУ 6, n┬атИИ┬аN. Show that the first threeterms of the sequence are zero and all other terms are positive.
Answer - 2 : -
Given:
an┬а= n3┬атАУ6n2┬а+ 11n тАУ 6, n┬атИИ┬аN
By using the values n= 1, 2, 3 we can find the first three terms.
When n = 1:
a1┬а=(1)3┬атАУ 6(1)2┬а+ 11(1) тАУ 6
= 1 тАУ 6 + 11 тАУ 6
= 12 тАУ 12
= 0
When n = 2:
a2┬а=(2)3┬атАУ 6(2)2┬а+ 11(2) тАУ 6
= 8 тАУ 6(4) + 22 тАУ 6
= 8 тАУ 24 + 22 тАУ 6
= 30 тАУ 30
= 0
When n = 3:
a3┬а=(3)3┬атАУ 6(3)2┬а+ 11(3) тАУ 6
= 27 тАУ 6(9) + 33 тАУ 6
= 27 тАУ 54 + 33 тАУ 6
= 60 тАУ 60
= 0
This shows that thefirst three terms of the sequence is zero.
Now, letтАЩs check forwhen n = n:
an┬а= n3┬атАУ6n2┬а+ 11n тАУ 6
= n3┬атАУ6n2┬а+ 11n тАУ 6 тАУ n + n тАУ 2 + 2
= n3┬атАУ6n2┬а+ 12n тАУ 8 тАУ n + 2
= (n)3┬атАУ3├Ч2n(n тАУ 2) тАУ (2)3┬атАУ n + 2
By using the formula,{(a тАУ b)3┬а= (a)3┬атАУ (b)3┬атАУ 3ab(aтАУ b)}
an┬а=(n тАУ 2)3┬атАУ (n тАУ 2)
Here, n тАУ 2 willalways be positive for n > 3
тИ┤┬аan┬аisalways positive for n > 3
Question - 3 : - Find the first four terms of the sequencedefined by a1┬а= 3 and an┬а= 3anтАУ1┬а+2, for all n > 1.
Answer - 3 : -
Given:
a1┬а= 3and an┬а= 3anтАУ1┬а+ 2, for all n > 1
By using the values n= 1, 2, 3, 4 we can find the first four terms.
When n = 1:
a1┬а= 3
When n = 2:
a2┬а=3a2тАУ1┬а+ 2
= 3a1┬а+2
= 3(3) + 2
= 9 + 2
= 11
When n = 3:
a3┬а=3a3тАУ1┬а+ 2
= 3a2┬а+2
= 3(11) + 2
= 33 + 2
= 35
When n = 4:
a4┬а=3a4тАУ1┬а+ 2
= 3a3┬а+2
= 3(35) + 2
= 105 + 2
= 107
тИ┤┬аFirst four termsof sequence are 3, 11, 35, 107.
Question - 4 : - Write the first five terms in each of the following sequences:
(i) a1┬а= 1, an┬а= anтАУ1┬а+ 2, n> 1
(ii) a1┬а= 1 = a2, an┬а= anтАУ1┬а+anтАУ2, n > 2
(iii) a1┬а= a2┬а=2, an┬а= anтАУ1┬атАУ1, n > 2
Answer - 4 : -
(i)┬аa1┬а=1, an┬а= anтАУ1┬а+ 2, n > 1
By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.
Given:
a1┬а= 1
When n = 2:
a2┬а= a2тАУ1┬а+2
= a1┬а+2
= 1 + 2
= 3
When n = 3:
a3┬а= a3тАУ1┬а+2
= a2┬а+2
= 3 + 2
= 5
When n = 4:
a4┬а= a4тАУ1┬а+2
= a3┬а+2
= 5 + 2
= 7
When n = 5:
a5┬а= a5тАУ1┬а+2
= a4┬а+2
= 7 + 2
= 9
тИ┤┬аFirst five termsof the sequence are 1, 3, 5, 7, 9.
(ii)┬аa1┬а= 1= a2, an┬а= anтАУ1┬а+ anтАУ2, n> 2
By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.
Given:
a1┬а= 1
a2┬а= 1
When n = 3:
a3┬а= a3тАУ1┬а+a3тАУ2
= a2┬а+a1
= 1 + 1
= 2
When n = 4:
a4┬а= a4тАУ1┬а+a4тАУ2
= a3┬а+a2
= 2 + 1
= 3
When n = 5:
a5┬а= a5тАУ1┬а+a5тАУ2
= a4┬а+a3
= 3 + 2
= 5
тИ┤┬аFirst five termsof the sequence are 1, 1, 2, 3, 5.
(iii)┬аa1┬а= a2┬а=2,an┬а= anтАУ1┬атАУ 1, n > 2
By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.
Given:
a1┬а= 2
a2┬а= 2
When n = 3:
a3┬а= a3тАУ1┬атАУ1
= a2┬атАУ1
= 2 тАУ 1
= 1
When n = 4:
a4┬а= a4тАУ1┬атАУ1
= a3┬атАУ1
= 1 тАУ 1
= 0
When n = 5:
a5┬а= a5тАУ1┬атАУ1
= a4┬атАУ1
= 0 тАУ 1
= -1
тИ┤┬аFirst five termsof the sequence are 2, 2, 1, 0, -1.
Question - 5 : - The Fibonacci sequence is defined by a1┬а= 1┬а= a2,an┬а= anтАУ1┬а+ anтАУ2┬аfor n > 2.Find (an+1)/an┬аfor n = 1, 2, 3, 4, 5.
Answer - 5 : -
Given:
a1┬а= 1
a2┬а= 1
an┬а= anтАУ1┬а+anтАУ2
When n = 1:
(an+1)/an┬а=(a1+1)/a1
= a2/a1
= 1/1
= 1
a3┬а= a3тАУ1┬а+a3тАУ2
= a2┬а+a1
= 1 + 1
= 2
When n = 2:
(an+1)/an┬а=(a2+1)/a2
= a3/a2
= 2/1
= 2
a4┬а= a4тАУ1┬а+a4тАУ2
= a3┬а+a2
= 2 + 1
= 3
When n = 3:
(an+1)/an┬а=(a3+1)/a3
= a4/a3
= 3/2
a5┬а= a5тАУ1┬а+a5тАУ2
= a4┬а+a3
= 3 + 2
= 5
When n = 4:
(an+1)/an┬а=(a4+1)/a4
= a5/a4
= 5/3
a6┬а= a6тАУ1┬а+a6тАУ2
= a5┬а+a4
= 5 + 3
= 8
When n = 5:
(an+1)/an┬а=(a5+1)/a5
= a6/a5┬а=8/5
тИ┤ Value of (an+1)/an┬аwhenn = 1, 2, 3, 4, 5 are 1, 2, 3/2, 5/3, 8/5