Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8a Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8a
I. Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are given.
Question 1.
an = n(n + 2)
Solution:
Given that an = n(n + 2).
Substituting n = 1, 2, 3, 4 and 5, we have
a1 = 1(1 + 2) = 3; a2 = 2(2 + 2) = 8; a3 = 3(3 + 2) = 15; a4 = 4(4 + 2) = 24; a5 = 5(5 + 2) = 35
Hence, the required terms are 3, 8, 15, 24 and 35.
Question 2.
an = \(\frac{\mathrm{n}}{\mathrm{n}+1}\)
Solution:
Given that an = \(\frac{\mathrm{n}}{\mathrm{n}+1}\)
Substituting n = 1, 2, 3, 4 and 5 we have,
a1 = \(\frac{1}{1+1}=\frac{1}{2}\); a2 = \(\frac{2}{2+1}=\frac{2}{3}\); a3 = \(\frac{3}{3+1}=\frac{3}{4}\); a4 = \(\frac{4}{4+1}=\frac{4}{5}\); a5 = \(\frac{5}{5+1}=\frac{5}{6}\)
Hence, the required terms are \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}\) & \(\frac{5}{6}\).
Question 3.
an = 2n
Solution:
Given that an = 2n
Substituting n = 1, 2, 3, 4 and 5 we have
a1 = 21 = 2, a2 = 22 = 4, a3 = 23 = 8, a4 = 24 = 16, a5 = 25 = 32
Hence, the required terms are 2, 4, 8, 16 and 32.
Question 4.
an = \(\frac{2 n-3}{6}\)
Solution:
Given that an = \(\frac{2 n-3}{6}\)
Substituting n = 1, 2, 3, 4 and 5 we have
a1 =(-1)1-1.51+1 = (-1)0.52 = 25,
a2 =(-1)2-1.52+1 = (-1)1.53 = 25,
a3 =(-1)3-1.53+1 = (-1)2.54 = 25,
a4 =(-1)4-1.54+1 = (-1)3.55 = 25,
a5 =(-1)5-1.55+1 = (-1)4.56 = 25,
Hence, the required terms are 25, -125, 625, -3125 & 15625
Question 6.
an = n\(\frac{n^2+5}{4}\)
Solution:
Given that an = n\(\frac{n^2+5}{4}\)
Substituting n = 1, 2, 3, 4 and 5 we have
a1 = 1 × \(\left[\frac{1^2+5}{4}\right]\) = 1 × \(\left[\frac{2^2+5}{4}\right]\)
a2 = 2 × \(\left[\frac{3^2+5}{4}\right]\) = 1 × \(\frac{9}{4}=\frac{9}{2}\)
a3 = 3 × \(\left[\frac{4^2+5}{4}\right]\) = 1 × \(\frac{14}{4}=\frac{21}{2}\)
a4 = 4 × \(\left[\frac{5^2+5}{4}\right]\) = 1 × \(\frac{21}{4}\) = 21
a5 = 5 × \(\left[\frac{5^2+5}{4}\right]\) = 1 × \(\frac{30}{4}=\frac{75}{2}\)
Hence, the required terms are \(\frac{3}{2}, \frac{9}{2}, \frac{21}{2}\), 21 and \(\frac{75}{2}\)
Find the indicated terms in each of the sequences In Exercises 7 to 10 whose nth terms are given.
Question 7.
an = 4n – 3;(a17, a24)
Solution:
Given that an = 4n – 3
Substituting n = 17, we have a17 = 4(17) – 3 = 68 – 3 = 65
Substituting n = 24, we have a24 = 4(24) – 3 = 96 – 3 = 93
Question 8.
an = \(\frac{\mathrm{n}^2}{2^{\mathrm{n}}}\), a7.
Solution:
Given that an = \(\frac{\mathrm{n}^2}{2^{\mathrm{n}}}\).
Substituting n = 7,we have a7 = \(\frac{7^2}{2^7}=\frac{49}{128}\)
Question 9.
an = (-1)n-1n3, a9
Solution:
Given that an = \(\frac{\mathrm{n}(\mathrm{n}-2)}{\mathrm{n}+3}\)
Substituting n = 9, we have, a9 = (-1)9-1. 93 = (-1)8 . 93 = 93 = 729
Question 10.
an = \(\frac{n(n-2)}{n+3}\);a20
Solution:
Given that a = \(\frac{n(n-2)}{n+3}\)
Substituting n = 20, we have a20 = \(\frac{20(20-2)}{20+3}=\frac{20 \times 18}{23}=\frac{360}{23}\)
II. Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series.
Question 1.
a1 = 3, an = 3an-1 + 2 for all n > 1
Solution:
Given that a1 = 3, an = 3an-1 + 2 for all n> 1
Substituting n = 2, 3, 4 and 5, we have
a2 = 3a2-1 + 2 = 3a1 + 2 = 3 × 3 + 2 = 11,
a3 = 3a3-1 + 2 = 3a2 + 2 = 3 × 11 + 2 = 11,
a4 = 3a4-1 + 2 = 3a3 + 2 = 3 × 35 + 2 = 11,
a5 = 3a5-1 + 2 = 3a4 + 2 = 3 × 107 + 2 = 11,
Therefore, the required terms are 3, 11, 35, 107 and 323.
Hence, the corresponding series is 3 + 11 +35 + 107 + 323 + ………..
Question 2.
a1 = -1, an = \(\frac{a_{n-1}}{n}\); n ≥ 2
Solution:
Given that a1 = -1, an = \(\frac{\mathrm{a}_{\mathrm{n}-1}}{\mathrm{n}}\), n ≥ 2
Substituting n = 2, 3, 4 and 5 we have,
a2 = \(\frac{a_{2-1}}{2}=\frac{a_1}{2}=-\frac{1}{2}\)
a3 = \(\frac{a_{3-1}}{3}=\frac{a_2}{3}=\frac{-\frac{1}{2}}{3}=\frac{-1}{6}\)
a4 = \(\frac{a_{4-1}}{4}=\frac{a_3}{4}=\frac{-\frac{1}{6}}{4}=\frac{-1}{24}\)
a5 = \(\frac{a_{5-1}}{5}=\frac{a_4}{5}=\frac{-\frac{1}{24}}{5}=\frac{-1}{120}\)
Hence, the required terms are -1, \(-\frac{1}{2},-\frac{1}{6},-\frac{1}{24}\) & \(-\frac{1}{120}\)
Hence, the corresponding series is -1 + \(\left(-\frac{1}{2}\right)+\left(-\frac{1}{6}\right)+\left(-\frac{1}{24}\right)+\left(\frac{-1}{120}\right)+\ldots \ldots\)
Question 3.
a1 = a2 = 2; an = an-1 – 1, n > 2
Solution:
Given that a1 = a2 = 2; an = an-1 – 1, n > 2
Substituting n = 3, 4 and 5 we have
a3 = a3-1 – 1 = a2 – 1 = 2 – 1 = 1
a4 = a4-1 – 1 = a3 – 1 = 1 – 1 = 1
a5 = a5-1 – 1 = a4 – 1 = 0 – 1 = 1
Hence, the required terms are 2, 2, 1,0 and -1
Hence, the corresponding series is 2 + 2+ 1 + 0 + (-1) ….
Question 4.
The Fibonacci sequence is defined by 1 = a1 = a2 and an = an-1 + an-2, n > 2.
Find \(\frac{a_{n+1}}{a_n}\), for n = 1, 2, 3, 4, 5.
Solution:
Given that 1 = a1 = a2 and an = an-1 + an-2
Substituting n = 3, 4 and 5, we have
a3 = a3-1 + a3-2 = a2 + a1 = 1 + 1 = 2
a4 = a4-1 + a4-2 = a3 + a2 = 2 + 1 = 2
a5 = a5-1 + a5-2 = a4 + a3 = 3 + 2 = 2
a6 = a6-1 + a6-2 = a5 + a4 = 5 + 3 = 2
Putting n = 1 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}\) = 1
Putting n = 2 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}\) = 2
Putting n = 3 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}\)
Putting n = 4 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}\)
Putting n = 5 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}\)