Inter 1st Year Maths Sequences and Series Solutions Exercise 8a

Practicing the AP Board Solutions Class 11 Maths and Chapter 8 Inter 1st Year Maths Sequences and Series Solutions Exercise 8a Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Sequences and Series Solutions Exercise 8a

Sequences and Series Exercise 8a Solutions

Sequences and Series Class 11 Exercise 8a Solutions – Sequences and Series 8a Exercise Solutions

I. Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

Question 1.
a_n = n(n + 2)
Solution:
Given, nth term of a sequence a_n = n(n + 2)
on substituting n = 1, 2, 3, 4, and 5,
We get the first five terms
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.
a_n = \(\frac{\mathrm{n}}{\mathrm{n}+1}\)
Solution:
Given the nth term, a_n = \(\frac{\mathrm{n}}{\mathrm{n}+1}\)
On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}\) and \(\frac {5}{6}\).

Question 3.
a_n = 2ⁿ
Solution:
Given the nth term, a_n = 2ⁿ
On substituting n = 1, 2, 3, 4, 5, we get
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.
a_n = \(\frac{2 n-3}{6}\)
Solution:
Given the nth term a_n = \(\frac{2 n-3}{6}\)
On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are \(-\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}\) and \(\frac {7}{6}\)

Question 5.
a_n = (-1)^n-1 5ⁿ⁺¹
Solution:
Given the nth term, a_n = (-1)^n-1 5ⁿ⁺¹
On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 25, -125, 625, -3125, and 15625.

Question 6.
a_n = \(\mathrm{n} \frac{\mathrm{n}^2+5}{4}\)
Solution:
Given the nth term, a_n = \(\mathrm{n} \frac{\mathrm{n}^2+5}{4}\)
On substituting n = 1, 2, 3, 4, 5
We get the first 5 terms.

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:

Question 7.
a_n = 4n – 3; a₁₇, a₂₄
Solution:
Given, the nth term of the sequence is a_n = 4n – 3
On substituting n = 17, we get
a₁₇ = 4(17) – 3 = 68 – 3 = 65
Next, on substituting n = 24 we get
a₂₄ = 4(24) – 3 = 96 – 3 = 93

Question 8.
a_n = \(\frac{n^2}{2^n}\); a₇
Solution:
Given, The nth term of the sequence is a_n = \(\frac{n^2}{2^n}\)
Now, on substituting n = 7, we get
a₇ = \(\frac{7^2}{2^7}=\frac{49}{128}\)

Question 9.
a_n = (-1)^n-1 n³; a₉
Solution:
Given, the nth term of the sequence is a_n = (-1)^n-1 n³
on substituting n = 9 we get
a₉ = (-1)^9-1 (9)³ = 1 × 729 = 729

Question 10.
a_n = \(\frac{n(n-2)}{n+3}\); a₂₀
Solution:
Given, The nth term of the sequence is a_n = \(\frac{n(n-2)}{n+3}\)
On substituting n = 20, we get
a₂₀ = \(\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}\)

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, a_n = 3a_n-1 + 2 for all n > 1
Solution:
Given, a_n = 3a_n-1 + 2 and a₁ = 3 Then,
a₂ = 3a₁ + 2 = 3(3) + 2 = 11
a₃ = 3a₂ + 2 = 3(11) + 2 = 35
a₄= 3a₃ + 2 = 3(35) + 2 = 107
a₅ = 3a₄ + 2 = 3(107) + 2 = 325
Thus, the first 5 terms of the sequence are 3, 11, 35, 107, and 323.
Hence, the corresponding series is 3 + 11 + 35 + 107 + 323……..

Question 2.
a₁ = -1, a_n = \(\frac{a_{n-1}}{n}\), n ≥ 2
Solution:
Given, a_n = \(\frac{a_{n-1}}{n}\) and a₁ = -1, then,

Thus, the first 5 terms of the sequence are \(-1,-\frac{1}{2},-\frac{1}{6}, \frac{-1}{24}\) and \(\frac {-1}{20}\).
Hence, the corresponding series is \(-1+\left(-\frac{1}{2}\right)+\left(-\frac{1}{6}\right)+\left(-\frac{1}{24}\right)+\left(-\frac{1}{120}\right)+\ldots\)

Question 3.
a₁ = a₂ = 2, a_n = a_n-1 – 1, n > 2
Solution:
a₁ = a₂, a_n = a_n-1 – 1
Then, a₃ = a₂ – 1 = 2 – 1 = 1
a₄ = a₃ – 1 = 1 – 1 = 0
a₅ = a₄ – 1 = 0 – 1 = -1
Thus, the first 5 terms of the sequence are 2, 2, 1, 0, and -1.
The corresponding series is 2 + 2 + 1 + 0 + (-1) + …………

Question 4.
The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_n-1 + a_n-2, n > 2. Find \(\frac{a_{n+1}}{a_n}\), for n = 1, 2, 3, 4, 5
Solution:
Given, 1 = a₁ = a₂
a_n = a_n-1 + a_n-2, n > 2
so, a₃ = a₂ + a₁ = 1 + 1 = 2
a₄ = a₃ + a₂ = 2 + 1 = 3
a₅ = a₄ + a₃ = 3 + 2 = 5
a₆ = a₅ + a₄ = 5 + 3 = 8
Thus,