Class 9 Advanced Maths Chapter 6 Exploring Some More Progressions Ex 6.2 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 6 Exploring Some More Progressions Ex 6.2 to verify their answers.

Advanced Maths Class 9 Exercise 6.2 Solutions

Class 9 Advanced Maths Ex 6.2 Solutions

Question 1.
Find the n^th term and the sum of the first n terms of the series
1 + 9 + 24 + 46 + 75 + …
Solution:
Here, we have
Series: 1, 9, 24, 46, 75,…
To Find: t_n and S_n
The given sequence can be written as
Original series: 1, 9, 24, 46, 75
1^st differences: 8, 15, 22, 29
2^nd differences: 7, 7, 7
The 2^nd differences are constant = 7, so 1^st differences form an AP
b = 1 (first term of series)
a = 8 (first term of 1st differences)
d = 7 (common difference of 1st differences = first term of 2^nd differences)
Find the n^th term.
t_n = b + a(n – 1)+ \(\frac{d(n-1)(n-2)}{2}\)
⇒ t_n = 1 + 8(n – 1) + \(\frac{7(n-1)(n-2)}{2}\)
⇒ t_n = 1 + 8n – 8 + \(\left(\frac{7}{2}\right)\)(n² – 3n + 2)
⇒ t_n = 1 + 8n – 8 + \(\frac{7 n^2-21 n+14}{2}\)
⇒ t_n = \(\frac{2+16 n-16+7 n^2-21 n+14}{2}\)
⇒ t_n = \(\frac{7 n^2-5 n}{2}\)
Find sum to n terms.
S_n = bn + a.\(\frac{n(n-1)}{2}\) + d.\(\frac{n(n-1)(n-2)}{6}\)
= 1.n + 8.\(\frac{n(n-1)}{2}\) + 7.\(\frac{n(n-1)(n-2)}{6}\)
= n + 4n(n – 1) + \(\left(\frac{7}{6}\right)\)n(n – 1)(n -2)
=n[1 + 4(n – 1) + \(\left(\frac{7}{6}\right)\)(n – 1)(n -2)]
\(\begin{aligned}
& =n \frac{6+24 n-24+7\left(n^2-3 n+2\right)}{6} \\
& =n \frac{\left[6+24 n-24+7 n^2-21 n+14\right]}{6} \\
& =n \frac{\left[7 n^2+3 n+20-24\right]}{6} \\
& =n \frac{\left[7 n^2+3 n-4\right]}{6} \\
& =n \frac{\left[7 n^2+7 n-4 n-4\right]}{6} \\
& =n \frac{[7 n(n+1)-4(n+1)]}{6}
\end{aligned}\)
S_n = \(\frac{n(7 n-4)(n+1)}{6}\)

Question 2.
Find the 10^th term and the sum of the first 10 terms of the series
4 + 5 + 9 + 16 + 26 + …
Solution:
Here, we have Series: 4, 5, 9, 16, 26,…
To Find: t₁₀ and S₁₀.
The given sequence can be written as
Original Series: 4, 5, 9, 16, 26
1^st differences: 1, 4, 7, 10
2^nd differences: 3, 3, 3
The 2^nd differences are constant = 3.
b = 4, a = 1, d = 3
Find t₁₀.
⇒ t_n = b + a(n – 1) + \(\frac{d(n-1)(n-2)}{2}\)
⇒ t₁₀ = 4 + 1(9) + \(\frac{3(9)(8)}{2}\)
⇒ t₁₀ = 4 + 9 + 3 × 36
⇒ t₁₀ = 4 + 9 + 108 = 121
⇒ t₁₀ = 121
Find S₁₀
S_n = bn + a.\(\frac{n(n-1)}{2}\) + d.\(\frac{n(n-1)(n-2)}{6}\)
= 4(10) + 1.\(\frac{(10)(9)}{2}\) + 3.\(\frac{(10)(9)(8)}{6}\)
= 40 + 45 + 360
S₁₀ = 445

Question 3.
Find the n^th term and the sum of the first 12 terms of the series
3 + 6 + 11 + 18 + 27 + …
Solution:
Here, we have
Series: 3, 6, 11, 18, 27,…
To Find: t_n and S₁₂
The given sequence can be written as
Original Series: 3, 6, 11, 18, 27
1^st differences: 3, 5, 7
2^nd differences: 2, 2, 2
2^nd differences are constant = 2.
b = 3, a = 3, d = 2
Find n^th term.
t_n = b + a(n – 1) + \(\frac{d(n-1)(n-2)}{2}\)
= 3 + 3(n – 1) + \(\frac{2(n-1)(n-2)}{2}\)
= 3 + 3n – 3 + (n- 1)(n – 2)
= 3n + (n² – 3n + 2)
t_n = n² + 2
Find S₁₂.
S_n = bn + a.\(\frac{n(n-1)}{2}\) + d.\(\)
= 3n + 3.\(\frac{n(n-1)}{2}\) + 2.\(\frac{n(n-1)(n-2)}{6}\)
= 3n + \(\left(\frac{3}{2}\right)\)n(n – 1) + \(\left(\frac{1}{3}\right)\)n(n – 1)(n – 2)
For S₁₂, substitute n = 12:
S₁₂ = 3(12) + \(\left(\frac{3}{2}\right)\)(12)(11) + \(\left(\frac{1}{3}\right)\)(12)(11)(10)
= 36 + \(\left(\frac{3}{2}\right)\)(132) + \(\left(\frac{1}{3}\right)\)(1320)
= 36 + 198 + 440
S₁₂ = 674
Alternatively, using t_n = n² + 2:
S_n = ∑(n² + 2) = \(\frac{n(n+1)(2 n+1)}{6}\) + 2n
S₁₂ = 12 × 13 × \(\frac{25}{6}\) + 2 × 12
= 650 + 24 = 674

Question 4.
Find the 8^th term and the sum of the first 8 terms of the series
4 + 13 + 28 + 49 + 76 + …
Solution:
Here, we have
Series: 4, 13, 28, 49, 76,…
To Find: t₈ and S₈
The given sequence can be written as
Original Series: 4, 13, 28, 49, 76
1^st differences: 9, 15, 21, 27
2^nd differences: 6, 6, 6
2^nd differences are constant = 6.
b = 4, a = 9, d = 6
Find t₈.
t_n = b + a(n – 1) + \(\frac{d(n-1)(n-2)}{2}\)
= 4 + 9(7) + \(\frac{6(7)(6)}{2}\)
= 4 + 63 + 126
t₈ = 193
Find S₈.
S_n = bn + a.\(\frac{n(n-1)}{2}\) + d.\(\frac{n(n-1)(n-2)}{6}\)
= 4(8) + 9.\(\frac{(8)(7)}{2}\) + 6.\(\frac{(8)(7)(6)}{6}\)
= 32 + 9 × 28 + 6 × 56
= 32 + 252 + 336
S₈ = 620