Mathematics solution NCERT
Class 9 - Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions
Q1: Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Given AP: 3, 8, 13, 18, ...
\(First term, a = 3\)
\(Common difference, d = 8 - 3 = 5\)
\(Formula: t_{n} = a + (n - 1)d\)
10th Term
\(t_{10} = 3 + (10 - 1)(5)\)
\(= 3 + 45\)
\(= 48\)
Answer: t10 = 48
26th Term
\(t_{26} = 3 + (26 - 1)(5)\)
\(= 3 + 125\)
\(= 128\)
Answer: t26 = 128
Q2: Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.
Given AP: 21, 18, 15, ...
\(a = 21\)
\(d = -3\)
\(t_{n} = a + (n - 1)d\)
\(t_{n} = 21 + (n - 1)(-3)\)
\(t_{n} = 24 - 3n\)
For −81
\(24 - 3n = -81\)
\(-3n = -105\)
\(n = 35\)
Answer: −81 is the 35th term.
Checking Whether 0 is a Term
\(24 - 3n = 0\)
\(3n = 24\)
\(n = 8\)
Since n is a whole number, 0 is a term of the AP.
Answer: 0 is the 8th term.
Q3: Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
Given AP: 11, 8, 5, 2, ...
\(a = 11\)
\(d = -3\)
nth Term
\(t_{n} = a + (n - 1)d\)
\(= 11 + (n - 1)(-3)\)
\(= 11 - 3n + 3\)
\(= 14 - 3n\)
Answer:
\(t_{n} = 14 - 3n\)
Recursive Rule
\(t_{1} = 11\)
\(t_{n+1} = t_{n} - 3\)
Recursive Rule:
\(t_{1} = 11, t_{n+1} = t_{n} - 3\)
Q4: An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Number of terms = 50
\(3^{rd} term = 12\)
\(Last term = 106\)
\(a + 2d = 12\)
\(a + 49d = 106\)
Subtracting:
\(47d = 94\)
\(d = 2\)
\(a + 2(2) = 12\)
\(a = 8\)
29th Term
\(t_{29} = a + 28d\)
\(= 8 + 28(2)\)
\(= 8 + 56\)
\(= 64\)
Answer: t29 = 64
Q5: How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
\(Smallest 2-digit multiple of 3 = 12\)
\(Largest 2-digit multiple of 3 = 99\)
AP: 12, 15, 18, ..., 99
\(a = 12, d = 3, l = 99\)
Number of Terms
\(99 = 12 + (n - 1)3\)
\(87 = 3(n - 1)\)
\(29 = n - 1\)
\(n = 30\)
Answer: 30 numbers
Sum of All Numbers
\(S_{n} = n/2 [a + l]\)
\(= \frac{30}{2} (12 + 99)\)
\(= 15 \times 111\)
\(= 1665\)
Answer: Sum = 1665
Q6: Harish started work at an annual salary of `5,00,000 and received an increment of `20,000 each year. After how many years did his income reach `7,00,000?
\(Initial salary = ₹5,00,000\)
\(Annual increment = ₹20,000\)
\(Target salary = ₹7,00,000\)
AP: 5,00,000, 5,20,000, 5,40,000, ...
\(a = 5,00,000\)
\(d = 20,000\)
\(t_{n} = 7,00,000\)
\(7,00,000 = 5,00,000 + (n - 1)(20,000)\)
\(2,00,000 = (n - 1)(20,000)\)
\(10 = n - 1\)
\(n = 11\)
The 11th salary is ₹7,00,000.
\(Years required = 10\)
Answer: After 10 years.
Q7: A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
Rows contain: 1, 2, 3, 4, ..., 25
This is an AP with:
\(a = 1\)
\(d = 1\)
\(n = 25\)
Total Marbles
\(S_{25} = \frac{25}{2} [2(১৷) + (25 - 1)(1)]\)
\(= \frac{25}{2} [2 + 24]\)
\(= \frac{25}{2} \times 26\)
\(= 25 \times 13\)
\(= 325\)
Answer: The child uses 325 marbles in all.
Summary of Answers
| Question | Answer |
|---|---|
| Q1 | t10 = 48, t26 = 128 |
| Q2 | −81 is 35th term, 0 is 8th term |
| Q3 | tn = 14 − 3n; t1 = 11, tn+1 = tn − 3 |
| Q4 | 29th term = 64 |
| Q5 | 30 numbers; Sum = 1665 |
| Q6 | After 10 years |
| Q7 | 325 marbles |