Class 9 Maths Ganita Manjari Chapter 8 MCQ – Predicting What Comes Next: Exploring Sequences and Progressions

A sequence is an ordered list of numbers, and each number is a term. This list contains exactly five numbers and stops at 96. Therefore, it is finite, with 6 as its first term and 96 as its fifth term.

An explicit formula connects a term’s value to its position number n. Substituting the required position gives that term directly. Unlike a recursive rule, it does not require us to work through earlier terms before finding a later one.

Substitute n = 53 into the explicit formula. We get t₅₃ = 2 × 53 − 1 = 106 − 1 = 105. The number 53 tells us the position of the term, rather than the value of the term.

The sum of the first n positive odd numbers is n². This connects the odd-number sequence with the square-number sequence. For 12 odd numbers, the sum is 12² = 144. The last number, 23, is the twelfth positive odd number.

To find the sixth term, replace every n in the formula with 6. Then t₆ = 6² − 2 × 6 + 3 = 36 − 12 + 3 = 27. The square applies to n, not to the whole expression.

For a number x to occur, n = (x + 3)/5 must be a positive integer. For 98, this gives 101/5 = 20.2, which is not a valid position. The other numbers give positions 20, 35 and 122, respectively.

Apply the recursive rule one step at a time. Starting with 1, the next term is 2 × 1 + 3 = 5. Then we obtain 13 and finally 29. Thus, the first four terms are 1, 5, 13 and 29.

Using the given starting terms, the sequence is 1, 2, 3, 5, 8, 13, 21, 34, 55. Each new term is the sum of the two immediately before it. Therefore, V₉ = V₈ + V₇ = 34 + 21 = 55.

First find T₄ = 1 + 2 + 4 = 7. Next, T₅ = 2 + 4 + 7 = 13. Therefore, T₆ = 4 + 7 + 13 = 24. At each step, use only the three immediately preceding terms.

In the list −7, −3, 1, 5, 9, each term is obtained by adding 4 to the previous term. Its consecutive differences are all 4, so it is an AP. The differences in each of the other lists change.

The common difference is found by subtracting a term from the next term. Here, d = 7 − 11 = −4. Checking again, 3 − 7 = −4. A negative common difference means the terms decrease as their positions increase.

Starting at the first term a, we add d once to reach the second term and twice to reach the third. Reaching the nth term requires n − 1 additions. Therefore, its value is a + (n − 1)d.

There are five equal steps from the 3rd term to the 8th term. Hence, 5d = 27 − 12 = 15, so d = 3. The 10th term is two steps after the 8th, giving 27 + 2 × 3 = 33.

Here, a = 21 and d = −3. Set 21 + (n − 1)(−3) = −81. This simplifies to −3(n − 1) = −102, so n − 1 = 34 and n = 35. Therefore, −81 is the 35th term.

The common difference is 5/2 − 1/2 = 2. Using tₙ = a + (n − 1)d gives t₁₀ = 1/2 + 9 × 2 = 1/2 + 18 = 37/2. Fractional terms can form an AP just like integer terms.

Explanation: In an AP, each increase of 1 in the term position changes the term value by the same amount d. This constant change produces points on a straight line. The line may rise, fall or be horizontal, depending on d.

The booking fee is paid once, and the distance charge is 40 × 10 = ₹400. Adding the fixed fee gives ₹200 + ₹400 = ₹600. The fares for 1 km, 2 km and 3 km form the AP 240, 280, 320.

The total is the sum of the first 25 natural numbers: 1 + 2 + … + 25. Use n(n + 1)/2 to obtain 25 × 26/2 = 325. This total is also the 25th triangular number.

Subtract the sum from 1 to 24 from the sum from 1 to 58. The required total is (58 × 59)/2 − (24 × 25)/2 = 1711 − 300 = 1411. Subtracting through 24 keeps 25 as the first included number.

For n = 44, the sum is 44 × 45/2 = 990, which is below 1000. For n = 45, the sum is 45 × 46/2 = 1035. Since these sums increase with n, the smallest suitable value is 45.

Divide a term by the previous term to find the common ratio. Here, (15/4) ÷ 5 = 3/4, and (45/16) ÷ (15/4) = 3/4. The ratio is constant. Since it is between 0 and 1, these positive terms decrease.

The first term is 3 and the common ratio is 2. The GP formula is tₙ = a × rⁿ⁻¹, so tₙ = 3 × 2ⁿ⁻¹. The exponent is n − 1 because the first term has undergone no multiplication by 2.

The common ratio is −1/2, so t₆ = 3 × (−1/2)⁵ = −3/32. Multiplying by a negative ratio makes the signs alternate. The 6th term is negative because five multiplications by the negative ratio follow the positive first term.

From the 8th term to the 12th term, there are four steps. Each step multiplies the value by 2, giving t₁₂ = 192 × 2⁴ = 192 × 16 = 3072. We do not multiply by the term number itself.

The nth term is 2 × 3ⁿ⁻¹. Setting this equal to 4374 gives 3ⁿ⁻¹ = 2187 = 3⁷. Therefore, n − 1 = 7 and n = 8. The exponent counts the multiplications after the first term.

The counts at the end of hours 1, 2, 3 and 4 are 60, 120, 240 and 480. Four hours mean four doublings from the initial count. Thus, the required number is 30 × 2⁴ = 480.

The first bounce reaches 24 × 3/4 = 18 m, and the second reaches 18 × 3/4 = 13.5 m. The third reaches 13.5 × 3/4 = 10.125 m. The initial release height is not counted as a bounce.

The initial fall contributes 16 m. Before the second ground impact, the ball rises 8 m and falls 8 m. Before the third, it rises 4 m and falls 4 m. The total is 16 + 8 + 8 + 4 + 4 = 40 m.

At each step, the number of shaded triangles is multiplied by 3, while their total area is multiplied by 3/4. Stage 4 is four steps after Stage 0. Therefore, the count is 3⁴ = 81 and the total area is (3/4)⁴ = 81/256 square unit.

Each step retains eight of the nine equal parts, so the total area is multiplied by 8/9. After two steps, the remaining area is (8/9)² = 64/81 square unit. There are 64 smallest retained squares, each with area 1/81 square unit.