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CBSE · Class 9 · Mathematics

Exploring some more Progressions (Advanced)

Standard difficulty · 20 questions · 40 marks · Time: 34 minutes

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The sum of the first 8 terms of 1, 3, 9, … is:
  1. 6,560
  2. 6,561
  3. 9,841
  4. 3,280
2.
The sum to infinity of the GP 19, −76/5, 304/25, … is:
  1. 171/5
  2. 95
  3. 104/9
  4. 95/9
3.
The sum of the first n terms of a GP with first term a and ratio r ≠ 1 is:
  1. n(a + arⁿ) ÷ 2
  2. arⁿ ÷ (1 − r)
  3. a(1 − r) ÷ (1 − rⁿ)
  4. a(1 − rⁿ) ÷ (1 − r)
4.
The sequence 9, 22, 41, 66, 97, … has constant second differences. Its 11th term is:
  1. 409
  2. 482
  3. 342
  4. 283
5.
The sum of the first n terms of 3, 9, 27, … is 3,279. Then n is:
  1. 9
  2. 8
  3. 6
  4. 7
6.
The first differences of 3, 7, 13, 21, 31 are:
  1. 10, 20, 34, 52
  2. 3, 4, 6, 8
  3. 2, 2, 2, 2
  4. 4, 6, 8, 10
7.
An infinite GP has first term 5 and sum 25/3. Its common ratio is:
  1. 5/3
  2. −2/5
  3. 2/5
  4. 3/5
8.
A ball dropped from 70 m rebounds to 4/5 of the height it falls each time. The total distance it travels before coming to rest is:
  1. 560 m
  2. 700 m
  3. 350 m
  4. 630 m
Section B — Short answer8 × 2 = 16 marks
9.
Find the exact sum of the first 8 terms of the GP 24, 12, 6, …
10.
Use differences to find the nth term of 10, 24, 46, 76, 114, …, and hence its 12th term.
11.
Find the sum of the first 7 terms of 7 + 77 + 777 + …
12.
The sum to infinity of a GP is 5/2, and the sum to infinity of the squares of its terms is 25/6. Find the first term a and the common ratio r.
13.
Find the sum of the first 14 terms of 4 + 12 + 22 + 34 + 48 + …
14.
Find the 8th term of 4, 6, 13, 27, 50, … (the third differences are constant).
15.
Find the value of 64^(1/4) × 64^(1/16) × 64^(1/64) × … to infinity.
16.
For the infinite GP 8, 24/5, 72/25, 216/125, …, find (t1 + t3 + t5 + …) ÷ (t2 + t4 + t6 + …).
Section C — Word problems4 × 4 = 16 marks
17.
A square of side 24 cm is cut from card. The midpoints of its sides are joined to make a second square, which is cut and pasted on it, and so on. Find (a) the area of the 8th square (b) the total area of card for the first 8 squares (c) the total area if this went on forever.
18.
The midpoints of the sides of an equilateral triangle of side 36 cm are joined to make a smaller triangle, and this is repeated forever. Find (a) the side of the 6th triangle (b) the sum of the perimeters of all the triangles.
19.
A ball is dropped from a height of 216 m. Each time it hits the floor it rebounds to 2/3 of the height it fell. Find (a) the height it reaches after the 3rd bounce (b) the total distance it travels before coming to rest.
20.
For the series −1 + 8 + 23 + 44 + 71 + …, find (a) the nth term (b) the 8th term (c) the sum of the first 8 terms.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 9 · Mathematics

Answer key

Exploring some more Progressions (Advanced) · AW-9M-APG-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 3,280a(rⁿ − 1)/(r − 1) = 1(38 − 1)/2
2(d) 95/9a ÷ (1 − r) = 19 ÷ (1 − (−4/5))
3(d) a(1 − rⁿ) ÷ (1 − r)subtract rSn from Sn
4(a) 409tn = 3n2 + 4n + 2; t11 = 409
5(d) 73(3ⁿ − 1)/2 = 3,279 gives 3ⁿ = 2,187
6(d) 4, 6, 8, 10subtract each term from the next
7(c) 2/525/3 = 5 ÷ (1 − r), so r = 1 − 5 ÷ 25/3
8(d) 630 m70 + 2 × 70 × (4/5) ÷ (1 − 4/5) = 630
9765/16a(1 − rⁿ)/(1 − r) = 24(1 − (1/2)8) ÷ (1 − (1/2))
10tn = 4n2 + 2n + 4; t12 = 604b = 10, a = 14, d = 8; tn = b + a(n − 1) + d(n − 1)(n − 2)/2
118,641,9697/9 × (9 + 99 + 999 + …) = 7/9 × [10(107 − 1)/9 − 7]
12a = 2, r = 1/5S2 ÷ S2 = (1 + r)/(1 − r) = 3/2, so r = 1/5; a = S(1 − r) = 2
131,512b = 4, a = 8, d = 2; Sn = bn + a·n(n − 1)/2 + d·n(n − 1)(n − 2)/6
14193b = 4, a = 2, d = 5, d1 = 2; t8 = 4 + 2 × 7 + 5 × C(7, 2) + 2 × C(7, 3)
154the powers add to 1/4 ÷ (1 − 1/4) = 1/3; 64^(1/3) = 4
165/3both are GPs with ratio r2; the ratio of their sums is t1 ÷ t2 = 1/r = 5/3
17(a) 9/2 cm2; (b) 2295/2 cm2; (c) 1,152 cm2the areas form a GP 576, 288, … with r = 1/2; (b) 576(1 − (1/2)8) ÷ (1/2); (c) 576 ÷ (1 − 1/2)
18(a) 9/8 cm; (b) 216 cm(a) 36 × (1/2)5; (b) perimeters 108, 54, …: 108 ÷ (1 − 1/2) = 216
19(a) 64 m; (b) 1080 m(a) 216 × (2/3)3; (b) 216 + 2 × [144 ÷ (1 − 2/3)]
20(a) tn = 3n2 − 4; (b) 188; (c) 580first differences 9, 15, 21, 27; second differences 6; b = −1, a = 9, d = 6