Figure it Out — Number Pyramids — Algebra Play

1. 13 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 140, Q1

   Without building the entire pyramid, find the topmost number of a three-row pyramid whose bottom row is (i) 4, 13, 8 (ii) 7, 11, 3 (iii) 10, 14, 25.

   Hint. Work out the top of a three-row pyramid in letters once, then substitute.

   Derive the formula once. Take a bottom row a, b, c.

       a + 2b + c
      a+b     b+c
     a    b     c
   

   Second row: a + b and b + c Top: (a + b) + (b + c) = a + 2b + c

   The middle number is counted twice because it feeds into both boxes of the second row. That single observation replaces all the addition.

   (i) 4, 13, 8 → 4 + 2(13) + 8 = 4 + 26 + 8 = 38 (ii) 7, 11, 3 → 7 + 2(11) + 3 = 7 + 22 + 3 = 32 (iii) 10, 14, 25 → 10 + 2(14) + 25 = 10 + 28 + 25 = 63

   Check (i) the long way. Second row 4 + 13 = 17 and 13 + 8 = 21; top 17 + 21 = 38 ✓

   A useful shortcut inside the shortcut. a + 2b + c = (a + b + c) + b, so the top is the sum of the whole bottom row plus the middle number again. For (iii): 10 + 14 + 25 = 49, and 49 + 14 = 63 ✓

   ✦ (i) 38 (ii) 32 (iii) 63, all from the formula a + 2b + c — the middle number counts twice.

2. 23 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 140, Q2

   Write an expression for the topmost number of a pyramid with 4 rows in terms of the values in the bottom row.

   Hint. Build it up row by row from a bottom row a, b, c, d, and watch the coefficients.

   Build it up from the bottom row a, b, c, d.

   Third row: a + b, b + c, c + d Second row: (a + b) + (b + c) = a + 2b + c, and (b + c) + (c + d) = b + 2c + d Top: (a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d

           a + 3b + 3c + d
        a+2b+c      b+2c+d
      a+b     b+c      c+d
    a     b      c       d
   

   The pattern in the coefficients. Look at the tops of pyramids of different heights:

   Rows	Top
   2	a + b
   3	a + 2b + c
   4	a + 3b + 3c + d
   5	a + 4b + 6c + 4d + e

   The coefficients are 1, 1 then 1, 2, 1 then 1, 3, 3, 1 then 1, 4, 6, 4, 1 — the rows of Pascal's triangle. This is no coincidence: Pascal's triangle is itself built by exactly the same rule of adding the two entries above, so the coefficient of a bottom entry counts the number of paths from that entry up to the top box.

   Check with numbers. Bottom row 1, 2, 3, 4: formula gives 1 + 6 + 9 + 4 = 20. Building up: 3, 5, 7 → 8, 12 → 20 ✓

   ✦ Top = a + 3b + 3c + d. The coefficients 1, 3, 3, 1 are a row of Pascal's triangle, because each counts the number of upward paths from that bottom box to the top.

3. 33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 140, Q3

   Without building the entire pyramid, find the topmost number of a four-row pyramid whose bottom row is (i) 8, 19, 21, 13 (ii) 7, 18, 19, 6 (iii) 9, 7, 5, 11.

   Hint. Use the formula from the previous question.

   Use a + 3b + 3c + d.

   (i) 8, 19, 21, 13 = 8 + 3(19) + 3(21) + 13 = 8 + 57 + 63 + 13 = 141

   (ii) 7, 18, 19, 6 = 7 + 3(18) + 3(19) + 6 = 7 + 54 + 57 + 6 = 124

   (iii) 9, 7, 5, 11 = 9 + 3(7) + 3(5) + 11 = 9 + 21 + 15 + 11 = 56

   Check (iii) the long way. Third row 9 + 7 = 16, 7 + 5 = 12, 5 + 11 = 16; second row 16 + 12 = 28, 12 + 16 = 28; top 28 + 28 = 56 ✓

   A faster grouping. a + 3b + 3c + d = (a + d) + 3(b + c), so add the two ends, add the two middles, treble that, and add. For (i): (8 + 13) + 3(19 + 21) = 21 + 120 = 141 ✓. For (ii): (7 + 6) + 3(18 + 19) = 13 + 111 = 124 ✓.

   ✦ (i) 141 (ii) 124 (iii) 56.

4. 43 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 140, Q4

   The Virahāṅka-Fibonacci sequence is 1, 2, 3, 5, … where each number is the sum of the two before it. Write the first three of these numbers in the bottom row of a three-row pyramid and fill it in. What numbers appear? What is at the top? Are they all Virahāṅka-Fibonacci numbers?

   Hint. Fill it in and compare every entry against the sequence 1, 2, 3, 5, 8, 13, …

   The sequence. 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … — each term is the sum of the two before it. Call them V₁ = 1, V₂ = 2, V₃ = 3, V₄ = 5, V₅ = 8, and so on.

   Build the pyramid on 1, 2, 3.

   Second row: 1 + 2 = 3 and 2 + 3 = 5 Top: 3 + 5 = 8

           8
         3   5
       1   2   3
   

   What numbers appear? 1, 2, 3 in the bottom row; 3 and 5 in the middle; 8 at the top. Written out, the pyramid contains 1, 2, 3, 3, 5, 8.

   Are they all Virahāṅka-Fibonacci numbers? Yes — 1, 2, 3, 5 and 8 are all in the sequence, and the only repeat is the 3.

   Why it had to happen. The pyramid rule adds two neighbours; the sequence rule also adds two neighbours. Since the bottom row is three consecutive terms of the sequence, adding neighbouring pairs simply produces the next terms:

   • 1 + 2 = 3 = V₃
   • 2 + 3 = 5 = V₄

   So the second row is V₃, V₄ — again consecutive terms — and the same argument applies once more: V₃ + V₄ = V₅ = 8.

   The top. 8 = V₅, the fifth Virahāṅka-Fibonacci number, from a bottom row of the first three.

   ✦ The pyramid is 1, 2, 3 / 3, 5 / 8. Every entry is a Virahāṅka-Fibonacci number, and the top is 8 = V₅. It works because the pyramid rule and the sequence rule are the same rule — adding two neighbours — so consecutive terms stay consecutive terms all the way up.

5. 54 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 140, Q5

   What can you say about the numbers in the pyramid and the number at the top when (i) the first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four-row pyramid, and (ii) the first 29 are written in the bottom row of a 29-row pyramid?

   Hint. Do the four-row case in full, look at which terms of the sequence each row uses, then generalise.

   (i) Bottom row 1, 2, 3, 5 in a four-row pyramid.

   Third row: 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8 Second row: 3 + 5 = 8, 5 + 8 = 13 Top: 8 + 13 = 21

             21
          8     13
        3    5     8
      1   2    3     5
   

   Every entry — 1, 2, 3, 5, 3, 5, 8, 8, 13, 21 — is a Virahāṅka-Fibonacci number. The top is 21 = V₇.

   The pattern in the rows. Writing each row as terms of the sequence:

   • bottom row = V₁, V₂, V₃, V₄
   • third row = V₃, V₄, V₅
   • second row = V₅, V₆
   • top = V₇

   Each row is again a run of consecutive terms, and each row starts two places further along than the row below. The starting terms are V₁, V₃, V₅, V₇ — the odd-numbered terms.

   (ii) Bottom row V₁ to V₂₉ in a 29-row pyramid.

   The same argument runs all the way up. Row k counted from the bottom starts at V₂ₖ₋₁, so:

   • row 1 starts at V₁, row 2 at V₃, row 3 at V₅, … , row 29 at V₅₇

   Row 29 is the top and holds a single number, so the top is V₅₇.

   Every entry of the whole 29-row pyramid is a Virahāṅka-Fibonacci number, because each row is built by adding neighbouring consecutive terms, which always lands on the next term of the sequence.

   The value at the top. V₅₇ = 591,286,729,879 — nearly six hundred billion, from a bottom row whose largest entry is only V₂₉ = 832,040. The pyramid magnifies enormously, which is what the coefficients 1, 3, 3, 1 and their larger cousins do.

   ✦ (i) Every number in the pyramid is a Virahāṅka-Fibonacci number and the top is 21 = V₇. (ii) The same holds for 29 rows: every entry is a Virahāṅka-Fibonacci number, row k from the bottom starts at V₂ₖ₋₁, and the top is V₅₇ = 591,286,729,879.

6. 64 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 140, Q6

   If the bottom row of an n-row pyramid contains the first n Virahāṅka-Fibonacci numbers, what can we say about the numbers in the pyramid, and what can we say about the number at the top?

   Hint. Show that each row is again a run of consecutive terms, and track where each row starts.

   The claim. Every number in the pyramid is a Virahāṅka-Fibonacci number, row k from the bottom is the run V₂ₖ₋₁, V₂ₖ, …, V_{n+k−1}, and the top is V₂ₙ₋₁.

   The key step. Suppose some row is a run of consecutive terms

   V_m, V_{m+1}, V_{m+2}, … , V_M

   Adding neighbouring pairs gives

   V_m + V_{m+1} = V_{m+2}, V_{m+1} + V_{m+2} = V_{m+3}, … , V_{M−1} + V_M = V_{M+1}

   because that is precisely the defining rule of the sequence. So the row above is

   V_{m+2}, V_{m+3}, … , V_{M+1}

   which is again a run of consecutive terms — starting two places later and ending one place later, so it is one entry shorter, exactly as a pyramid row must be.

   Run it from the bottom. Row 1 is V₁, …, V_n. Applying the step repeatedly:

   Row (from bottom)	Contents	Length
   1	V₁ … V_n	n
   2	V₃ … V_{n+1}	n − 1
   3	V₅ … V_{n+2}	n − 2
   k	V₂ₖ₋₁ … V_{n+k−1}	n − k + 1
   n	V₂ₙ₋₁	1

   Setting k = n gives a single box holding V₂ₙ₋₁.

   Check against the cases already done.

   n	Predicted top	Actual
   2	V₃ = 3	1, 2 → 3 ✓
   3	V₅ = 8	✓
   4	V₇ = 21	✓
   5	V₉ = 55	1,2,3,5,8 → 3,5,8,13 → 8,13,21 → 21,34 → 55 ✓
   6	V₁₁ = 144	✓

   Two consequences worth stating.

   1. Every entry in the pyramid is a Virahāṅka-Fibonacci number — not merely the top.
   2. The top is the (2n − 1)th term, so doubling the height of the pyramid roughly doubles the index, and since the sequence grows by a factor of about 1.618 each step, the top grows extremely fast.

   ✦ Every number in the pyramid is a Virahāṅka-Fibonacci number, row k from the bottom is the consecutive run V₂ₖ₋₁ … V_{n+k−1}, and the number at the top is V₂ₙ₋₁. The reason is that the pyramid rule and the sequence rule are the same rule, so a run of consecutive terms always produces another run of consecutive terms starting two places further on.

Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp206.pdf), Chapter 6 'Algebra Play', pages 135-147. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every answer was derived from first principles and independently recomputed in Python. FIGURES READ OFF HIGH-DPI RENDERS AND RE-SOLVED: the three p.138 pyramids (bottom rows 6,2 / 3,4,3 / 5,4,5,0 giving tops 8, 14 and 32); the p.138 top-down pyramid 10 / 4,6 / 1,3,3; the p.139 letter-number pyramid 60 / 32,28 / 12,20,8; the three p.139 four-row pyramids, whose bottom rows solve to 4,9,6,1 (top 50), 5,14,7,2 (top 70) and 3,5,5,2 (top 35); the p.140 Figure it Out bottom rows 4,13,8 / 7,11,3 / 10,14,25 (tops 38, 32, 63) and 8,19,21,13 / 7,18,19,6 / 9,7,5,11 (tops 141, 124, 56); and the two p.142 algebra grids, which solve to square 11 and circle 5 with a third-row total of 21, and circle 4 and diamond 7 with a third-row total of 15 (the fourth row of that grid is printed blank for the student to design). The Virahanka-Fibonacci result was proved and then checked by computation for n = 2 to 6: every entry of the pyramid is a Virahanka-Fibonacci number, row k from the bottom is the consecutive run V(2k-1) to V(n+k-1), and the top is V(2n-1) - so a 29-row pyramid tops out at V57 = 591286729879. All six arrangements were enumerated for each Largest Product question: 32 x 5 = 160, 31 x 7 = 217 and 53 x 9 = 477, each beating its runner-up by exactly the margin the algebra predicts (4, 4 and 12). The date trick decodes as 100M + 165 + D, giving 4 November, 29 February and 31 January for 1269, 394 and 296. The puzzle answers were each verified by substitution: 7 flowers with 8 per shrine, 20 horses and 35 hens, a daughter of 6 and mother of 30, Gauri 6 cows and Naina 12, a dosa price of Rs 80 or a target of 175 dosas, the fraction sequence equal to 1/3 because n^2/(4n^2 - n^2) = 1/3, and Karim starting with 7 coins with the genie's general ruinous charge being c = 2^k x / (2^k - 1).. Questions are referenced from the NCERT textbook for identification.