Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
An arithmetic progression (AP) is a list of numbers in which each term, except the first, is obtained by adding a fixed number d to the term before it. The fixed number d is the common difference. It can be positive, negative or zero.
An AP with a last term is finite (like the heights of the students in a line). One that goes on for ever is infinite. A finite AP has a last term l.
Find a₂ − a₁, a₃ − a₂, a₄ − a₃, … If they are all equal, the list is an AP with that common difference. One unequal difference is enough to say it is not.
Note the order: d = a₂ − a₁ (later term minus earlier term), not a₁ − a₂.
If a, b, c are in AP, then b − a = c − b, so
b is called the arithmetic mean of a and c.
The nth term (general term) of an AP with first term a and common difference d is
Example: the 10th term of 2, 7, 12, … is a₁₀ = 2 + 9 × 5 = 47.
How many three-digit numbers are divisible by 7? They are 105, 112, …, 994, an AP with d = 7:
Counting from the last term l backwards gives an AP with common difference −d, so the nth term from the end is
Writing the sum forwards and backwards in the same way for any AP gives
Use the second form when the last term is known.
If the sum Sₙ is given for every n:
Example: if Sₙ = 4n − n², then S₁ = 3 = a₁, and aₙ = (4n − n²) − [4(n − 1) − (n − 1)²] = 5 − 2n.
Solving Sₙ = given value leads to a quadratic in n. Keep only natural-number roots, and check the situation: 24 + 21 + 18 + … gives 78 for n = 4 and for n = 13, because terms 10 to 13 are negative and cancel terms 5 to 9.
| Situation | a | d |
|---|---|---|
| Salary with a fixed yearly increment | starting salary | increment |
| Savings rising by a fixed amount | first saving | increase |
| Rows of plants, seats or logs | first row | change per row |
| Penalty rising each day | first day's penalty | daily rise |
| Prizes each ₹20 less than the previous | largest prize | −20 |
| Multiples of k in a range | first multiple | k |
Money at compound interest (10000, 10800, 11664, …) and air left after removing ¼ each time (V, ¾V, 9V/16, …) multiply by a fixed number, so they are not APs.
| Quantity | Formula |
|---|---|
| Common difference | d = aₙ₊₁ − aₙ |
| nth term | aₙ = a + (n − 1)d |
| nth term from the end | l − (n − 1)d |
| Sum of n terms | Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l) |
| nth term from sums | aₙ = Sₙ − Sₙ₋₁ |
| Arithmetic mean of a, c | (a + c)/2 |
Try each problem first, then open the solution to check your working.
Is 4, 10, 16, 22, … an AP? If so, write the next three terms.
Use: Check the differences
10 − 4 = 16 − 10 = 22 − 16 = 6, so it is an AP with d = 6.
Next terms: 28, 34, 40
Answer: Yes; 28, 34, 40
Find the 25th term of 5, 9, 13, …
Given: a = 5, d = 4, n = 25
Use: aₙ = a + (n − 1)d
a₂₅ = 5 + 24 × 4 = 101
Answer: 101
Which term of 7, 11, 15, … is 151?
Given: a = 7, d = 4
Use: Solve aₙ = 151
7 + (n − 1)4 = 151 ⇒ n − 1 = 36 ⇒ n = 37
Answer: The 37th term
Is 301 a term of 5, 11, 17, 23, …?
Given: a = 5, d = 6
Use: n must be a natural number
5 + (n − 1)6 = 301 ⇒ n − 1 = 296/6 = 148/3
n = 151/3 is not a natural number.
Answer: No
How many two-digit numbers are divisible by 3?
Given: 12, 15, …, 99
Use: Solve aₙ = 99
12 + (n − 1)3 = 99 ⇒ n − 1 = 29 ⇒ n = 30
Answer: 30
The 7th term of an AP is −1 and the 16th term is 17. Find the general term.
Given: a + 6d = −1; a + 15d = 17
Use: Subtract
9d = 18 ⇒ d = 2; a = −1 − 12 = −13
aₙ = −13 + (n − 1)2 = 2n − 15
Answer: aₙ = 2n − 15
Three numbers in AP add up to 24 and their product is 440. Find them.
Given: take a − d, a, a + d
Sum: 3a = 24 ⇒ a = 8
Product: (8 − d) × 8 × (8 + d) = 440 ⇒ 64 − d² = 55 ⇒ d = ±3
Answer: 5, 8, 11 (or 11, 8, 5)
Find the 10th term from the end of 4, 9, 14, …, 254.
Given: l = 254, d = 5
Use: l − (n − 1)d
254 − 9 × 5 = 209
Answer: 209
Find x if 2x, x + 10 and 3x + 2 are in AP.
Use: Middle term = (first + third) ÷ 2
2(x + 10) = 2x + 3x + 2 ⇒ 2x + 20 = 5x + 2 ⇒ x = 6
The terms are 12, 16, 20.
Answer: x = 6
Find the sum of the first 20 terms of 1, 4, 7, …
Given: a = 1, d = 3, n = 20
Use: Sₙ = n/2 [2a + (n − 1)d]
S₂₀ = 10 (2 + 57) = 590
Answer: 590
Find 3 + 11 + 19 + … + 803.
Given: a = 3, d = 8, l = 803
Use: Find n, then Sₙ = n/2 (a + l)
803 = 3 + (n − 1)8 ⇒ n = 101
S = 101/2 × 806 = 40703
Answer: 40703
Find the sum of all odd numbers between 100 and 200.
Given: 101, 103, …, 199
199 = 101 + (n − 1)2 ⇒ n = 50
S = 50/2 (101 + 199) = 7500
Answer: 7500
How many terms of 24, 21, 18, … must be taken to make the sum 78?
Given: a = 24, d = −3
Use: Sₙ = 78
n/2 [48 − 3(n − 1)] = 78 ⇒ n(51 − 3n) = 156 ⇒ n² − 17n + 52 = 0 ⇒ (n − 4)(n − 13) = 0
Both work: the 10th to 13th terms (−3, −6, −9, −12) cancel the 5th to 9th (12, 9, 6, 3, 0).
Answer: 4 terms or 13 terms
The sum of the first n terms of an AP is 3n² + 5n. Find the nth term and the 25th term.
Given: Sₙ = 3n² + 5n
Use: aₙ = Sₙ − Sₙ₋₁
aₙ = 3n² + 5n − [3(n − 1)² + 5(n − 1)] = 6n + 2
a₂₅ = 152
Answer: aₙ = 6n + 2; a₂₅ = 152
Find the sum of all multiples of 9 between 300 and 700.
Given: 306, 315, …, 693
Use: Find n, then Sₙ = n/2 (a + l)
693 = 306 + (n − 1)9 ⇒ n − 1 = 43 ⇒ n = 44
S = 22 (306 + 693) = 21978
Answer: 21978
Show that the sum of the first n even natural numbers is n(n + 1), and find 2 + 4 + … + 20.
Given: a = 2, d = 2
Sₙ = n/2 [4 + 2(n − 1)] = n/2 (2n + 2) = n(n + 1)
For 2 + 4 + … + 20, n = 10: 10 × 11 = 110
Answer: 110
Ravi saves ₹50 in the first month and increases his saving by ₹20 every month. How much does he save in a year?
Given: a = 50, d = 20, n = 12
Use: Total = S₁₂
S₁₂ = 6 (100 + 220) = 1920
Answer: ₹1920
A hall has 20 seats in the first row and each row has 2 more seats than the row in front. How many seats are there in 25 rows?
Given: a = 20, d = 2, n = 25
S₂₅ = 25/2 (40 + 48) = 1100
Answer: 1100 seats
A man repays a loan of ₹3250 by paying ₹20 in the first month and then ₹15 more each month than the month before. In how many months is the loan cleared?
Given: a = 20, d = 15, Sₙ = 3250
n/2 [40 + 15(n − 1)] = 3250 ⇒ 15n² + 25n − 6500 = 0 ⇒ 3n² + 5n − 1300 = 0
n = −5 + √25 + 156006 = −5 + 1256 = 20
Answer: 20 months
A clock strikes once at 1 o'clock, twice at 2 o'clock, and so on up to 12 times at 12 o'clock. How many times does it strike in a day?
Given: 1 + 2 + … + 12, twice a day
1 + 2 + … + 12 = 12 × 13/2 = 78
In 24 hours: 2 × 78 = 156
Answer: 156 times
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(i) Fares: 15, 23, 31, 39, … Each differs from the previous by 8. AP, d = 8.
(ii) If the air is V, the amounts are V, 3V4, 9V16, … Differences −V4, −3V16 are not equal. Not an AP.
(iii) Costs: 150, 200, 250, 300, … Each rises by 50. AP, d = 50.
(iv) Amounts: 10000, 10800, 11664, … Differences 800, 864 are not equal. Not an AP.
Answer: (i) and (iii) are APs; (ii) and (iv) are not
(i) 10, 20, 30, 40
(ii) −2, −2, −2, −2
(iii) 4, 1, −2, −5
(iv) −1, −12, 0, 12
(v) −1.25, −1.50, −1.75, −2.00
Answer: as listed
d = second term − first term.
(i) a = 3, d = 1 − 3 = −2
(ii) a = −5, d = −1 − (−5) = 4
(iii) a = 13, d = 53 − 13 = 43
(iv) a = 0.6, d = 1.7 − 0.6 = 1.1
Answer: (i) 3, −2 (ii) −5, 4 (iii) 1/3, 4/3 (iv) 0.6, 1.1
(i) Differences 2, 4, 8: not an AP.
(ii) Differences all 12: AP, d = 12; next 4, 92, 5.
(iii) Differences all −2: AP, d = −2; next −9.2, −11.2, −13.2.
(iv) Differences all 4: AP, d = 4; next 6, 10, 14.
(v) Differences all √2: AP, d = √2; next 3 + 4√2, 3 + 5√2, 3 + 6√2.
(vi) Differences 0.02, 0.002, 0.0002: not an AP.
(vii) Differences all −4: AP, d = −4; next −16, −20, −24.
(viii) Differences all 0: AP, d = 0; next −12, −12, −12.
(ix) Differences 2, 6, 18: not an AP.
(x) Differences all a: AP, d = a; next 5a, 6a, 7a.
(xi) Differences a² − a, a³ − a², … are not equal in general: not an AP.
(xii) The terms are √2, 2√2, 3√2, 4√2: AP, d = √2; next 5√2 = √50, 6√2 = √72, 7√2 = √98.
(xiii) √6 − √3 ≠ √9 − √6: not an AP.
(xiv) 1, 9, 25, 49: differences 8, 16, 24: not an AP.
(xv) 1, 25, 49, 73: differences all 24: AP, d = 24; next 97, 121, 145.
Answer: APs: (ii), (iii), (iv), (v), (vii), (viii), (x), (xii), (xv). Not APs: (i), (vi), (ix), (xi), (xiii), (xiv)
Use aₙ = a + (n − 1)d.
(i) a₈ = 7 + 7 × 3 = 28
(ii) 0 = −18 + 9d ⇒ d = 2
(iii) −5 = a + 17(−3) ⇒ a = 46
(iv) 3.6 = −18.9 + (n − 1)2.5 ⇒ n − 1 = 9 ⇒ n = 10
(v) a₁₀₅ = 3.5 + 104 × 0 = 3.5
Answer: (i) 28 (ii) 2 (iii) 46 (iv) 10 (v) 3.5
(i) a = 10, d = −3: a₃₀ = 10 + 29(−3) = −77.
(ii) a = −3, d = −12 − (−3) = 52: a₁₁ = −3 + 10 × 52 = 22.
Answer: (i) (C) −77 (ii) (B) 22
(i) 2 + 2d = 26 ⇒ d = 12: 14.
(ii) a + d = 13, a + 3d = 3 ⇒ d = −5, a = 18: 18, 13, 8, 3.
(iii) 5 + 3d = 192 ⇒ d = 32: 5, 612, 8, 912.
(iv) −4 + 5d = 6 ⇒ d = 2: −4, −2, 0, 2, 4, 6.
(v) a + d = 38, a + 5d = −22 ⇒ 4d = −60, d = −15, a = 53: 53, 38, 23, 8, −7, −22.
Answer: (i) 14 (ii) 18, 8 (iii) 6½, 8 (iv) −2, 0, 2, 4 (v) 53, 23, 8, −7
a = 3, d = 5. 3 + (n − 1)5 = 78 ⇒ n − 1 = 15 ⇒ n = 16.
Answer: The 16th term
(i) a = 7, d = 6: 7 + (n − 1)6 = 205 ⇒ n − 1 = 33 ⇒ n = 34.
(ii) a = 18, d = −52: 18 − 52(n − 1) = −47 ⇒ 52(n − 1) = 65 ⇒ n − 1 = 26 ⇒ n = 27.
Answer: (i) 34 (ii) 27
a = 11, d = −3. 11 − 3(n − 1) = −150 ⇒ 3(n − 1) = 161 ⇒ n = 164/3, which is not a natural number.
Answer: No, −150 is not a term
a + 10d = 38 and a + 15d = 73 ⇒ 5d = 35, d = 7, a = −32.
a₃₁ = −32 + 30 × 7 = 178.
Answer: 178
a + 2d = 12 and a + 49d = 106 ⇒ 47d = 94, d = 2, a = 8.
a₂₉ = 8 + 28 × 2 = 64.
Answer: 64
a + 2d = 4 and a + 8d = −8 ⇒ 6d = −12, d = −2, a = 8.
8 + (n − 1)(−2) = 0 ⇒ n − 1 = 4 ⇒ n = 5.
Answer: The 5th term
a₁₇ − a₁₀ = (a + 16d) − (a + 9d) = 7d = 7 ⇒ d = 1.
Answer: d = 1
a = 3, d = 12. a₅₄ = 3 + 53 × 12 = 639. Required term = 639 + 132 = 771.
3 + (n − 1)12 = 771 ⇒ n − 1 = 64 ⇒ n = 65.
(Quicker: 132 more means 132/12 = 11 more steps, so 54 + 11 = 65.)
Answer: The 65th term
Let the first terms be a and b, common difference d.
(a + 99d) − (b + 99d) = a − b = 100.
(a + 999d) − (b + 999d) = a − b = 100.
Answer: 100
The smallest is 105 and the largest is 994: AP 105, 112, …, 994 with d = 7.
105 + (n − 1)7 = 994 ⇒ n − 1 = 127 ⇒ n = 128.
Answer: 128
The multiples are 12, 16, …, 248 (d = 4).
12 + (n − 1)4 = 248 ⇒ n − 1 = 59 ⇒ n = 60.
Answer: 60
63 + (n − 1)2 = 3 + (n − 1)7 ⇒ 60 = 5(n − 1) ⇒ n = 13.
Check: both 13th terms are 87.
Answer: n = 13
a₇ − a₅ = 2d = 12 ⇒ d = 6.
a + 2d = 16 ⇒ a = 4.
Answer: 4, 10, 16, 22, …
Reading from the end: 253, 248, 243, … with d = −5.
20th term = 253 + 19(−5) = 158.
Answer: 158
(a + 3d) + (a + 7d) = 24 ⇒ a + 5d = 12.
(a + 5d) + (a + 9d) = 44 ⇒ a + 7d = 22.
Subtract: 2d = 10, d = 5, a = −13.
Answer: −13, −8, −3
Salaries: 5000, 5200, 5400, … (a = 5000, d = 200).
5000 + (n − 1)200 = 7000 ⇒ n − 1 = 10 ⇒ n = 11.
The 11th year counting 1995 as the first is 2005.
Answer: In 2005
a = 5, d = 1.75. 5 + (n − 1)1.75 = 20.75 ⇒ (n − 1)1.75 = 15.75 ⇒ n − 1 = 9 ⇒ n = 10.
Answer: n = 10
Use Sₙ = n/2 [2a + (n − 1)d].
(i) a = 2, d = 5: S₁₀ = 5(4 + 45) = 245.
(ii) a = −37, d = 4: S₁₂ = 6(−74 + 44) = −180.
(iii) a = 0.6, d = 1.1: S₁₀₀ = 50(1.2 + 108.9) = 5505.
(iv) a = 115, d = 112 − 115 = 160: S₁₁ = 112(215 + 1060) = 112 × 310 = 3320.
Answer: (i) 245 (ii) −180 (iii) 5505 (iv) 33/20
(i) a = 7, d = 312, l = 84: 7 + (n − 1)72 = 84 ⇒ n = 23. S = 232(7 + 84) = 20932 = 104612.
(ii) a = 34, d = −2, l = 10: n = 13. S = 13/2 (34 + 10) = 286.
(iii) a = −5, d = −3, l = −230: −5 − 3(n − 1) = −230 ⇒ n = 76. S = 38(−5 − 230) = −8930.
Answer: (i) 1046½ (ii) 286 (iii) −8930
(i) 5 + 3(n − 1) = 50 ⇒ n = 16; S₁₆ = 8(5 + 50) = 440.
(ii) 7 + 12d = 35 ⇒ d = 7/3; S₁₃ = 13/2 (7 + 35) = 273.
(iii) a + 33 = 37 ⇒ a = 4; S₁₂ = 6(4 + 37) = 246.
(iv) a + 2d = 15 and 5(2a + 9d) = 125 ⇒ 2a + 9d = 25. With a = 15 − 2d: 30 + 5d = 25 ⇒ d = −1, a = 17; a₁₀ = 17 − 9 = 8.
(v) 92(2a + 40) = 75 ⇒ 2a + 40 = 503 ⇒ a = −353; a₉ = −353 + 40 = 853.
(vi) n/2 [4 + 8(n − 1)] = 90 ⇒ 4n² − 2n − 90 = 0 ⇒ 2n² − n − 45 = 0 ⇒ (n − 5)(2n + 9) = 0 ⇒ n = 5; a₅ = 2 + 32 = 34.
(vii) n/2 (8 + 62) = 210 ⇒ n = 6; 8 + 5d = 62 ⇒ d = 54/5.
(viii) a = 4 − 2(n − 1) = 6 − 2n. n/2 (a + 4) = −14 ⇒ n(5 − n) = −14 ⇒ n² − 5n − 14 = 0 ⇒ n = 7; a = −8.
(ix) 4(6 + 7d) = 192 ⇒ 6 + 7d = 48 ⇒ d = 6.
(x) 9/2 (a + 28) = 144 ⇒ a + 28 = 32 ⇒ a = 4.
Answer: (i) n = 16, S = 440 (ii) d = 7/3, S = 273 (iii) a = 4, S = 246 (iv) d = −1, a₁₀ = 8 (v) a = −35/3, a₉ = 85/3 (vi) n = 5, aₙ = 34 (vii) n = 6, d = 54/5 (viii) n = 7, a = −8 (ix) d = 6 (x) a = 4
a = 9, d = 8. n/2 [18 + 8(n − 1)] = 636 ⇒ n(4n + 5) = 636 ⇒ 4n² + 5n − 636 = 0.
n = −5 ± √25 + 101768 = −5 ± 1018 ⇒ n = 12 (the other root is negative).
Answer: 12 terms
n/2 (5 + 45) = 400 ⇒ n = 16.
45 = 5 + 15d ⇒ d = 40/15 = 8/3.
Answer: 16 terms; d = 8/3
350 = 17 + (n − 1)9 ⇒ n − 1 = 37 ⇒ n = 38.
S = 38/2 (17 + 350) = 19 × 367 = 6973.
Answer: 38 terms; sum 6973
149 = a + 21 × 7 ⇒ a = 2.
S₂₂ = 11(2 + 149) = 1661.
Answer: 1661
d = 18 − 14 = 4, a = 10.
S₅₁ = 51/2 [20 + 50 × 4] = 51/2 × 220 = 5610.
Answer: 5610
7/2 (2a + 6d) = 49 ⇒ a + 3d = 7.
17/2 (2a + 16d) = 289 ⇒ a + 8d = 17.
So d = 2, a = 1.
Sₙ = n/2 [2 + 2(n − 1)] = n².
Answer: Sₙ = n²
(i) aₙ₊₁ − aₙ = [3 + 4(n + 1)] − (3 + 4n) = 4, the same for all n, so it is an AP with a₁ = 7, d = 4. S₁₅ = 15/2 (7 + a₁₅) = 15/2 (7 + 63) = 525.
(ii) aₙ₊₁ − aₙ = [9 − 5(n + 1)] − (9 − 5n) = −5, constant, so it is an AP with a₁ = 4, d = −5. S₁₅ = 15/2 (4 + a₁₅) = 15/2 (4 − 66) = −465.
Answer: (i) AP, S₁₅ = 525 (ii) AP, S₁₅ = −465
S₁ = 4 − 1 = 3, so the first term is 3.
S₂ = 8 − 4 = 4, so a₂ = S₂ − S₁ = 1.
S₃ = 12 − 9 = 3, so a₃ = S₃ − S₂ = −1.
S₁₀ = 40 − 100 = −60 and S₉ = 36 − 81 = −45, so a₁₀ = −15.
aₙ = Sₙ − Sₙ₋₁ = (4n − n²) − [4(n − 1) − (n − 1)²] = 5 − 2n.
Answer: S₁ = 3; S₂ = 4; a₂ = 1; a₃ = −1; a₁₀ = −15; aₙ = 5 − 2n
6, 12, …, 240: a = 6, d = 6, n = 40.
S₄₀ = 20 (6 + 240) = 4920.
Answer: 4920
8, 16, …, 120: S₁₅ = 15/2 (8 + 120) = 960.
Answer: 960
1, 3, 5, …, 49: 49 = 1 + 2(n − 1) ⇒ n = 25.
S = 25/2 (1 + 49) = 625.
Answer: 625
a = 200, d = 50, n = 30.
S₃₀ = 15 [400 + 29 × 50] = 15 × 1850 = 27750.
Answer: ₹27750
a = largest prize, d = −20, n = 7.
7/2 [2a − 120] = 700 ⇒ 2a − 120 = 200 ⇒ a = 160.
Answer: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40
Trees per class (3 sections): 3, 6, 9, …, 36: a = 3, d = 3, n = 12.
S₁₂ = 6 (3 + 36) = 234.
Answer: 234 trees
Length of a semicircle of radius r is πr. So the lengths are π(0.5), π(1.0), π(1.5), …, π(6.5).
Sum of radii = 13/2 (0.5 + 6.5) = 45.5 cm.
Total length = 227 × 45.5 = 143 cm.
Answer: 143 cm
a = 20, d = −1. n/2 [40 − (n − 1)] = 200 ⇒ n(41 − n) = 400 ⇒ n² − 41n + 400 = 0 ⇒ (n − 16)(n − 25) = 0.
n = 25 gives a₂₅ = 20 − 24 = −4, which is impossible. So n = 16.
Top row: a₁₆ = 20 − 15 = 5.
Answer: 16 rows; 5 logs in the top row
Distances of the potatoes from the bucket: 5, 8, 11, …, 32 m. Each is run twice (there and back).
Total = 2 × 10/2 (5 + 32) = 2 × 185 = 370 m.
Answer: 370 m
a = 121, d = −4. aₙ = 121 − 4(n − 1) = 125 − 4n < 0 ⇒ n > 31¼.
So n = 32: a₃₂ = −3 (and a₃₁ = 1).
Answer: The 32nd term
(a + 2d) + (a + 6d) = 6 ⇒ a + 4d = 3. So a₃ = 3 − 2d and a₇ = 3 + 2d.
(3 − 2d)(3 + 2d) = 8 ⇒ 9 − 4d² = 8 ⇒ d = ±½.
If d = ½: a = 1, S₁₆ = 8 (2 + 15 × ½) = 76.
If d = −½: a = 5, S₁₆ = 8 (10 − 15 × ½) = 20.
Answer: 76 or 20
Number of rungs = 250/25 + 1 = 11.
Lengths form an AP from 45 cm to 25 cm: S = 11/2 (45 + 25) = 385 cm.
Answer: 385 cm (3.85 m)
Sum before x: 1 + 2 + … + (x − 1) = (x − 1)x/2.
Sum after x: (1 + … + 49) − (1 + … + x) = 1225 − x(x + 1)/2.
Setting them equal: (x² − x)/2 = 1225 − (x² + x)/2 ⇒ x² = 1225 ⇒ x = 35.
Check: 1 + … + 34 = 595 and 36 + … + 49 = 1225 − 630 = 595 ✓
Answer: x = 35
Step k is a solid block k4 m high, 12 m deep and 50 m long: volume = k4 × 12 × 50 = 6.25k m³.
The volumes 6.25, 12.5, 18.75, … form an AP with a = 6.25, d = 6.25, n = 15.
Total = 15/2 (6.25 + 93.75) = 750 m³.
Answer: 750 m³