Question

A watermelon vendor arranged the watermelons similar to shown in the adjoining picture: The number of watermelons in subsequent rows differ by ‘d’. The bottommost row has 101 watermelons and the topmost row has 1 watermelon. There are 21 rows from bottom to top.

Based on the above information, answer the following questions:

(i) Find the value of ‘d’.   [1]

(ii) How many watermelons will be there in the 15th row from the bottom?   [1]

(iii) (a) Find the total number of watermelons from bottom to top.   [2]

OR

(iii) (b) If the number of watermelons in the nth row from top is equal to number of watermelons in the nth row from bottom, find the value of n.   [2]

Answer 1

This forms an Arithmetic Progression. 

Bottom row (first term) a = 101

Top row (Last term) l = 1

Number of rows (n) = 21

Common difference = d.

(i) l = a + (n – 1)d

1 = 101 + (21 – 1)d

⇒ `d = (-100)/20`

d = –5

(ii) a_n = a + (n – 1)d

a₁₅ = 101 + (15 – 1)(–5)

a₁₅ = 101 + 14(–5)

a₁₅ = 101 – 70

a₁₅ = 31

There will be 31 watermelons in the 15th row from the bottom.

(iii) (a) `S_n = n/2 (a + l)`

`S_21 = 21/2 (101 + 1)`

= `21/2 (102)`

= 21 × 51

= 1071

∴ The total number of watermelons from bottom to top is 1071.

OR

(iii) (b) Sequence from bottom to top is an AP.

101, 96, 91, ... 1

n^th row from bottom

a_n = 101 + (n – 1)(–5)

= 101 – 5(n – 1)

n^th row from top

If there are 21 rows total, then n^th row from top = (21 – n + 1) = (22 – n)^th term from bottom

a_22–n = 101 – 5((22 – n) – 1)

= 101 – 5(21 – n)

101 – 5(n – 1) = 101 – 5(21 – n)

106 – 5n = –4 + 5n

106 + 4 = 5n + 5n

110 = 10n

n = 11