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'.
(ii) How many watermelons will be there in the 15th row from the bottom? (iii) (a) Find the total number of watermelons from bottom to top. OR
(iii) (b) If the number of watermelons in the nth row from top is equal to the number of watermelons in the nth row from bottom, find the value of n.
Show Hint
The middle row of any AP with an odd number of terms is always the average of the first and last terms: $(101+1)/2 = 51$.
Step 1: Understanding the Concept:
The arrangement forms an Arithmetic Progression (AP). Let's define the sequence from bottom to top. The first term ($a$) is 101, the number of terms ($n$) is 21, and the last term ($l$) is 1. Step 2: Key Formula or Approach:
1. $a_n = a + (n-1)d$
2. $S_n = \frac{n}{2}(a + l)$ Step 3: Detailed Explanation:
1. (i) Find 'd': - $a = 101, a_{21} = 1, n = 21$.
- $1 = 101 + (21 - 1)d$
- $-100 = 20d \implies d = -5$.
2. (ii) 15th row from bottom:
- $a_{15} = 101 + (15 - 1)(-5)$
- $a_{15} = 101 - 70 = 31$ watermelons.
3. (iii) (a) Total number:
- $S_{21} = \frac{21}{2}(101 + 1) = \frac{21}{2}(102) = 21 \times 51 = 1071$.
4. (iii) (b) OR:
- Let row $n$ from bottom be $a_n = 101 + (n-1)(-5)$.
- Let row $n$ from top be $b_n = 1 + (n-1)(5)$.
- $101 - 5n + 5 = 1 + 5n - 5 \implies 106 - 5n = 5n - 4$
- $110 = 10n \implies n = 11$. Step 4: Final Answer:
(i) $d = -5$. (ii) 31 watermelons. (iii)(a) 1071 watermelons or (iii)(b) $n = 11$.
