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This section comprises $3$ case study-based questions of $4$ marks each
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 $15$th 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 $n$th row from top is equal to number of watermelons in the $n$th row from bottom, find the value of $n$.
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 $15$th 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 $n$th row from top is equal to number of watermelons in the $n$th row from bottom, find the value of $n$.
Show SolutionHide Solution↓
(i) $101 + 20d = 1$ (1/2 Mark)
$d=-5$ (1/2 Mark)
(ii) $a_{15} = 101 + 14(-5)$ (1/2 Mark)
$= 31$ (1/2 Mark)
(iii) (a) $S_{21} = \frac{21}{2} [202 + 20(-5)]$ (1 Mark)
$= 1071$ (1 Mark)
OR
(iii) (b) $1 + (n-1) 5 = 101 + (n-1) (-5)$ (1 1/2 Mark)
$n = 11$ (1/2 Mark)
$d=-5$ (1/2 Mark)
(ii) $a_{15} = 101 + 14(-5)$ (1/2 Mark)
$= 31$ (1/2 Mark)
(iii) (a) $S_{21} = \frac{21}{2} [202 + 20(-5)]$ (1 Mark)
$= 1071$ (1 Mark)
OR
(iii) (b) $1 + (n-1) 5 = 101 + (n-1) (-5)$ (1 1/2 Mark)
$n = 11$ (1/2 Mark)