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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
number of watermelons in the nth row from bottom, find the
value of n.
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
number of watermelons in the nth row from bottom, find the
value of n.
Show SolutionHide Solution↓
Ans.
(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
frac{1}{2} Mark)
$= 1071$ (1/2 Mark)
OR
(iii) (b) $1 + (n-1)5 = 101 + (n-1)(-5)$ (1
frac{1}{2} Mark)
$n = 11$ (1/2 Mark)
(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
frac{1}{2} Mark)
$= 1071$ (1/2 Mark)
OR
(iii) (b) $1 + (n-1)5 = 101 + (n-1)(-5)$ (1
frac{1}{2} Mark)
$n = 11$ (1/2 Mark)