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Case Study - 2
In the month of September, villagers of Ankurhut were falling ill with high temperature. Paracetamol was one of the highest sold medicines during that phase. A survey was conducted to estimate the overall sale of Paracetamol of each pharmacy during the last $7$ days. It was observed that the number of Paracetamol sold in different shops were all 3-digit numbers, divisible by $13$, taken in order.
Based on the information given above, answer the following questions :
(i) How many Paracetamols were sold by the $7^{th}$ pharmacy?
(ii) What was the difference between the number of Paracetamols sold by the $14^{th}$ and the $9^{th}$ pharmacy?
(iii) (a) How many Paracetamols were sold by the $9^{th}$ pharmacy from the last?
OR
(iii) (b) What was the total number of Paracetamols sold in that week?
In the month of September, villagers of Ankurhut were falling ill with high temperature. Paracetamol was one of the highest sold medicines during that phase. A survey was conducted to estimate the overall sale of Paracetamol of each pharmacy during the last $7$ days. It was observed that the number of Paracetamol sold in different shops were all 3-digit numbers, divisible by $13$, taken in order.
Based on the information given above, answer the following questions :
(i) How many Paracetamols were sold by the $7^{th}$ pharmacy?
(ii) What was the difference between the number of Paracetamols sold by the $14^{th}$ and the $9^{th}$ pharmacy?
(iii) (a) How many Paracetamols were sold by the $9^{th}$ pharmacy from the last?
OR
(iii) (b) What was the total number of Paracetamols sold in that week?
Show SolutionHide Solution↓
(i) A.P. formed is
$104, 117, 130, \dots$ with $a = 104$ and $d = 13$
$a_7 = 104 + 6 \times 13 = 182$
(ii) $a_{14} - a_9 = 5 \times 13 = 65$
(iii) (a) Last term of A.P. is $988$ and $d = -13$
$a_9 = 988 + 8 \times (-13) = 884$
OR
(b) Last term of A.P. is $988$ and $d = -13$
$988 = 104 + (n - 1) \times 13$
$\Rightarrow n = 69$
$S_{69} = \frac{69}{2} \times (104+988)$
$= 37674$
$104, 117, 130, \dots$ with $a = 104$ and $d = 13$
$a_7 = 104 + 6 \times 13 = 182$
(ii) $a_{14} - a_9 = 5 \times 13 = 65$
(iii) (a) Last term of A.P. is $988$ and $d = -13$
$a_9 = 988 + 8 \times (-13) = 884$
OR
(b) Last term of A.P. is $988$ and $d = -13$
$988 = 104 + (n - 1) \times 13$
$\Rightarrow n = 69$
$S_{69} = \frac{69}{2} \times (104+988)$
$= 37674$