Answer
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Hint – In order to solve this problem use the formula of the nth term of an AP and make the equations according to the conditions given, from that you will find the first term and common difference and then you can obtain the temperature of each day.
Complete step-by-step answer:
Let us assume the variables first,
Temperature of Monday = ${{\text{a}}_{\text{1}}}$
Temperature of Tuesday = ${{\text{a}}_{\text{2}}}$
Temperature of Wednesday = ${{\text{a}}_{\text{3}}}$
Temperature of Thursday = ${{\text{a}}_{\text{4}}}$
Temperature of Friday = ${{\text{a}}_{\text{5}}}$
Temperature of Saturday = ${{\text{a}}_{\text{6}}}$
It is given that the sum of the temperature of Monday and Saturday was ${5^0}$C more than the sum of Tuesday and Saturday.
So, we can do,
(${{\text{a}}_{\text{1}}}$+${{\text{a}}_{\text{6}}}$)-(${{\text{a}}_{\text{2}}}$+${{\text{a}}_{\text{6}}}$)=5
${{\text{a}}_{\text{1}}}$-${{\text{a}}_2}$= 5 ……(i)
We know ${{\text{a}}_{\text{n}}}$= a+(n-1)d
Using the above formula in equation (i) we get,
a+(1-1)d-(a+(2-1)d) = 5
a-a-d=5
d=-5 …..(ii)
It is also given that the temperature of Wednesday was -30 degrees.
So, we can do ${{\text{a}}_{\text{3}}}$= a+2d=-30
Therefore, a+2d=-30 ……….(iii)
From equation (ii) and (iii) we get,
a+2(-5)=-30
Therefore, a=-20
So, we found first term as -20 and common difference as -5
Therefore the temperature of Monday is ${{\text{a}}_{\text{1}}}$=a=-20 degrees
Temperature of Tuesday = ${{\text{a}}_{\text{2}}}$ =a +d=-20+(-5)=-25 degrees
Temperature of Wednesday = ${{\text{a}}_{\text{3}}}$ = -30 degrees (given)
Temperature of Thursday = ${{\text{a}}_{\text{4}}}$ =a+3d=-20+3(-5)=-35 degrees
Temperature of Friday = ${{\text{a}}_{\text{5}}}$ = a+4d=-20+4(-5)=-40 degrees
Temperature of Saturday = ${{\text{a}}_{\text{6}}}$= a+5d = -20+5(-5)=-45 degrees
Hence, we find the temperature of all the five days.
Note – Whenever you face such a type of problem and it is said that the series is in AP then use the formulas of AP as per the given conditions in the questions. Then find the unknown terms like first term and common difference as unknown here after finding those variables you can get the right solution to the problem.
Complete step-by-step answer:
Let us assume the variables first,
Temperature of Monday = ${{\text{a}}_{\text{1}}}$
Temperature of Tuesday = ${{\text{a}}_{\text{2}}}$
Temperature of Wednesday = ${{\text{a}}_{\text{3}}}$
Temperature of Thursday = ${{\text{a}}_{\text{4}}}$
Temperature of Friday = ${{\text{a}}_{\text{5}}}$
Temperature of Saturday = ${{\text{a}}_{\text{6}}}$
It is given that the sum of the temperature of Monday and Saturday was ${5^0}$C more than the sum of Tuesday and Saturday.
So, we can do,
(${{\text{a}}_{\text{1}}}$+${{\text{a}}_{\text{6}}}$)-(${{\text{a}}_{\text{2}}}$+${{\text{a}}_{\text{6}}}$)=5
${{\text{a}}_{\text{1}}}$-${{\text{a}}_2}$= 5 ……(i)
We know ${{\text{a}}_{\text{n}}}$= a+(n-1)d
Using the above formula in equation (i) we get,
a+(1-1)d-(a+(2-1)d) = 5
a-a-d=5
d=-5 …..(ii)
It is also given that the temperature of Wednesday was -30 degrees.
So, we can do ${{\text{a}}_{\text{3}}}$= a+2d=-30
Therefore, a+2d=-30 ……….(iii)
From equation (ii) and (iii) we get,
a+2(-5)=-30
Therefore, a=-20
So, we found first term as -20 and common difference as -5
Therefore the temperature of Monday is ${{\text{a}}_{\text{1}}}$=a=-20 degrees
Temperature of Tuesday = ${{\text{a}}_{\text{2}}}$ =a +d=-20+(-5)=-25 degrees
Temperature of Wednesday = ${{\text{a}}_{\text{3}}}$ = -30 degrees (given)
Temperature of Thursday = ${{\text{a}}_{\text{4}}}$ =a+3d=-20+3(-5)=-35 degrees
Temperature of Friday = ${{\text{a}}_{\text{5}}}$ = a+4d=-20+4(-5)=-40 degrees
Temperature of Saturday = ${{\text{a}}_{\text{6}}}$= a+5d = -20+5(-5)=-45 degrees
Hence, we find the temperature of all the five days.
Note – Whenever you face such a type of problem and it is said that the series is in AP then use the formulas of AP as per the given conditions in the questions. Then find the unknown terms like first term and common difference as unknown here after finding those variables you can get the right solution to the problem.
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