Answer
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Hint: Let the temperatures of 5 days of the week in Ooty starting from Monday be $\left( a-2d \right){}^\circ C,\text{ }\left( a-d \right){}^\circ C,\text{ }a{}^\circ C,\text{ }\left( a+d \right){}^\circ C,\text{ }\left( a+2d \right){}^\circ C$ . Form the equations using the conditions given in the question and solve the equations to get the answer.
Complete step-by-step solution -
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{r}}$, and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$
Now let us start the solution to the above question. As it is given that the temperatures are in A.P., we let the temperatures of the consecutive 5 days of the week in Ooty starting from Monday be $\left( a-2d \right){}^\circ C,\text{ }\left( a-d \right){}^\circ C,\text{ }a{}^\circ C,\text{ }\left( a+d \right){}^\circ C,\text{ }\left( a+2d \right){}^\circ C$ .
Now it is given that the sum of temperatures from Monday to Wednesday is $0{}^\circ C$ .
$\therefore \left( a-2d \right)+\left( a-d \right)+a=0$
$\Rightarrow 3a-3d=0$
$\Rightarrow a=d...........(i)$
Also, it is given that the sum of the temperatures from Wednesday to Friday is $18{}^\circ C$
$\therefore a+\left( a+d \right)+\left( a+2d \right)=18$
$\Rightarrow 3a+3d=18$
Now, if we substitute the value of ‘a’ from equation (i), we get
$3d+3d=18$
$\Rightarrow 6d=18$
$\Rightarrow d=3$
Therefore, if we put the value in equation (i), we get the value of a to be 3 as well.
Therefore, the temperatures of 5 consecutive days of the week in Ooty starting from Monday is $-3,\text{ 0}{}^\circ C,\text{ 3}{}^\circ C,\text{ 6}{}^\circ C,\text{ 9}{}^\circ C$ .
Note: In a question related to an arithmetic progression, the most important thing is to find the common difference and the first term. You also need to learn the basic properties of all the standard sequences like the arithmetic progression and geometric progressions.
Complete step-by-step solution -
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{r}}$, and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$
Now let us start the solution to the above question. As it is given that the temperatures are in A.P., we let the temperatures of the consecutive 5 days of the week in Ooty starting from Monday be $\left( a-2d \right){}^\circ C,\text{ }\left( a-d \right){}^\circ C,\text{ }a{}^\circ C,\text{ }\left( a+d \right){}^\circ C,\text{ }\left( a+2d \right){}^\circ C$ .
Now it is given that the sum of temperatures from Monday to Wednesday is $0{}^\circ C$ .
$\therefore \left( a-2d \right)+\left( a-d \right)+a=0$
$\Rightarrow 3a-3d=0$
$\Rightarrow a=d...........(i)$
Also, it is given that the sum of the temperatures from Wednesday to Friday is $18{}^\circ C$
$\therefore a+\left( a+d \right)+\left( a+2d \right)=18$
$\Rightarrow 3a+3d=18$
Now, if we substitute the value of ‘a’ from equation (i), we get
$3d+3d=18$
$\Rightarrow 6d=18$
$\Rightarrow d=3$
Therefore, if we put the value in equation (i), we get the value of a to be 3 as well.
Therefore, the temperatures of 5 consecutive days of the week in Ooty starting from Monday is $-3,\text{ 0}{}^\circ C,\text{ 3}{}^\circ C,\text{ 6}{}^\circ C,\text{ 9}{}^\circ C$ .
Note: In a question related to an arithmetic progression, the most important thing is to find the common difference and the first term. You also need to learn the basic properties of all the standard sequences like the arithmetic progression and geometric progressions.
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