Find five consecutive terms in an A.P. such that their sum is 60 and the product of the third and fourth term exceeds the fifth by 172.

Hint: Arithmetic progression is a series of numbers that increase or decrease successively by the same amount each time i.e. every two consecutive numbers have a common difference between them.

Complete step-by-step solution -
\[a,\text{ }a+d,\text{ }a+2d,\text{ }a+3d,\text{ }a+4d\text{ }\ldots .\] and so on. Where \[a\] is the first term of the A.P. and \[d\] is the common difference.
When the sum of the odd number of terms is asked we take the terms in such a way that common difference cancels out.
Example:
If sum of three consecutive numbers is given we take numbers as
\[a-d,a,a+d\] so that when we add them the common difference cancels out, \[\left( a-d \right)+a+\left( a+d \right)=3a\]
Similarly if of four consecutive numbers is given we take numbers as
\[a-3d,\text{ }a-d,\text{ }a+d,\text{ }a+3d\] with common difference as \[2d\].
Explanation:
Step 1: Let the five consecutive numbers of the A.P. be
\[a-2d,\text{ }a-d,\text{ }a,\text{ }a+d,\text{ }a+2d\] where d is the common difference.
Step 2: By first statement we know that their sum is 60.
\[\left( a-2d \right)+\left( a-d \right)+a+\left( a+d \right)+\left( a+2d \right)=60\]
\[5a=60\] therefore we get,\[a=12\]
Step 3: By second statement,
Product of third and fourth term is \[a\times \left( a+d \right)\]
As given it exceeds fifth term which is \[\left( a+2d \right)\] by 172 this implies,
\[a\left( a+d \right)\left( a+2d \right)=172\]
From step 2 we know \[a=12\] so substituting in above equation we get,
\[12\left( 12+d \right)\left( 12+2d \right)=172\]
\[144 + 12d - 12 - 2d=172\]
\[10d\text{ }=\text{ }40\] this implies,
\[d=4\]
Step 4: Now we have to find the numbers by simply substituting the values of a and d in the five numbers we took,
\[a-2d=12-2\times 4=4\]
\[a-d=12-4=8\]
\[a=12\]
\[a+d=12+4=16\]
\[a+2d=12+2\times 4=20\]
Hence we get five consecutive numbers of the A.P. as \[4,\text{ }8,\text{ }12,\text{ }16,\text{ }20\].

Note: In an arithmetic progression question whenever we are given the sum of numbers we assume numbers to be such that on adding them the common difference always cancels out. Sometimes students make mistakes by taking the terms as the sum of a and d consecutively which makes our solution complicated. so we take the terms in this way that we remain with only one variable.