The sum of 5 numbers in an arithmetic progression is 75 and the product of the least and the greatest is 161 find the numbers

Step-by-step explanation:

Suppose the first and the last term of an A.P. are aand l respectively.

Since the product of the greatest and the least is 161, so we have;

al = 161       ...(i)

Also the sum of five numbers of an A.P. is 75, so we have;

s5 = 52(a+l) = 75⇒(a+l) = 75×25⇒a+l = 30 ...(ii)

So from (i) and (ii) we have;

a+161a = 30⇒a2+161a = 30⇒a2+161 = 30a⇒a2−30a+161 = 0⇒a2−23a−7a+161 = 0⇒a(a−23)−7(a−23) = 0⇒(a−7)(a−23) = 0⇒(a−7) = 0 and (a−23) = 0⇒a = 7 and  a = 23

So from (i) we get, when a = 23, then l = 7 and when a= 7, then l = 23

Therefore the greatest number is 23 and the least number is 7.

Suppose d is the commomdn difference of an A.P.

And we know the formula for the fifth term = t5 = a+(5−1)d

Considering 7 as the first term and 23 as the last term i.e. 5th term we have;

⇒23 = 7+(5−1)d⇒23−7 = 4d⇒16 = 4d

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