Question

The cost of digging a well, after every meter of digging when it costs ₹150 for the First mette and rises by ₹50 for each subsequent meter .is arithmetic progression or not why​

Answer 1

Answer:

n/2(300+(n-1)50)

or n(125+25n)

Step-by-step explanation:

a=150 d=50 N=n

Sn=n/2(2a+(n-1)d)

Sn=n/2(300+(n-1)50)

Sn=n(125+25n)

Answer 2

Solution

We have ,

Cost of digging well for 1st meter =Rs150

cost of digging for subsequent dist. =Rs50

A/q

let a1=150

d=50

a1 =150

a2=a+d =150 +50=200

a3=a+2d=150+2(50)=250

a4=a+3d=150+3(50)=150+150=300

Therefore , we see that cost for each dist. is

150, 200, 250, 300

By this we can conclude that the following is the form of an A.P.