Question

a child puts one five rupee coin of her saving in the piggy bank on the first day.she increases her saving by one rupee coin daily.if the piggy bank can hold 190 coins in 5 rupee in all,find the number of days she can continue to put the coins into it and find the total money she saved

Answer 1

money added to money bank on first day = rs 5
                                         on second day = rs 10
                                        0n third day  =  rs 15


and so on
                                    n = 190

thus

AP = 5, 10. 15, ...........195

thus   a =5
          d =5
         n= 190


Thus  S =  n/2[2a+(n-1)d]
             = 95[ 10 + 189*5]
             = 90725


thus total money is 90725 Rs

Answer 2

Solution :-

The child puts in her piggy bank on first day = Rs. 5 coin 

On second day she puts = 5 + 5 = Rs. 10 

On third day she puts = 5 + 5 + 5 = Rs. 15

Her piggy bank can hold 190 5 rupee coin in all.

So, this is the case of an Arithmetic Progression.

5, 10, 15, ......................190

a = 5, d = 5 and n = 190

S = n/2[2a + (n - 1)d]

⇒ 190/2[2*5 + (190 - 1)5

⇒ 95[10 + 189*5]

⇒ 95[10 + 945]

⇒ 95*955

= Rs. 90725

Her total saving is Rs. 90725