thchild puts one five-rupee coin of her saving in the piggy bank onfir

1.	thchild puts one five-rupee coin of her saving in the piggy bank onfirst day. She increases her saving by onc five-rupec coin dailyDigEy bank can hold 190 coins of five rupees in all, find the numberdays she can continue to put the five-rupce coins into it and find the tmoney she saved.Write your views on the habit of saving.
Answer» 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[25 + (190 - 1)5 ⇒ 95[10 + 1895] ⇒ 95[10 + 945] ⇒ 95*955 = Rs. 90725 Her total saving is Rs. 90725

thchild puts one five-rupee coin of her saving in the piggy bank onfirst day. She increases her saving by onc five-rupec coin dailyDigEy bank can hold 190 coins of five rupees in all, find the numberdays she can continue to put the five-rupce coins into it and find the tmoney she saved.Write your views on the habit of saving.

Answer»

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