Sir please give me a detailed solution of this question from arithmetic progression chapter in which we have been asked to find the number of days a child can contribute to put the five-rupee coins into piggy bank and also we are asked to find the total money she saved if she puts one five rupee coin her saving in the piggy bank on the first day. She increases her saving by one five rupee-coin daily. If the piggy bank can hold 190 coins of five rupees in all.
Book RS Aggarwal, Class 10, chapter 5C, question no 49
Saving of the child on the first day = Rs 5
Saving on the second day = Rs 5 + Rs 5 = Rs 10
Saving on the third day = Rs 5 + 2 × Rs 5 = Rs 15 and so on
The saving of the child on different days are Rs 5, Rs 10, Rs 15, ….
Since the savings of the child for each succeeding day is Rs 5 more than for the preceeding day, therefore the savings for different days forms an AP with first term a = Rs 5 and common difference d = Rs 5.
Suppose the number of days she continued to put the five-rupees coin in the piggy bank be n.
It is given that the total number of five-rupees coins in the piggy bank is 190.
So, the total sum of money saved by the child in n days = 190 × 5 = Rs 950
∴��=190×5=950⇒�22×5+�-1×5=950⇒�2�+1=190⇒�2+�=380⇒�2+�-380=0⇒�2+20�-19�-380=0⇒��+20-19�+20=0⇒�+20�-19=0⇒�+20=0or�-19=0⇒�=-20or�=19
Since the number of days cannot be negative, so n = 19.
So, the number of days she continued to put the five-rupees coin in the piggy bank is 19.
Also,
Total sum of money saved by her = Rs 950
→ Child’s daily input is 5, 10, 15, —
& the number of coins on each days are
1, 2, 3, _ _ _ → A.P with a = 1, d = 1 where Sn⩽190
⇒2n(2+(n−1)1)⩽190
⇒n+2n^2−2n⩽190
⇒n^2+n⩽380
⇒n^2+n−380⩽0
⇒n^2+20n−19n−380⩽0
⇒n(n+20)−19(n+20)⩽0
⇒(n−19)(n+20)⩽0
⇒nϵ[−20,19]
⇒ she can
put coins for 19 days
For amount, 5,10,15,...→AP=5 (1, 2, 3,_ _ _ )
Sum of 19 terms =[5×19(20)]/2 =Rs.950