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
Answers
Answered by
122
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
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
Answered by
122
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
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
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