On 1st january 2016, sanika decides to save ₹10,₹11bon second day, ₹12 on the third day. If she decides to save like this, then on 31st December 2016, what would be her total savings?
Answers
Answered by
13
Step-by-step explanation:
t1=a=10
t2=11
t3= 12
t2 -t1
11-10= 1
2016 was leap yr.
sn= n/2 ( 2a +(n-1) d)
=n/2 (20+(n-1) 1)
= 366/2 (20+365)
=183 × 385
= 70455
70455 rps was her Total savings
Answered by
3
Answer:
Sanika saves rs 10 on the 1st day.
a=10
The saving will increase by rs 1 after each additional day.
d=1.
To find the saving from 1st jan'16 to 31st dec'16 we need to find S366 (as it was a leap year)
we know, Sn= n÷2[2a+( n-1 d)d]
S36=366÷2[2(10)+(366-1)1]
=183×[20+365]
=183×385
=70,455rs
i.e Sanika saves rs 70,455 by the end of the year.
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