On 1st Jan 2018, Sanikadecides to save Rupees 10,Rupees 11 on the second day, Rupees 12 on the third day . She decides to save like this . What would be her total savings at the end of the year ?

Answer

Sanika saved Rs. 70,455.

Explanation

Sanika saves Rs. 10 on the first day, Rs. 11 on the second day, Rs, 12 on the third day….
`therefore`10, 11, 12, … is a sequence. The common difference d = 11 -10 = 12 - 11 =1 which is constant.
`therefore` 10, 11, 12, … is sn A.P.
Here, a = 10 and d = 1.
The year 2016 was a leap year.
A leap year has 366 days. `thereforen=366.`
The total savings in 366 days is `S_(366)`
`S_(n)=(n)/(2)[2a+(n-1)d]` ...(Formula)
`thereforeS_(366)=(366)/(2)[2xx10+(366-1)xx1]` ...(Substituting the values)
`=183(20+365)`
`183xx385`
`thereforeS_(366)=70455`.