According to the question we can form an A.P.
10, 11, 12, 13, ……
Hence, the first term a = 10
Second term t1 = 11
Third term t2 = 12
Thus, common difference d = t2 – t1 = 12 – 11 = 1
Here, number of terms from 1st Jan 2016 to 31st Dec 2016 is,
n = 366
We need to find S366
Now, by using sum of nth term of an A.P. we will find its sum
![Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter3 Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 21](https://cdn1.byjus.com/wp-content/uploads/2020/04/maharashtra-board-solutions-for-class-10-maths-part-1-chapter-2-image-211.png)
Where, n = no. of terms
a = first term
d = common difference
Sn = sum of n terms
Thus, on substituting the given value in formula we get,
![Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter3 Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 22](https://cdn1.byjus.com/wp-content/uploads/2020/04/maharashtra-board-solutions-for-class-10-maths-part-1-chapter-2-image-221.png)
⇒S366 = 183 [ 20 + 365]
⇒S366 = 183 × 385
⇒S366 = Rs 70,455