Question

(3) On 1st January 2016, Sanika decides to save 10, 11 on second day, 12 on
third day. If she decides to save like this, then on 31st December 2016, what
would be her total savings?​

Answer 1

Answer:

2016 is a leap year

hence total no. of days=366

sanika saves

₹10 on 1st Jan 2016

₹20 on 2nd Jan 2016

₹30 on 3rd Jan 2016

and so on

hence A.P= 10,20,30,40......

here,

a= t1= 10

d= 10

To find: S366 = ?

Solution:

By formula,

Sn= n/2 [ 2a+(n-1)d ]

S366 = 366/2 [ 2×10+(366-1)10]

=366/2 [ 20×365×10]

= 366/2[200×365]

=366/2[73000]

=53436000

hence sanika saved, ₹53436000 on 31st Dec 2016