(3) On 1st January 2016, Sanika decides to save? 10,ă11 on second

1.	(3) On 1st January 2016, Sanika decides to save? 10,ă11 on second day,12 onthird day. If she decides to save like this, then on 31st December 2016, whatwould be her total savings?
Answer» On 1st Jan,She saved 10 Rs. Next Day 11 Rs. Next Day 12 rs So money saved by her= 10 + 11 + 12 + 13 + .........................+ (till 31st Dec) Now, Above Sum is the sum of a Arithmetic Progression witha = 10d = 1n = 366 ( since year 2016 is a Leap Year) =>Sum of an A.P. = (n/2){ 2a + (n-1)d} = (366/2){ 20 + 365 } = 70455 So, she saved Rs. 70455 till the end of the year Like my answer if you find it useful!

(3) On 1st January 2016, Sanika decides to save? 10,ă11 on second day,12 onthird day. If she decides to save like this, then on 31st December 2016, whatwould be her total savings?

Answer»

On 1st Jan,She saved 10 Rs. Next Day 11 Rs. Next Day 12 rs

So money saved by her= 10 + 11 + 12 + 13 + .........................+ (till 31st Dec)

Now, Above Sum is the sum of a Arithmetic Progression witha = 10d = 1n = 366 ( since year 2016 is a Leap Year)

=>Sum of an A.P. = (n/2){ 2a + (n-1)d} = (366/2){ 20 + 365 } = 70455

So, she saved Rs. 70455 till the end of the year

Like my answer if you find it useful!