Question

2) on 1st Jan 2016, Sanika decide to save Rs. 10, Rs. 11 on second day, Rs. 12 on third day. If
she decide to save like this, then on 31" Dec 2016 what would be her total saving ?​

Answer 1

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 with

a = 10

d = 1

n = 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

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Hope it will help you!!!!

Answer 2

Answer:

It is a case of A P

1st number is 10, common diff 1 and number of terms 366

put in the formula

sum = (366/2)(2×10+ (366-1)×1)

=183×(20+365)

=183×385

=70455 Ans