Question

this question from arethmetic progression

On 1st June 2016, Sarika decides to save ₹ 10. On First day, 11₹ on Second day,
12₹ on thirds day. If she decides to save like this, then on 31st December 2016.
What would be her total saving?
-
-​

Answer 1

Answer:

Sanika saved rs.70,455.

Step-by-step explanation:

Sanika saves rs.10 on the first day , rs.11 on

the second day , rs.12 on the third day , ...

so, 10 , 11 , 12, ... is a sequence.

The common difference d = 11-10 = 12-11 = 1

which is constant.

so, 10 , 11 , 12 , ... is an A.P.

Here, a = 10 and d = 1.

The year 2016 was a leap year.

A leap year has 366 days. so, n = 366.

The total saving in 366 days is S366

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

so, S366 = 366/2[2×10+(366-1)×1]...( Substituting the values )

=183(20+365)

=183×385

so, S366 = 70455.

Answer 2

Hope this solution helps you

And if you have any question ask I will forever solve your question without any hesitation

Thankyou

Solution