Question

On 1st Jan 2016, Sanika decides to save Rs 10, Rs 11 on second day, Rs 12 on third day. If she decides to save like this, then on 31^st Dec 2016 what would be her total saving?

Answer 1

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 1^st Jan 2016 to 31^st Dec 2016 is,

n = 366

We need to find S₃₆₆

Now, by using the sum of the n^th term of an A.P. we will find its sum

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

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, on substituting the given value in the formula we get,

⇒ S₃₆₆  = `366/2`[2 × 10 + (366 - 1) × 1]

⇒ S₃₆₆ = 183 [ 20 + 365]

⇒ S₃₆₆ = 183 × 385

⇒ S₃₆₆ = Rs 70,455