Question

on 1st jan 2016 sanika decides to save ₹10,11,and on second day ₹12 on third day .if she decides to save like this .then on 31st dec 2016 what would be her total saving ?​

Answer 1

= we have

= 1st January =

10+11+20= 41

then we do this same

so we get = 41 multiply by 41

= ans = 1681

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Answer 2

Answer :

›»› Sanika total saving would be ₹70455 till the end of the year.

Given :

• On 1st jan 2016 sanika decides to save ₹10, ₹11, and on second day ₹12 on third day.

To Find :

• On 31st dec 2016 what would be her total saving ?

Solution :

Sanika saves ₹10 on first day, ₹11 on second day, ₹12 on third day. This is an sequence.

→ 10 + 11 + 12 + ............ + (till 31 december).

The common difference will be,

→ d = a₂ - a₁

→ d = 11 - 10

→ d = 1

Therefore, 10, 11, 12 ..... is an AP.

Here,

• a = 10.
• d = 1.

The year 2016 is a leap year. A leap year has 366 days.

• n = 366.

The total saving in 366 days is s₃₆₆.

As we know that

→ sₙ = n/2 [2a + (n - 1)d]

→ s₃₆₆ = 366/2 [2 * 10 + (366 - 1)1]

→ s₃₆₆ = 183 [2 * 10 + (366 - 1)1]

→ s₃₆₆ = 183 [2 * 10 + 365 * 1]

→ s₃₆₆ = 183 [2 * 10 + 365]

→ s₃₆₆ = 183 [20 + 365]

→ s₃₆₆ = 183 * 385

→ s₃₆₆ = 70455.

Hence, the Sanika total saving would be ₹70455 till the end of the year.