A student of class 10 saves Rs 4 on day one Rs 7 on day two Rs 10 on day three.and so on. How much amount will the student save in month of February 2011?

Question

Answer

Answer

It is Feb 2011 therefore no. of days=28
he starts on Feb 1
ap: 4,7,10.....
a=4,d=7-4=3,n=28
to find: amount saved in 28 days
soln: Sn=n/2 (2a+(n-1)d)
S28= 28/2 (2×4+(28-1)3)
S28= 14 (8+81)
S28= 14×89=1246
therefore he will save Rs 1246 in the month of Feb 2011

Answer

When did he start?

Answer

From chapter arithmetic progression?

Answer

Savings Calculation for February 2011:

To calculate the amount the student will save in the month of February 2011, we first need to determine the number of days in February. In a non-leap year, February consists of 28 days. Therefore, we will consider the student's savings for 28 days.

The student saves Rs 4 on day one, Rs 7 on day two, Rs 10 on day three, and so on. This pattern indicates that the savings are increasing by Rs 3 each day.

Calculation:
1. Day one: Rs 4
2. Day two: Rs 4 + Rs 3 = Rs 7
3. Day three: Rs 7 + Rs 3 = Rs 10

The student saves an additional Rs 3 each day. We can observe that the savings on each day follow an arithmetic progression with a common difference of 3.

To calculate the total savings for February, we need to find the sum of the savings for 28 days. We can use the formula for the sum of an arithmetic progression:

Sum = (n/2) * (2a + (n-1)d)

Where:
- n is the number of terms (28 in this case)
- a is the first term (Rs 4)
- d is the common difference (Rs 3)

Substituting the values into the formula:
Sum = (28/2) * (2 * 4 + (28 - 1) * 3)
= 14 * (8 + 27 * 3)
= 14 * (8 + 81)
= 14 * 89
= Rs 1246

Therefore, the student will save a total of Rs 1246 in the month of February 2011.

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