Cricket Practice Dates

This is a simple series. The 1st date of practice = 1st, second date is thus 5th, third is 9th (since every FOURTH DAY mean a difference of 4 days in between)

Thus, we have a series: 1, 5, 9, 13, 17...

Common formula for this is: 4n - 3

We can use a trial and error method for the number of practices he has done:

for n = 6, 4(6) - 3 = 21st April (Nah!)

for n = 7, 4(7) - 3 = 25 (Can still do better!)

for n = 8, 4(8) - 3 = 29th April, and this is the last date since April has a maximum 30 days, so if he does another practice, that will be in May!

Hence, 29th April

I just listed out the first three dates of practice: 1, 5, 9 and realized a pattern each number was one more than a multiple of 4. So I found the multiple closest to 30 (the number of days in April) and got 28 so I added one to it resulting in 29.

Sanjeev has practice every 4 days. So the most cycles you can fit would be 7. 7x4 = 28th day. However, His first practice was on the 1st. So that would move the cycle by 1 day.

Hence his last practice is on the 29th.

Therefore , Starting date = 1 April

He played every 4th day

So the last date = 29 April

we can form a series as 1,5,9,13,17,21,25,29,33..... As a month does not have 33rd day, answer is obviously 29.

30/4=7,5

7*4=28

28+1=29

Let guess from the 1st day, 1, 5, 9, 13,.. it should be 4n+1=x, x<=30/31

april month is 30 days. the last digit of this month divided by 4 is 28. as he start practice at 1st april so the last day will be 29th april.

SIMPLE if the first date of practice is 1 then the second will be 5 and third will be 9 and so on........... and the last date becomes 29th april

This series goes on common difference of 4... so it follows A.P(Arithmetic Progression)... t(n) = a + (n-1)d a=1 ; d= 4 t(1) = 1 + 0x4 = 1 ; i.e., given first day t(2) = 1 + 1x4 = 5 ; next day t(3) = 1 + 2x4 = 9 ; third day . . . t(8) = 1 + 7x4= 29 ; last day of practice from these we clear, there are totally 8 practices and the last practice is on 29th April

Simple count. 1,5,9,13,17,21,25,29

Sanjeev has cricker practice every 4th day in the month of april starting with 1st april

there are 30 days in april month

thus, 4 * 7 = 28

but practice started on 1st april

therefore by adding 1 in 28 results 29 28 + 1 = 29

so last date of practice for the nonth is 29 april.

The first date is the 1st. The next is on the 5th. The third would be on the 9th. Next is simple work out the nth term which is 4n-3. Think of the largest multiple of 4 just over 30 which is 32 then minus 3= 29

on 1st April Sanjeev starts and he will practice at 4th day so that the last day should be multiple of four then after plus 1 in that,means the last day will have 4*7+1=29 date.....

First day of practise: 1st April 4th day means 3 days later... Counting from 1st we get our 4th day as 5th April... Keep on adding 4 days ... 9th,13th,17th,21st,25,29th.. Hence, the last of practise is 29th of April.! :)

The practice days are - 1,5,9,........,29.

So 29th will be his last date of practice for the month.

The dates of practice forms a sequence of $4n+1$ and 29th is the last date of such form in the month of April.

he plays cricket every 4th day starting fro 1. therefore he plays on following days: 5,9,13,17,21,25 and 29. since 29 is last day he played last in 29th.

in aprail 30 days . so 4*7=28 & 1st aprail is also include so answer is 29

The dates which are for cricket practice in April are $1,5,9,....$ which is an arithmetic progression (AP) with first term $(a)=1$ and common difference $(d)=4$ and term no. $n$ and $a_n$ be the $n$ th term of AP.

We see that if $n=8$ , then the 8th term of AP, $a_8=a+(n-1)d=1+(8-1)4=1+28=29$ . The next practice date will be after 4 days but April only has 30 days. So, last day of cricket practice in that month $=\boxed{29\text{ April }}$

TOTAL DAYS IN APRAIL IS 30 AND HE STARED 1ST APRAIL, HE PRACTICE EVERY 4TH DAY SO 28 IS THE LAST MULTIPLE OF 4 AND HE STARTED ON 1 THAT;S WHY 28+1=29

ITS SIMPLE JOB EVERY 4RH DAY MEANS MULTIPLE OF 4. STARTING FROM 1ST APRIL, SO 28 AND 32 IS MULTIPLE OF 4, BUT 32 APRIL CAN NOT BE POSSIBLE. SO (28+1) APRIL IS FINAL ANSWER.

sequences Un = U1 + (n-1)d. Since he train every 4th day, d = 4. He starts at 1st Apr, U1 = 1. Un is the day he finished. So Un = 1 + (n-1) 4 . Since 7 4=28, 8 4=32(April DON'T have 33rd day), we make n-1=7. So Un= 1 + 7 4=1 + 28=29.