GOOD Morning :-P 6.how many 3 digit no:have the property tht their digit taken from left to right form an AP or GP
Taking case by case 1st for AP 1 the common difference can be -0,1,2,3,4 Hence possibilities are 111 123 135 147 159=5 cases for 2 same way as 0,1,2,3 222 234 246 258 = 4 cases for 3 333 345 357 369= 4cases for 4 444 456 468 =3 cases for 5 555 567 579=3 cases for 6 666 678 =2 cases for 7 777 789 = 2 cases two cases for 8,9 total aps=25 aps are possible
For GPs with r=1 are same for all 1-9 hence we won't take that let with r=2 124 248 And with r=3 139 Total cases 3
GOOD Morning :-P 1. Find 28383rd term in series 1234567891011.. a.3 b.4 c.9 d.7 2. sum of 1/1*5+1/5*9+1/9*13+..+1/221*225 a.28/221 b.56/221, c.56/225 d. none 3. sum of 1/(2^1/2+1^1/2)+1/(2^1/2+3^1/2)+..+1/(120^1/2+121^1/2) a. 10 b.11 c.12 d.none 4. sum of infinite term 1/1+1/3+1/6+1/10+1/15. a.2 b.2.25 c.3 d.4 5. sum of +1/6+1/12+1/20+..+1/156+1/182 6.how many 3 digit no:have the property tht their digit taken from left to right form an AP or GP
Taking case by case 1st for AP 1 the common difference can be -0,1,2,3,4 Hence possibilities are 111 123 135 147 159=5 cases for 2 same way as 0,1,2,3 222 234 246 258 = 4 cases for 3 333 345 357 369= 4cases for 4 444 456 468 =3 cases for 5 555 567 579=3 cases for 6 666 678 =2 cases for 7 777 789 = 2 cases two cases for 8,9 total aps=25 aps are possible
For GPs with r=1 are same for all 1-9 hence we won't take that let with r=2 124 248 And with r=3 139 Total cases 3
GOOD Morning :-P ---------------- 5. sum of +1/6+1/12+1/20+..+1/156+1/182 ---------------
hi, +1/6+1/12+1/20+..+1/156+1/182 can be written as 1 - 1/2 + 1/2 -1/3 + 1/3 -1/4 + 1/4 -1/5 + .... + 1/12 -1/13 + 1/13 - 1/14 which comes out to be 1 - 1/14 = 13/14
Taking case by case 1st for AP 1 the common difference can be -0,1,2,3,4 Hence possibilities are 111 123 135 147 159=5 cases for 2 same way as 0,1,2,3 222 234 246 258 = 4 cases for 3 333 345 357 369= 4cases for 4 444 456 468 =3 cases for 5 555 567 579=3 cases for 6 666 678 =2 cases for 7 777 789 = 2 cases two cases for 8,9 total aps=25 aps are possible
For GPs with r=1 are same for all 1-9 hence we won't take that let with r=2 124 248 And with r=3 139 Total cases 3
Hence net cases possible for AP or GP =25+3 =28
Correct me if i am wrong
0 cant be taken as a common difference bro,,, also for GPS.. u have common ratio of 2/3 and 3/2 in that case you have nos. like 964 and 469 with common ratio 1/3 we have nos. 962 and with common ratio 1/2 we have 421, 842 etc...lot of gps are possible...thy'll be finite though:)
GOOD Morning :-P ----------- 4. sum of infinite term 1/1+1/3+1/6+1/10+1/15. a.2 b.2.25 c.3 d.4 -----------
hi, here the nth terms can be written as 1/(1+2+3+...+n) = 2/n(n+1) = 2 here the sum upto n terms of the series will be = 2 so as n tends to infinity, the term 1/(n+1) will tend to 0... hence the sum will become 2 = 2 hope it is correct...
How did you solve the first question please explain.
1---9 there are 9 digits 10-99 there are 180 digits 100-999 --->2700 digits....
similarly we find the the remaining digits and find out total 4 digit number present... the number will be 7371...here 2 will be 28381 digit.. 28383rd digit will be 3...
GOOD Morning :-P 1. Find 28383rd term in series 1234567891011.. a.3 b.4 c.9 d.7 2. sum of 1/1*5+1/5*9+1/9*13+..+1/221*225 a.28/221 b.56/221, c.56/225 d. none 3. sum of 1/(2^1/2+1^1/2)+1/(2^1/2+3^1/2)+..+1/(120^1/2+121^1/2) a. 10 b.11 c.12 d.none 4. sum of infinite term 1/1+1/3+1/6+1/10+1/15. a.2 b.2.25 c.3 d.4 5. sum of +1/6+1/12+1/20+..+1/156+1/182 6.how many 3 digit no:have the property tht their digit taken from left to right form an AP or GP
1. taking terms to be digits option a 45 + 90*2 + 900*3 + 4x = 28383 x = 6364, rem = 2 7363 7364 28383rd digit = 3 2. option c sum = (1/4) sum =(1/4)224/225 = 56/225 3. option a sum = root2 - root1 + root3 - root2 + root4 - root3 +.....+root121 - root120 sum = 11 - 1 = 10 4. option a nth term = 2/n(n+1) = 2 sum = 2 sum = 2 5. nth term = 1/n(n+1) = 1/n - 1/n+1 sum = 1 - 1/14 = 13/14
6. AP:- abc b = a+c/2 a+c = 2==>2 a+c = 4==>4 a+c = 6==>6 a+c = 8==>8 a+c = 10==>9 a+c = 12==>7 a+c = 14==>5 a+c = 16==>3 a+c = 18==>1 total = 45 GP:- b = root(ac) ac = 1==>1.......111 ac = 4==>3.......421,124,222 ac = 9==>3.......333,931,139 ac = 16==>3......444,842,248 ac = 25==>1......555 ac = 36==>3......666,964,469 ac = 49==>1......777 ac = 64==>1......888 ac = 81==>1......999 total = 17 common = 9 grand total = 53
p is a natural number,2p is having 28 divisors and 3p has 30 divisors.how many divisors of 6p will be there ?how to do such kind of questions?plzz explain.
p is a natural number,2p is having 28 divisors and 3p has 30 divisors.how many divisors of 6p will be there ?how to do such kind of questions?plzz explain.
Let p= 2^a*3^b*5^c..... =>No. of factor of p=(a+1)(b+1)(c+1).... =F 2p= 2^(a+1)*3^b*5^c....=>.No. of factor of 2p=(a+2)(b+1)(c+1).... =28 3p= 2^a*3^(b+1)*5^c.....=>No. of factor of 3p=(a+1)(b+2)(c+1).... =30
6p= 2^(a+1)*3^(b+1)*5^c+.... No. of factor of 6p=(a+2)(b+2)(c+1).... Put the value of (a+2)and (b+2) from the above eqns, u will get the answer
p is a natural number,2p is having 28 divisors and 3p has 30 divisors.how many divisors of 6p will be there ?how to do such kind of questions?plzz explain.
28=4*7 thus it is in form 2p=2^6*3^3 thus p= 2^5*3^3 6p=2^6*3^4 thus total factor = 7*5= 35
p is a natural number,2p is having 28 divisors and 3p has 30 divisors.how many divisors of 6p will be there ?how to do such kind of questions?plzz explain.
