what is expected in this ? Plz post the full question
if the Question is to get the sum to 15 terms . then this is my approach...
the series is of the form : n(n+1)^2 we need to find it sum from n = 3 to n = 15.
now what we can do is find its sum form n = 1 to n = 15 and then subtract the sum of the first two terms form the original sum.
soln: Tn = n(n^2 + 2n + 1) = n^3 + 2n^2 + n now, Sn = sum of n terms . Sn = Sum to n terms of Tn . = sum to n terms (n^3 ) + 2 * sum to n terms (n^2) + sum to n terms (n) = {n(n+1)/2}^2 + 2 * n(n+1)(2n+1)/6 + n(n+1)/2 here n = 15. so the expression becomes , S15 = {15(16)/2}^2 + 15*16*31/3 + 15*16/2 = (120)^2 + 5 * 16 *31 + 120 = 14400 + 2480 + 120 = 17000. now the Sum of first two terms : 1*2^2 + 2*3^2 = 4+ 18 = 22. subtract 22 from S15 . therefor answer is 16978.
Please correct me if I am wrong.
Firstly, yes we had to find the sum till 15 terms.
Secondly, what i can decipher from your post is that the solution is absolutely correct but i think there's a problem at the very beginning.
If you're calling Sn as a sum of n terms and got it's general form as n(n+1)^2 it should give us the sum of 3 terms if we put n=3( Keeping in mind that we will subtract S2 from S15 later).
instead we get the third term i.e 3.4^2 . Hence, the general terms doesn't serve our purpose. Hence, I think n(n+1)^2 isn't the sum of n terms.
Correct me if I'm wrong. I couldn't find a flaw in the solution.
Firstly, yes we had to find the sum till 15 terms.
Secondly, what i can decipher from your post is that the solution is absolutely correct but i think there's a problem at the very beginning.
If you're calling Sn as a sum of n terms and got it's general form as n(n+1)^2 it should give us the sum of 3 terms if we put n=3( Keeping in mind that we will subtract S2 from S15 later).
instead we get the third term i.e 3.4^2 . Hence, the general terms doesn't serve our purpose. Hence, I think n(n+1)^2 isn't the sum of n terms.
Correct me if I'm wrong. I couldn't find a flaw in the solution.
I couldn't devise a formula for this. Is it a GP or an AP?
Please tell if there's a formula for these kind of questions.
Firstly find out the last term of - 3x4^2 + 4x5^2 + 5x6^2 + ... (15 terms)
Doing it separately like - 1) 3, 4, 5,...( 15 terms ) -> It is an A.P. with a = 3, d = 1, n = 15. We know, l = a + (n - 1) *d ( Last term of an A.P.) = 3 + (15 - 1) *1 = 17
2) And as the general term is n * ((n + 1)^2). So, Last term( L ) will be 17 * ((17 + 1)^2) i.e 17 * (18^2).
Now considering the first two terms - 3 * (4^2) + 4 * (5^2) -> 4(3 * 4 + (5^2))( Take out 4 as common )
Now take the next two terms - 5 * (6^2) + 6 * (7^2) -> 6(5 * 6 + (7^2))( Take out 6 as common )
So we are now getting a sequence - 2n( (2n-1) * 2n + ((2n+1) ^2) ) where 2 = And answer will be in the form -
2n( (2n-1) * 2n + ((2n+1) ^2) ) where 2 =
Putting values of n = 2, 3,...,8 and adding all ( Dont forget to add Last term here ) -
Firstly, yes we had to find the sum till 15 terms.
Secondly, what i can decipher from your post is that the solution is absolutely correct but i think there's a problem at the very beginning.
If you're calling Sn as a sum of n terms and got it's general form as n(n+1)^2 it should give us the sum of 3 terms if we put n=3( Keeping in mind that we will subtract S2 from S15 later).
instead we get the third term i.e 3.4^2 . Hence, the general terms doesn't serve our purpose. Hence, I think n(n+1)^2 isn't the sum of n terms.
Correct me if I'm wrong. I couldn't find a flaw in the solution.
P.S - the correct answer is 27110
hi Kunal,
I have taken Tn = n(n+1)^2 not Sn. So my approach is correct but yes I should have taken n = 17 instead of n = 15. This thought came to my mind.
Anyways thanks to everyone for correcting my mistake.
Please help me with these 2 question and provide an approch to it:
a)An alloy of gold ,silver and bronze contains 90% bronze ,7% gold and 3% silver .A second alloy of bronze and silver only is melted with the first and the mixture contains 85% of bronze ,5% of gold and 10% of silver . Find the percentage of bronze in the second alloy
Ans 72.5%
b) A cask contains 12 gallons of mixture of wine and water in the ratio 3:1.How much of the mixture must be drawn off and water subsituted ,so that wine and water in the cask become haff and half.
Please help me with these 2 question and provide an approch to it:
a)An alloy of gold ,silver and bronze contains 90% bronze ,7% gold and 3% silver .A second alloy of bronze and silver only is melted with the first and the mixture contains 85% of bronze ,5% of gold and 10% of silver . Find the percentage of bronze in the second alloy
Ans 72.5%
b) A cask contains 12 gallons of mixture of wine and water in the ratio 3:1.How much of the mixture must be drawn off and water subsituted ,so that wine and water in the cask become haff and half.
Ans 4
a) Let first alloy be of 100 gms. then, bronze= 90gm, Gold= 7gm, Silver= 3gm Let second alloy be of y gm and percentage of bronze in it be x. amount of bronze in the combined melted mixture can be expressed as,
/ (100+y) = 85 --- (1) also since second alloy does not have gold.. so the gold coming in the mixture is only from the first alloy, hence
(7x100)/ (100+y) = 5 ---- (2) solving for y from here we get y= 40 , putting this in eq (1) we get x= 72.5%
b) 12 gallons of mixture contains 9 gallons= wine and 3 gallons= water let the required amount to be replaced be x gallons, then
replacing the x gallons of mixture with water, the following equation results.
wine in replaced mix= 3x/4 and water= x/4
(9-3x/4)/ (3-x/4=x)= 1 (since the resulting mix and has water and wine in equal proportions)
can some one please explain for the solution of below problem.
A set S is formed by indicating some of the first one thosand natural numbers.S contains the maximum number of numbers such that they satisfy the following conditions, No number of the set S is prime and when the numbers of the set S are selected two at a time, we always see co prime numbers.What is the number of elements in the set S.
Reference- Arun Sharma, Number System,LOD-II page Number-46,
can some one please explain for the solution of below problem.
A set S is formed by indicating some of the first one thosand natural numbers.S contains the maximum number of numbers such that they satisfy the following conditions, No number of the set S is prime and when the numbers of the set S are selected two at a time, we always see co prime numbers.What is the number of elements in the set S.
Reference- Arun Sharma, Number System,LOD-II page Number-46,
hey guys m new to this n m tryin 2 self study for mba mostly snap and nmat so please help me out with these questions and the logic (most importantly)
Progressions
1.The sum of 5 numbers in AP is 30 and the sum of their squares is 220.what is the 3rd term?
2.Find the sum of nos. between 10-50 excluding the number divisible by 8(include 10 and 50)?
3.find the no of terms of the series 1/81,-1/27,1/9,.....,-729 ?
NUMBERS
1.what will be the units digit of (1273)^122! . (in this i knw the concept of cycle and everythin jus have problem understand the factorial part in the power.what is the logic and how do i solve this?)
hey guys m new to this n m tryin 2 self study for mba mostly snap and nmat so please help me out with these questions and the logic (most importantly)
Progressions
1.The sum of 5 numbers in AP is 30 and the sum of their squares is 220.what is the 3rd term?
2.Find the sum of nos. between 10-50 excluding the number divisible by 8(include 10 and 50)?
3.find the no of terms of the series 1/81,-1/27,1/9,.....,-729 ?
NUMBERS
1.what will be the units digit of (1273)^122! . (in this i knw the concept of cycle and everythin jus have problem understand the factorial part in the power.what is the logic and how do i solve this?)
1) the 5 nos in AP are a-2d a-d a a+d a+2d Their sum is 130 So adding =>5a=30 a=6(3rd term)
1.what will be the units digit of (1273)^122! . (in this i knw the concept of cycle and everythin jus have problem understand the factorial part in the power.what is the logic and how do i solve this?)
The answer is The power 122! is divisible by 4 So it can be written as 4x =(1237)^4x 3^1=3 3^2=9 3^3=27 3^4=81
how can u say it s divisible by 4??the ans s the 4th in the cycle ie 1?? i mean its really confusing
try explanin it dept takin this as an eg
2312^123! x 32447^79! thank you..
see anything above 3! is always divisible by 4 n!= 1 * 2 * 3 * 4 * 5 * ... * n so whether 123! or 79!, both are divisible by 4 or can also be said as in 4n form cyclicity of 2 : 2 4 8 6 cyclicity of 7 : 7 9 3 1 so unit digit of 2312^123! will be 6 & of 32447^79! will be 1 so unit digit of the expression : 6*1 = 6
At a particular time in the 21st century there were seven bowlers in the indian cricket team's list of 16 players short listed to play the next world cup.Statisticians discovered that if you look at the wickets taken by any of the 7 bowlers of the current cricket team, the no. of wickets taken by them had a strange property. The numbers were such that for any team selection of 1 players (having 1 to 7 bowlers) by using the number of wickets taken by each bowler and attaching coefficients +1,0 or -1 to each value available and adding resultant values , any number from 1 to 1093, both included could be formed.If we denote W1,W2,W3,W4,W5,W6,W7 as the 7 values in ascending order what could be the answer to the following questions: Find the value of W1+2W2+3W3+4W4+5W5+6W6 a.2005 b.1995 c.1985 d.none
Find the index of the largest power of 3 contained in the product W1W2W3W4W5W6W7 a.15 b.10 c.21 d.6
If the sum of seven coeff is 0, find the smallest number that can be obtained a. -1067 b. -729 c. -1040 d. -1053
in a hurry im solved 2 qns: answer for the first: only when the two digits have a 1 or 0 as one of the digits then sum>product of factorial of nos. also if both digits r same then sum=product. both cases can be subtracted from 90 (no of 2 digit nos.) 1st case: 17+9=26 2nd case: only 1. so total=27. so total nos: 90-27=63. ans: c
what is the case 2 dear??
i think answer should be 63 (numbers will be excluded 10-19,20,21,30,31,40,41,50,51,60,61,70,71,80,81,90,91,22------total 27)
Now please solve this question Find the number of divisors of 1080 excluding the throghout divisors, which are perfect squares.
1080 = 2^3 * 3^3 * 5 so no of divisors = (3+1)(3+1)(1+1) = 32 out of these, squares are : 2^2, 3^2 & (2*3)^2 so excluding squares, no of factors are 32-3 = 29
