The sum of the even numbers between 1 and k is 79*80

The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159 






Sn=nA1+[n(n-1)d]/2

n=(k-1)/2, A1=2, d=2 ===>

2n+n(n-1)=n(n+1) ===>

[(k-1)/2]*[(k-1+2)/2]=(k-1)*(k+1)/4 ====>

(k-1)*(k+1)=79*2*80*2 ===>

k=159

Mark bought a set of 6 flower pots of different sizes at a total cost

Mark bought a set of 6 flower pots of different sizes at a total cost of 8.25. Each pot cost 0.25 more than the next one below in size. What is the cost of the largest pot?

8.25 = 6/2 (2*x + (6-1) * 0.25)

X = ¾

Now Largest Post price = ¾ + 1.25 = 2

Answer: $2.00

240,120,60,30......
In the sequence above, each term after the first is 1/2 of the preceding term. What is the least term of the squence that is greater than 1?



A) 32/15
B) 16/15
C) 15/8
D) 15/4
E) 15/2






240,120,60,30, 15, 15/2, 15/4, 15/8,

The next term cud be 15/16 but the denominator gets bigger than the numerator so 15/16 is less than 1. So 15/8 is the last # in sequence >1.

The sum of first 50 even numbers is 2550 .What is the sum of even numbers from 102 to 200 inclusive?

1)5100
2)7550
3)10100
4)15500
5)20100





One can apply the formula for Sum of series that are in arithmetic progression:
S = n/2 ( 2a + [n-1] *d)

n = 50, a = 102, d= 2....

or

We can break down the problem as

102 + 104 + 106 + ... + 200
=
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 100)
=
(100 + 100 + 100 + ... + 100) + (2 + 4 + 6 + ... + 100)
=
(fifty 100's) + (2550 as given)
=
5000 + 2550
=
7550

When a certain tree was first planted, it was 4 feet tall, and the heigth of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?


Answers: 3/10, 2/5, 1/2, 2/3, 6/5


Yr0 = 4
Yr1 = 4+x
Yr2 = 4+x+x=4+2x
Yr3 = 4+x+x+x=4+3x
Yr4 = 4+x+x+x+x=4+4x
Yr5 = 4+x+x+x+x+x=4+5x
Yr6 = 4+x+x+x+x+x+x=4+6x

We are told the amount at the end of Year 6 is 6/5 of the amount at the end of year 4. Thus we can write:

4+6x = 6/5 (4+4x)
5(4+6x) = 6(4+4x)
20+30x = 24+24x
6x=4
x=2/3

What is the first term of an arithmetic progression of positive integers ?

a)Sum of the squares of the first and second term is 116.
b)The seventh term is divisible by 10.




Explanation - let x be the first and y be the second term.
Hence 0 less than x less than y

From Statement 1: x^2 + y^2 = 116 only when x=2 and y=10.
Hence the first term is 4. This is the only combination that works - hence sufficient

Statement 2: let 10z be the 7th term (z is an integer)
Thus x+6(y-x)=10z => y=(10z+5x)/6. Hence the last digit of the numerator can be 5 or 0. Now y is also an integer => x must be even => least possible value of y =10.
Assuming different values of z we get
z=1; x=12 This is impossible (x should be less than y)
z=2; x=8; 8 10 12 14 16 18 20
z=3; x=6; 6 10 14 18 22 26 30 and so on -- hence insufficient

Series of A(n) is such that A(n) = A(n-1) / n. How many elements of the series are bigger than 1/2 ?

(1) A(2) = 5
(2) A(1) - A(2) = 5



From statement (1) --- A(2) = 5 . It is given that A(n) = A(n-1) / n.
Hence A(2) = A(1) / 2
=> A(1) = 10
A(2) = 5
=A(3) = A(2) / 3 = 5 /3
A(4) = A(3) / 4 = 5 /( 3 * 4) = 5 /12

Hence statement (1) alone is sufficient to aswer the question.

From statement (2) --- A(1) - A(2) = 5
=> A(1) - A(1) /2 = 5
=> A(1) = 10
Hence A(2) = 5
A(3) = A(2) / 3 = 5 /3 ...

Hence statement (2) alone is sufficient to answer the question.

What is the 999th term of the series S ?

(1) The first 4 four terms of S are (1 + 1)² , (2 + 1)² , (3 + 1)² , and (4 + 1)².
(2) For every x, the xth term of S is (x + 1)².



From statement (1) - It is given that the 1st 4 four terms of S are (1 + 1)² , (2 + 1)² , (3 + 1)² , and (4 + 1)².

We only know the first four terms. We cannot assume that terms following the first 4 terms will be like -- nth term

= (n+1)^2.
There can be a sequence that follows a different rule for any term in it

From statement (2) - It is given that for every x, the xth term of S is (x + 1)². Hence sufficient as it implies

"for any number x that refers to the term in S."