Question

6. For a positive real number z, the difference 1.-z- is called the fractional part of r. Given arbitrary positive real numbers a and b, state a condition in terms of the fractional parts of and that is necessary and sufficient for la + bl = lal + Ibl Prove that this equation is true if and only if your condition holds. 7. Evaluate 10002+4 23k+5 k- 2 algebraically and simplify as much as possible. You ust show all steps. Leave large powers (which is to mean expressions of the form a with b ะ 1000) un-evaluated in your final answer. Do not give a calculator approxiation. 8. Suppose {a,.) is arithmetic and a3 = 5, a11-87. What is the common difference? Justify your answer

Answer 1

6) The following condition is necessary and sufficient for $\left \lfloor a+b \right \rfloor = \left \lfloor a \right \rfloor + \left \lfloor b \right \rfloor$   is a follows,

$\left \lfloor x \right \rfloor \leq x \leq \left \lfloor x \right \rfloor +1$

where x is a positive real number

7.

000 22k+4 3 23k +5

=

1000 32k × 3 2 2 3 3k 5

=

00으 32k × 81 23k × 32

=

1-2 9-8

=

1000 gk 1000 8k × 32

=$\frac{81}{32} \times \frac{\sum_{k=1}^{1000} (9^{k}-9) }{\sum_{k=1}^{1000} (8^{k}-8)}$

$\sum_{k=1}^{1000} 9^{k}$ and $\sum_{k=1}^{1000} 8^{k}$ are in Geometric sequence,

$\sum_{k=1}^{1000} 9^{k}$ =  
9 x 9(1000-1)
=$9^{(1000)}$

Similarly, $\sum_{k=1}^{1000} 8^{k}$ = $8^{(1000)}$

Inserting the found values in the equation we get,

=$\frac{81}{32} \times \frac{(9^{1000}-9) }{(8^{1000}-8)}$

=$\frac{3^{4}}{2^{5}} \times \frac{(3^{2000}-3^{2}) }{(2^{3000}-2^{3})}$

=$\frac{3^{6}}{2^{8}} \times \frac{(3^{1998}-1) }{(2^{2997}-1)}$

8) Given values are in arithmetic progression(A.P.),

we know that in A.P. the n^th values a_n = a ₀ + (n-1) d

where a₀ is initial value and d is the common difference

a₃ =5= a ₀ + (3-1) d

5 = a ₀ + 2d

a₁₁ =87= a ₀ + (11-1) d

87 =   a ₀ + 10d

Deducting above two equations we get,

$d= \frac{82}{8}$

Common diffrence,$d= \frac{82}{8} =10.25$