Progression - 5 · KnowledgeBin.org

PROBLEMS (MEDIUM TYPE)

1) A sequence a₁, a₂, a₃, ------- a_n of real numbers is such that a₁= 0, |a₂| = |a₁+1|, |a₃| = |a₂+1|, ---------- |a_n| = |a_n-1+1|. Prove that arithmetic mean of a₁, a₂, -------- a_n can not be less than -1/2.

Solution:

Let us add one more number a_n+1 to the given sequence. The number a_n+1 is such that |a_n+1| = |a_n+1|.

Squaring all numbers we have,

Adding the above equalities we get,

Dumb Question:

1) Why ?

Ans: a_n+1 is given to be a real no in the question and hence is going to be a positive number.

There fore .

2) There are (4n+1) terms in a certain sequence of which the first (2n+1) terms form an A.P of common difference 2 and the last (2n+1) terms are in G.P of common ratio 1/2. If the middle terms of both A.P and G.P be the same what is the mid term of this sequence.

Solution: d=2, r=1/2

If there be odd number of terms then mid term = .

T_2n+1 is the mid term of sequence of (4n+1) odd terms a+2nd = a+4n ------ (1)

This mid term is the last term of A.P and first term of following G.P. Each of (2n+1) terms with this term being common to both.

T_n+1 and t_n+1 are mid terms of A.P and G.P

T_n+1 = a+nd = a+2n.

By given condition T_n+1= t_n+1.

Hence mid term of sequence by (1) is

Dumb Question:

1) Why middle term of the G.P is ?

Ans: Well the first term of the G.P is the last term of the A.P

So, the first term = a+2(2n+1-1)

= a+4n.

Now the (n+1)^th term of the G.P is

3) If then find .

Solution:

Putting n = 1, 2, 3, ------n

4) Evaluate sum of n terms of the series

Solution:

5) Find the sum of

Solution:

If S be the sum then consider

6) If 2a+b+3c = 1 and a>0, b>0, c>0 find the greatest value of and obtain the corresponding values of a, b, c.

Solution:

Consider the positive numbers {As there is a⁴, take 4 equal parts of 2a; as there is b² take 2 equal parts of b; as there is c² take 2 equal parts of 3c}.