1.

Find some sequence which have no algebraic form

Answer»

Step-by-step explanation:

Given arithmetic sequence is 9, 15, 21,.

First term = a = 9

Common difference = d =15 9 = 6

(A) For natural number N

nth term = [(n−1)X6]+9=6n+3 where n=1,2,3,

Hence algebraic form of the sequence 9,15,21,. is x

n

=6n+3.

(B) nth term =6n+3

25

th

term =6(25)+3=150+3=153

(C) Sum of first n terms is given by,

S

n

=

2

1

n[X

1

+X

n

]

Sum of 24 terms is

S

24

=

2

1

n[X

1

+X

2

4]=

2

25

[9+147]=1950

Sum of 50 terms is

S

50

=

2

1

n[X

1

+X

50

]=

2

50

[9+6(50)+3]=7800

Sum of terms from twenty FIFTH to fiftieth

=S

50

−S

24

=7800−1950

=5850

(D) Let sum of first n terms is 2015,

S

n

=

2

1

n[X

1

+X

n

]

2015=

2

1

n[9+6n+3]

2015=3n

2

+6n

3n

2

+6n−2015=0

Solving above equation for n, the value of n is not a natural number.

Hence 2015 can not be the sum of some terms of this sequence.


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