Which of the following sequences are in G.P.? (vii) 16, 4, 1, 1/4,..........

Which of the following sequences are in G.P.?

(vii)  16, 4, 1, 1/4,..........

Solution :

First term (a)  = 16, common ratio

r  = t2/t1   =  4/16 r  = 1/4

r  = t2/t1  r  = 1/4

Since the common ratios are same, it is geometric sequence.

Which term of an A.P. 16, 11, 6, 1,... is -54 ?

Which term of an A.P. 16, 11, 6, 1,... is -54 ?

Solution :

Let -54 be the nth term  of the sequence

tn  = -54

a + (n - 1)d  = -54

a  = 16 , d = 11 - 16  ==> -5

16 + (n - 1)(-5)  = -54

(n - 1)(-5)  = -54 - 16

 (n - 1)(-5)  = -70

 n - 1  = -70/(-5)  ==> 14

n  = 14 + 1 

n  = 15

Hence 15th term of the given sequence is -54.

Check whether the following sequences are in A.P. (v) 1, -1, 1, -1, 1, -1,............

Check whether the following sequences are in A.P.

(v) 1, -1, 1, -1, 1, -1,............

Solution :

d = t2 - t1 d = -1 - 1 d  = -2

d = t3 - t2 d = 1 - (-1) =  1 + 1   =  2

The given sequence is not arithmetic progression.

Find the next three terms of the following sequence. (ii) 5, 1,-3,…

Find the next three terms of the following sequence.

(ii) 5, 1,-3,…

Solution :

Pattern :

By subtracting 4 from the 1st term, we get second term. By subtracting 4 from 2nd term, we get 3rd term.

4th term  = -3 - 4  = -7

5th term  = -7 - 4  = -11

6th term  = -11 - 4  = -15

Hence the next three terms are -7, -11, -15.

Let f : A -> B be a function defined by f (x) = (x/2) - 1 where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9} .

Let f : A -> B be a function defined by f (x) =  (x/2) - 1 where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9} .

Represent f by

(i) set of ordered pairs; (ii) a table; (iii) an arrow diagram; (iv) a graph

Solution :

(i) set of ordered pairs

f (x) =  (x/2) - 1

Set of ordered pairs  = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5)}

(ii) a table

(iii) an arrow diagram;

(iv) a graph