In each of the following arithmetic sequences, find () the 100th term; (i) the nth term

Question

a. 0, -5, 10,..

b. 1-√2, 1+2√2 ,.....

c. √3 +0.5, √3+3.5, √3+6.5,....

Answer 1

a)

0, -5, 10, ........

t1 = a = 0

t2 = -5

t3 = 10

Common difference d = t2 - t1 or t3 - t2

t2 - t1  =  -5 - 0 = -5

t3 - t2  =  10 - (-5)  =  10 + 5  =  15

From the above calculations it is clear that the common difference is not equal

Hence the series is not in AP.

b)

 1-√2, 1+2√2 ,........

t1  =  a  =  1-√2

t2  =  1+2√2

d  =  t2 - t1

    =  1+2√2  -  (1-√2)

    =  1 + 2√2  - 1 + √2)

d  =  3√2

nth term tn = a + (n-1)d --------------> (1)

Substitute a  =  1-√2 and d  =  3√2 in Eq (1)

tn  =   1 - √2 + (n - 1)(3√2)

     =   1 - √2 + 3n√2 - 3√2

tn  =   1 - 4√2 + 3n√2

Substitute n = 100 in above equation

t100  =   1 - 4√2 + 3(100)√2

         =   1 - 4√2 + 300√2

t100  =   1 + 296√2

c)

√3 +0.5, √3+3.5, √3+6.5,....

t1  =  a  = √3 +0.5

t2  =  √3+3.5

d  =  t2 - t1

    =  √3 + 3.5 - (√3 + 0.5)

    =  √3 + 3.5  -  √3 - 0.5

d  =  3

nth term tn = a + (n-1)d --------------> (2)

Substitute  a  = √3 +0.5 and d  =  3 in Eq (2)

tn  =  √3 +0.5 + (n - 1)(3)

     =  √3 +0.5 + 3n- 3  

tn  =  √3 - 2.5 + 3n

Substitute n = 100 in above equation

t100  =   √3 - 2.5 + 3(100)

         =   √3 - 2.5 + 300

t100  =   √3 + 297.5

Answer :

a) The series is not in AP..

b)  tn  =   1 - 4√2 + 3n√2

    t100  =   1 + 296√2

c)  tn  =  √3 - 2.5 + 3n

     t100  =   √3 + 297.5