Question

If 100 times the 100th term of an AP with non zero common difference equals the 50 times its 50th term, then the 150th term of this AP is :
(a) – 150
(b) 150 times its 50th term
(c) 150
(d) Zero

Answer 1

Answer:

0

Step-by-step explanation:

a = 100 and 50

d = 100-1 = 99

d = 50-1 = 49

The general term of an A.P. -

Tn = a+ ( n-1 )d

wherein  a is the first term  n is the number of term  and d is the common difference

Thus,

100(a +99d)= 50(a + 49d)  (Given)

100a + 9900d = 50a + 2450d  

50a + 7450d = 0

a + 149d = 0  

So 150 th term is 0

Thus the 150th term of the given AP is 0

Answer 2

Solution:

Let a and d are first term and

common difference of an a.p

$\boxed { n^{th}\: term = a+(n-1)d}$

i ) 100th term = $a_{100}=a+(100-1)d$

= $a+99d$ ----(1)

ii) 50th term = $a_{50}=a+49d$ ----(2)

iii) 150th term = $a_{150}=a+149d$----(3)

Now ,

According to the problem given,

According to the problem given,100 times the 100th term of an AP with non zero common difference equals the 50 times its 50th term

100[a+99d]=50[a+49d]

/* from (1)&(2) */

Divide both sides by 50 ,we get

=> 2(a+99d) = a+49d

=> 2a+198d-a-49d = 0

=> a + 149d = 0

$\implies a_{150}=0$

/* from (3) */

Therefore,

Option (d) is correct.

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