Question

5,8,11,14......an A.P. if so then what will be the 100th term ? check whether 92 is in this A.P. is number 61 in this A.P.

Answer 1

a=5(first term)
d=3(common difference)
n=100(nth term)
an=?
an=a+(n-1)d
a100=5+(100-1)3
a100=5+297
a100=302
Therefore 100th term is 302
an=92
92=5+(n-1)3
87=(n-1)3
29=(n-1)
n=30
The number 92 is in this A.P
an=61
61=5+(n-1)3
56=(n-1)3
56/3=n-1
56/3+1=n
n=59/3
The number 61 is not present in this A.P because the number shouldn't be in fraction.

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Answer 2

Answer:

The given sequence is  5, 8,11,14,…..  

Here, a1 = 5, a2 = 8, a3 = 11, a4 = 14  

Therefore, a2 – a1 = 8 – 5 = 3 ,

a3 – a2 = 11 – 8 = 3  

a4 – a3 = 14 – 11 = 3

a2 – a1 = a3 – a2 = a4 – a3 = 3 = d = constant

The difference between two consecutive terms is constant.

The common difference of the series is 3.

Therefore, the given sequence is an A.P(Arithmetic progression)

an = a + (n – 1)d .

Therefore,

$a_{100}$ = 5 + (100 – 1)3   since a=5, d=3.  

          = 5 + 99 × 3 = 5 + 297  

          = 302  

Hence, the 100th term of the given A.P. is 302.

To check whether 92 is in given A.P

let an = 92  

Therefore,  an = a + (n – 1)d

                   92 = 5 + (n – 1)3  

                   92 = 5 + 3n – 3

                   92 = 2 + 3n

                   90 = 3n  

                   n = 90/3 = 30  

92 is the 30th term of given A.P.

To check whether 61 is in given A.P

let an = 61

61 = 5 + (n – 1)3  

61 = 5 + 3n – 3  

61 = 2 + 3n  

61 – 2 = 3n  

59 = 3n

n = 59/3

But, n is natural number 59  

Therefore, n ≠ 59/3

Hence, 61 is not in the given A.P.