Question

Is 101 a term of the arithmetic sequence 13,24,35.....?what about 1001?

Answer 1

Answer:

Step-by-step explanation:

a=13   d=11

101=13+(n-1)11

101=13+11n-11

101=2+11n

99=11n

n=9

since n can be 9, therefore 101 is one of the terms of the given AP

1001=13+(n-1)11

1001=13+11n-11

1001=2+11n

11n=999

n=999/11=90.81...

n can'tbe a fraction

therefore, 1001 is not a term of the given AP

Answer 2

Given,

AP series: 13,24,35.....

To Find,

101 and 1001 are in AP or not =?

Solution,

Common difference = $a_2 -a_1$

Common difference (d) = 24 - 13 = 11

From  the formula of  the nth term in Ap, We know that

$a_n = a + (n-1)d$

For term 101,

$101 = a + (n-1)d$

$101 = 13 + (n-1)11$

$101 - 13 = 11(n-1)$

$88 / 11 = n-1\\n-1 = 8\\n =8+1 = 9$

$a_9$ Term of the AP is 101.

Similarly, for 1001,

$1001 = 13 + (n-1)11$

$1001- 13 = (n-1)11$

$1001- 13 = (n-1)11\\n-1 = 988 /11$

$n-1 = 89.8$

$n = 90.8$

n is in decimal which is not valid. It doesn't exist in the given AP.

Hence, 101 is a term in a given sequence but 1001 is not a term in a given sequence.