Sequence And Series Question 371
Question: If, 8, -4 and 13 be three (not necessarily consecutive term) of an A.P., how many such A.P. s are possible?
Options:
1
2
C) Infinitely many instances
D) No such A.P. is possible
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Since A.P. is either increasing or decreasing, if possible let -4 be the first term of an A.P., whose mth and nth terms are respectively 8 and 13. Then $ 8=-4+(m-1)d $ and $ 13=-4+(n-1)d $
$ \Rightarrow \frac{12}{m-1}=\frac{17}{n-1}=d $ Let $ \frac{m-1}{12}=\frac{n-1}{17}=k, $ then $ m=12,k+1 $ , and $ n=17,k+1 $
$ \therefore $ for $ k=1,2,3,\ldots $ we get different pairs of values of m and n, which shows that infinite number of A.P.s can be obtained
BETA
