A) 1
B) 2
C) Infinitely many
D) No such A.P. is possible
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 \[k=1,2,3.........\] we get different pairs of values of m and n, which shows that infinite number of A.P.?s can be obtained
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