Q.
If the fourth term of an arithmetic progression is 6, the mth term is 18 and if the A.P. has integral terms only, then the number of such A.P.s is
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answer is 9.
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Detailed Solution
Given that T4 =a+3d=6 and Tm=a+(m-1)d=18 (m-4)d=12 Since terms are integers, we have m-4=±1,±2,±3,±4,±6,±12 but m>0 m-4=±1,±2,±3,4,6,12. So, m has nine possible values
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