A) 4
B) 3
C) 2
D) 1
Correct Answer: D
Solution :
Let \[{{A}_{1}},{{A}_{2}},{{A}_{3}}\] and \[{{A}_{4}}\] are four numbers in A.P. \[{{A}_{1}}+{{A}_{4}}=8\] ?..(i) and \[{{A}_{2}}.\,{{A}_{3}}=15\] ?..(ii) The sum of terms equidistant from the beginning and end is constant and is equal to sum of first and last terms. Hence, \[{{A}_{2}}+{{A}_{3}}={{A}_{1}}+{{A}_{4}}=8\] ?..(iii) From (ii) and (iii), \[{{A}_{2}}+\frac{15}{{{A}_{2}}}=8\] Þ \[A_{2}^{2}-8{{A}_{2}}+15=0\] \[{{A}_{2}}=3\,\,\text{or}\,\,5\] and \[{{A}_{3}}=5\,\,\,\text{or}\,\,\text{3}\]. As we know, \[{{A}_{2}}=\frac{{{A}_{1}}+{{A}_{3}}}{2}\] Þ \[{{A}_{1}}=2{{A}_{2}}-{{A}_{3}}\] Þ \[{{A}_{1}}=2\times 3-5=1\] and \[{{A}_{4}}=8-{{A}_{1}}=7\] Hence the series is, 1, 3, 5, 7. So that least number of series is 1.
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