A) 65
B) 75
C) 55
D) 45
Correct Answer: A
Solution :
[a] Let the 4 terms in A.P are \[a-3d,\,\,a-d,\,\,a+d,\,\,a+3d\] According to question \[a-d+a+d=110\] ... (1) \[(a-3d)\,\,(a+3d)=2125\] ? (2) From equation (1) \[a-\,d+a+d=110\] \[2a=110\Rightarrow a=55\] From equation (2) \[(a-3d)\,\,(a+3d)=2125\] \[\Rightarrow \,\,\,{{a}^{2}}-9{{d}^{2}}=2125\] \[\Rightarrow \,\,\,{{(55)}^{2}}-\,9{{d}^{2}}=2125\] \[\Rightarrow 3025-\,9{{d}^{2}}=2125\] \[\Rightarrow 900=9{{d}^{2}}\Rightarrow 2125\] \[\Rightarrow 900=9{{d}^{2}}\Rightarrow {{d}^{2}}=100\Rightarrow d=10\] \[\therefore \,\,\,a=55,\,\,d=+10\] Series would be: 25, 45, 65, 85 .IIIrd term would be 65.
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