| (i) \[\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},....\] |
| (ii) \[\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},....\] |
| (iii) \[{{1}^{2}},{{3}^{2}},{{5}^{2}},{{7}^{2}},.....\] |
| (iv) \[{{1}^{2}},{{5}^{2}},{{7}^{2}},73,.....\] |
A) (i) & (ii)
B) (i) & (iii)
C) (i) & (iv)
D) (ii) & (iv)
Correct Answer: C
Solution :
(c): (i) \[\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32}\],........ Here, \[{{a}_{1}}=\sqrt{2},{{a}_{2}}=\sqrt{8}=2\sqrt{2},{{a}_{3}}=\sqrt{18}=3\sqrt{2},{{a}_{4}}\] \[=\sqrt{32}=4\sqrt{2};\,\,{{a}_{2}}-{{a}_{1}}=2\sqrt{2}-\sqrt{2}=\sqrt{2};\] \[{{a}_{3}}-{{a}_{2}}=3\sqrt{2}-2\sqrt{2}=\sqrt{2};\] \[{{a}_{4}}-{{a}_{3}}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}\] Clearly, difference of successive terms is constant, therefore list of numbers form an AP. So, common difference, \[d=\sqrt{2}\] (ii) \[\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},.....\] Here, \[{{a}_{1}}=\sqrt{3},{{a}_{2}}=\sqrt{6}.{{a}_{3}}=\sqrt{9},{{a}_{4}}=\sqrt{12}\,\,{{a}_{2}}-{{a}_{1}}\] \[=\sqrt{6}-\sqrt{3}=\sqrt{3\times 2}-\sqrt{3}=\sqrt{3}\left( \sqrt{2}-1 \right)\] And \[{{a}_{3}}-{{a}_{2}}=\sqrt{9}-\sqrt{6}=3-\sqrt{3\times 2}=\sqrt{3}\left( \sqrt{3}-\sqrt{2} \right)\] Clearly, \[{{a}_{2}}-{{a}_{1}}\ne {{a}_{3}}-{{a}_{2}}\] Hence, it is not an A.P. (iii) \[{{1}^{2}},{{3}^{2}},{{5}^{2}},{{7}^{2}},......\] Here, \[{{a}_{1}}={{1}^{2}}=1,{{a}_{2}}={{3}^{2}}=9.\,\,{{a}_{3}}={{5}^{2}}=25;\] \[{{a}_{2}}-{{a}_{1}}=9-1=8;\,\,{{a}_{3}}-{{a}_{2}}=25-9=16\] Clearly, \[{{a}_{2}}-{{a}_{1}}\ne {{a}_{3}}-{{a}_{2}}\] Hence, it is not an A.P. (iv) \[{{1}^{2}},{{5}^{2}},{{7}^{2}},73\] Here, \[{{a}_{1}}={{1}^{2}}=1,{{a}_{2}}={{5}^{2}}=25,{{a}_{3}}={{7}^{2}}=49,\] \[{{a}_{4}}=73;\,\,{{a}_{2}}-{{a}_{1}}=25-1=24\]; \[{{a}_{3}}-{{a}_{2}}=49-25=24;\,\,{{a}_{4}}-{{a}_{3}}=73-49=24\] Hence, it is an A.P.
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