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InterviewSolution
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In a G.P the sum of the first and last terms is 66, the product of the second and the last but one is 126, and the sum of the terms is 128 In any case, the difference of the least and greatest terms isA. 78B. 126C. 126D. none of these |
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Answer» Correct Answer - D Let a be the term and r the common ratio of the given G.P. Further, let there be n terms in the given G.P. Then `a_(1)+a_(n)=66rArra+ar^(n-1)=66` `a_(2)xxa_(n-1)=128` `rArrarxxar^(n-2)=128` or `a^(2)r^(n-1)=128` or `axx(ar^(n-1))=128` or `ar^(n-1)=128/a` Putting this value of `ar^(n-1)` in (1), we get `a+128/a=66` or `a^(2)-66a+128=0` or `(a-2)(a-64)=0` or a=2,64 Putting a=2 in (1), we get `2+2xxr^(n-1)=66` or `r^(n-1)=32` Putting a=64 in (1), we get `64+64r^(n-1)=66` or `r^(n-1)=1/32` For an increasing G.P., `rgt1`. Now, `S_(n)=126` `rArr2((r^(n)-1)/(r-1))=126` or `(r^(n)-1)/(r-1)=63` or `(r^(n-1)xxr-1)/(r-1)=63` or `(32r-1)/(r-1)=63` or r=2 `thereforer^(n-1)=32` `rArr2^(n-1)=32=2^(5)` `rArrn-1=5` `rArrn=6` For decreasing G.P., a=64 and r=1/2. Hence, the sum of infiite terms is 64/{1-(1/2)}=128. For a=2,r=2, terms are 2,4,8,16,32,64. For a=64,r=1/2 terms are 64,32,16,8,4,2. Hence, difference is 62. |
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