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Let a_1,a_2,a_3…… ,a_n be in ...

Updated On: 27-06-2022

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78126126none of these

Solution : 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 <br> a_(1)+a_(n)=66rArra+ar^(n-1)=66 <br> a_(2)xxa_(n-1)=128 <br> rArrarxxar^(n-2)=128 <br> or a^(2)r^(n-1)=128 <br> or axx(ar^(n-1))=128 or ar^(n-1)=128/a <br> Putting this value of ar^(n-1) in (1), we get <br> a+128/a=66 <br> or a^(2)-66a+128=0 <br> or (a-2)(a-64)=0 <br> or a=2,64 <br> Putting a=2 in (1), we get <br> 2+2xxr^(n-1)=66 or r^(n-1)=32 <br> Putting a=64 in (1), we get <br> 64+64r^(n-1)=66 or r^(n-1)=1/32 <br> For an increasing G.P., rgt1. Now, <br> S_(n)=126 <br> rArr2((r^(n)-1)/(r-1))=126 <br> or (r^(n)-1)/(r-1)=63 <br> or (r^(n-1)xxr-1)/(r-1)=63 <br> or (32r-1)/(r-1)=63 <br> or r=2 <br> thereforer^(n-1)=32 <br> rArr2^(n-1)=32=2^(5) <br> rArrn-1=5 <br> rArrn=6 <br> For decreasing G.P., a=64 and r=1/2. Hence, the sum of infiite terms is 64/{1-(1/2)}=128. <br> 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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