in an increasing GP , the sum of the first and the last term is 66 , the product of the second and the last but one term is 128 , and the sum of all the terms is 126. how many terms are there in the progression.

Let the G.P. are a,ar,ar2......upto n terms, where r>1So,T1+Tn=66a+arn-1=66         .....1And,T2Tn-1=128ararn-2=128a2rn-1=128arn-1=128a         .....2And,T1+....+Tn=126arn-1r-1=126         .....1Substitute value of arn-1 from equation2 in equation 1,a+128a=66a2-66a+128=0a-2a-64=0a=2,66In equation 1,When a=2, 2+2rn-1=66rn-1=32       ......4Substitute rn-1=32 and a=2 in equation 3, we have,ar×rn-1-1r-1=1262×32r-1r-1=126232r-1=126r-1r=2put r=2 in equation 4,2n-1=322n-1=25n-1=5n=6In equation 1,When a=64, 64+64rn-1=66rn-1=132not possible, because r>1Hence, In given G.P., there are 6 terms.

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