Question

find the sum of the infinitely decreasing gp whose third term ,three times the product of the first and fourth term and second term form an ap in the indicated order with common difference equal to 1/8

Answer 1

Let G.P. be a, ar, ar2, ar3, ........According to question:First term of A.P.=ar2Second term of A.P.=3a.ar3=3a2r3Third term of A.P.=arCommon difference(d)=18 Second term=First term+Third term2⇒3a2r3=ar2+ar2⇒6a2r3=ar2+arDividing each term by ar, we get,6ar2=r+1⇒a=r+16r2Now Third term−First term=2d⇒ar−ar2=2.18⇒(r+16r2)r−(r+16r2)r2=14⇒r+16r−(r+16)=14⇒r+1r−(r+1)=64=32⇒(r+1)(1r−1)=32⇒(1+r)(1−r)=3r2⇒2(1−r2)=3r⇒2−2r2=3r⇒2r2+3r−2=0⇒2r2+4r−r−2=0⇒2r(r+2)−1(r+2)=0⇒(r+2)(2r−1)=0⇒r=−2 or r=12But r is decreasing G.P.⇒r=12Now a=r+16r2⇒a=12+16(12)2⇒a=3264⇒a=1212=1Sum of infinite terms of G.P.=a1−r=11−12=112=2

⇒Sum of infinite terms of given G.P.=2 (Answer)

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