Given arithmetic progression of positive integers terms a1​,a2​,a3​,….. having common difference ' 2 ' and geometric progression of positive integers terms b1​,b2​,b3​,…. having common ratio ' 2 ' with a1​=b1​=c, such that2(a1​+a2​+a3​+…+an​)=b1​+b2​+b3​+…+bn​⇒2×2n​[2C+(n−1)2]=C(2−12n−1​)⇒2nC+2n2−2n=2n.C−C⇒C[2n−2n−1]=2n2−2n∵C∈N⇒2n2−2n≥22n−2n−1​⇒ and, also ​2n2+1≥2n⇒n≤6C>0⇒n>2​∴ The possible values of n are 3,4,5,6So, at n=3,C=8−6−1(2×9)−6​=12at, n=4,C=16−8−132−8​=924​=38​∈/Nat, n=5,C=32−10−150−10​=2140​∈/Nand at, n=6,C=64−12−172−12​=5160​∈/N∴ The required value of C=12 for n=3.