If the number of ways of selecting 3 numbers out of 1, 2, 3,...., 2n+1 such that they are in arithmetic progression is 441, then the sum of the…

Question

If the number of ways of selecting 3 numbers out of 1, 2, 3,...., 2n+1 such that they are in arithmetic progression is 441, then the sum of the divisors of n is equal to

MARKS Content Team · Accepted Answer

Correct answer: (B) 32. If a,b,c are in A.P., then a+c=2b i.e. the sum of two numbers is even ⇒ Both numbers are even or odd Odd numbers →1,3,5,......,2n+1 (total n+1 ) Even numbers →2,4,6,......,2n (total n ) ⇒ Required number of ways =^(n+1)C_2+^nC_2 ⇒^(n+1)C_2+^nC_2=441 ⇒(n+1n)/2+(nn-1)/2=441 ⇒n/2n+1+n-1=441 ⇒n^2=441⇒n=21=3×7 ⇒ Sum of divisors =1+31+7=4×8=32

If the number of ways of selecting $3$ numbers out of $1, 2, 3, . . . ., 2 n + 1$ such that they are in arithmetic progression is $441,$ then the sum of the divisors of $n$ is equal to

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