An arithmetic progression has 2n terms. The sum of all terms in odd positions is 33 , and the sum of all terms in even positions is 36 . If the first term a , the value n , and the common difference d form a geometric progression in that order, then the value of a + n + d is :

Step-by-step solution

Let the A.P. be a, a+d, a+2d, , a+(2n-1)d . The sum of the n odd-positioned terms is: S_ odd = a + (a+2d) + + (a+(2n-2)d) = n 2 [2a + (n-1)(2d)] = n(a + (n-1)d) = 33 The sum of the n even-positioned terms is: S_ even = (a+d) + (a+3d) + + (a+(2n-1)d) = n 2 [2(a+d) + (n-1)(2d)] = n(a + nd) = 36 Subtracting S_ odd from S_ even gives: S_ even - S_ odd = n(a+nd) - n(a+(n-1)d) = nd = 36 - 33 = 3 So, nd = 3 d = 3 n . Adding S_ odd and S_ even gives the sum of all 2n terms: S_ total = 33 + 36 = 69 2n 2 [2a + (2n-1)d] = 69