If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is:
❓ Question If the sum of the 2nd, 4th and 6th terms of a G.P. of positive terms is 21 and the sum of its 8th, 10th and 12th terms is 15309 , then find the sum of its first nine terms . 🖼️ Question Image ✍️ Short Solution Let the G.P. be a , a r , a r 2 , … a,\,ar,\,ar^{2},\dots with a > 0 , r > 0 a>0,\; r>0 Write the given sums: Sum of 2nd, 4th, 6th terms: S 1 = a r + a r 3 + a r 5 = a r ( 1 + r 2 + r 4 ) = 21. Sum of 8th, 10th, 12th terms: S 2 = a r 7 + a r 9 + a r 11 = a r 7 ( 1 + r 2 + r 4 ) = 15309. Divide S 2 S_2 by S 1 S_1 to eliminate the common factor ( 1 + r 2 + r 4 ) and a r a r : S 2 S 1 = a r 7 ( 1 + r 2 + r 4 ) a r ( 1 + r 2 + r 4 ) = r 6 = 15309 21 . Compute the right-hand side: 15309 21 = 729. So r 6 = 729 = 3 6 r^{6} = 729 = 3^{6} . Since r > 0 r>0 , we get r = 3. Now substitute r = 3 r=3 back into S 1 S_1 : a r ( 1 + r 2 + r 4 ) = a r ( 1 + 9 + 81 ) = a r ⋅ 91 = 21. Hence a r = 21 91...