the 10th term of an ap with positive terms is 3 times its third term and the product of 3rd and the 10th term is 147 find the AP
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Given, T10 = 3×T3 & T3 × T10 = 147
Taking first term of AP as A and common difference D,
A+9D = 3×(A+2D). Gives, 2A = 3D____(1)
(A+2D) × (A+9D) = 147.
Substituting (1) to this gives,
A = 3 & D = 2.
Thus AP is 3, 5, 7, 9,.....
Taking first term of AP as A and common difference D,
A+9D = 3×(A+2D). Gives, 2A = 3D____(1)
(A+2D) × (A+9D) = 147.
Substituting (1) to this gives,
A = 3 & D = 2.
Thus AP is 3, 5, 7, 9,.....
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