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The 4th term of an A.P. is equal to 3 times of the first term and the 7th term exceeds the 3rd term by 1. Find the first term and common difference. |
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Answer» GIVEN :- The 4th term of an A.P. is equal to 3 times of the first term and the 7th term EXCEEDS the 3rd term by 1. Solution :- Let first term is a and common difference is d . so, → 4th term = 3a → a + (4 - 1)d = 3a → a + 3D = 3a → 3d = 3a - a → 3d = 2a -------- Eqn.(1) and, → T(7) - T(3) = 1 → (a + 6d) - (a + 2d) = 1 → a - a + 6d - 2d = 1 → 4d = 1 → d = (1/4) PUTTING value of d in Eqn.(1) → 3 * (1/4) = 2a → 2a = 3/4 → a = (3/8) therefore, first term of given AP is (3/8) and common difference is (1/4) . Learn more :- EVALUATE the expression given by 83 - 81 + 87 - 85 +__________ + 395 - 393 + 399 - 397 If the nth term of an AP is (2n+5),the sum of first10 terms is |
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