Question

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.

Answer 1

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

https://brainly.in/question/14081691

If the nth term of an AP is (2n+5),the sum of first10 terms is

https://brainly.in/question/23676839