Question

The 4th term of an AP is equal to 3 times the first term and the 6th term exceeds twice the 3rd term by 1. find the 1st term and the common difference.​

Answer 1

Let 'a' be the first term & 'd' be the common difference of the AP referred to. Then, 4th term is 3 × 1st term ( as per the problem ). That is, a + 3d = 3a Or a + 3d -a = 3a - a. 7th term is 1 added to twice the third term ( as per the problem ).

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Answer 2

Answer:

Let a1,a2,a3 ,.........,a7 are in a.p. whose common difference is d

a4 = 3a1

a1 + 3d = 3a1

3d = 2a1

a7 = 2a3 + 1

a1 + 6d = 2a1 + 4d + 1

2d = a1 + 1

4a1/3 = a1 + 1

a1 = 3

d = 2