Question

4 term of an ap is thrice the first term and 7 term exceed twice the 3rd time by 1 find first term and CD​

Answer 1

Answer: d = 2 and a = 3.

Step-by-step explanation:

There is an AP, The 4th term is three times the first. So the relation is

3a = a + 3d, or

2a = 3d, or

a = (3/2)d …(1)

The 7th term exceeds twice the third term by 1. The relation is

a+6d = 2(a+2d)+1, or

a + 6d = 2a+4d+1, or

2a-a = 6d-4d-1, or

a = 2d-1 …(2).

Equate (1) and (2)

a = (3/2)d = 2d-1, or

3d = 4d-2, or

4d-3d=2, or d = 2 and a = 3.

The first term is 3 and the common difference is 3.

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Answer 2

I hope it will be answer of your question