Question

the fourth term of an ap is 3 times the first and the 7th term exceeds twice the third number by 1 find AP

Answer 1

please mark it as brainliest

Answer 2

Hi

Here is ur answer

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a4 = 3(a1)

a + (4-1)d = 3( a)      { since, a1 = a}

a + 3d = 3a

2a - 3d = 0 ---------- (1)

a7 = 2(a3) + 1

a + (7-1)d = 2( a + (3-1)d) + 1

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

a - 2d = -1 ------------ (2) ( dividing it by 2 for solving .... 2a - 4d = -2 )

Therefore, by solving equation 1 and 2

we get, ( elimination method)

hence,

d = 2

Then,  substitute d in any equation....

( I am substituting in equation 1)

2a - 3 (2) = 0

2a - 6 = 0

2a = 6

a = 3

Therefore , the AP are

3(a),5(a + d = 3 + 2),7(a+2d = 3 + 2(2)),9( a +3d),11(a +4d),13(a+5d)....and so on

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Hope it helps U