Question

the 4th term of an AP is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. find AP

Answer 1

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Answer 2

AP is 3, 5, 7, 9, ...

Step-by-step explanation:

Given:

The 4th term of an AP is equal to 3 times the first term,

$a_4 =3(a)$                 [1]

$a_4 = a+(4-1)d$                 [∵ $a_n=a+(n-1)d$]

3a = a + 3d                                       [from 1 ]

3d = 3a - a

3d = 2a

$d = \frac{2a}{3}$              [2]

Also,

The 7th term exceeds twice the 3rd term by 1,

$a_7 - 2(a_3) = 1$

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

a + 6d - [2 (a + 2d)] = 1

a + 6d - [2a + 4d] = 1

a + 6d -2a -4d =1

-a + 2d =1

a = 2d -1                                      [3]

Putting value of a in eq 2, we get

$d = \frac{2(2d-1)}{3}$

$d = \frac{4d-2)}{3}$

3d = 4d-2

3d - 4d = -2

-d = -2

d =2

Now putting value of d in eq 3, we get

a = 2(2) -1

a = 4 - 1

a = 3

Now AP is,

a, a+d, a+2d, a+3d, ...

⇒ 3, 3+2, 3+2(2), 3+3(2), ...

⇒ 3, 5, 7, 9, ...

Here AP is 3, 5, 7, 9, ...