Question

The 4th term of the ap is 3 times the first term and 7th term exceeds the twice of the third term by 1. Find its nth term

Answer 1

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Tn = a + ( n - 1 ).d

Now , according to question

T4 = 3 × T1 ....... (1)

T1 = a + ( 1 - 1 ) d
T1 = a

T4 = a + ( 4 - 1 ) d
T4 = a + 3d

From equation 1.

T4 = 3 × T1
a + 3d = 3 × a
a + 3d = 3a
2a = 3d
a = 3/2 . d

Now ,
T7 = 2 × T3 + 1 ....... (2)

T7 = a + ( 7 - 1 ) d
T7 = a + 6d

T3 = a + ( 3 - 1 ) d
T3 = a + 2d

From equation (2)...

T7 = 2 × T3 + 1
a + 6d = 2 × ( a + 2d ) + 1
a + 6d = 2a + 4d + 1
a = 2d - 1

Comparing the value of a

a = 3/2 . d .... and .... a = 2d - 1

3/2.d = 2d - 1
3/2.d - 2d = - 1
( 3d - 4d )/ 2 = - 1
- d / 2 = - 1
d = 2

:. a = 3/2.d
a = 3/2 × 2
a = 3

:. Tn = a + ( n - 1 ) d
= 3 + ( n - 1 ) 2
= 3 + 2n - 2
= 1 + 2n

Tn = 1 + 2n
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