Question

5) The fifth term of an A.P exceeds twice the second term by 1. The tenth term exceeds twice the fourth term by 3 Find (i) the first term. (ii)common difference (iii)the sum of first twenty-five terms.

Answer 1

Given : The 5th term of an A.P. exceeds twice the 2nd term by 1. The 10th term exceeds twice the 4

th term by 3.

To Find :  the first term and the common of the A.P.

Solution:

First term = a

Common difference = d

AP  a  , a + d , a + 2d , . . .

nth term = a + (n - 1) d

2nd term = a + d

5th term = a + 4d

a + 4d   = 2(a + d)  + 1

=> a + 4d = 2a + 2d + 1

=> 2d - a  = 1

1oth term = a + 9d

4th term = a + 3d

a + 9d =  2(a + 3d) + 3

=> a + 9d = 2a + 6d + 3

=> 3d  - a  =  3

2d - a  = 1

3d  - a  =  3

=> d = 2

a = 3

First term = 3

common difference = 2

AP

3  5  7  9  11  13 and so on

Answer 2

Answer:

Xn = a + (n-1)d

X5 = a + 4 d

X5 - 1 = 2 [ a + d ]

X10 - 3 = 2 [ a + 3 d ]

but X10 = a+9d.

So,

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

a = 2 d - 1.

a + 9 d - 3 = 2 a + 6 d

a = 3 d -3.

combine

3 d - 3 = 2 d - 1

d = 4.

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