Question

the 3rd term of an AP is a and 4th term is b find the 10th term and the general form​

Answer 1

Answer :

10th term , a(10) = 7b - 6a &

General term , a(n) = (n - 3)b - (n - 4)a

Note :

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

★ If a1 , a2 , a3 , . . . , an are in AP , then

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ The common difference of an AP is given by ; d = a(n) - a(n-1) .

★ The nth term of an AP is given by ;

a(n) = a1 + (n - 1)d .

Solution :

• Given : a(3) = a , a(4) = b
• To find : a(10) , a(n) = ?

We know that ,

The nth term of an AP is given by ;

a(n) = a1 + (n-1)d

Thus ,

=> a(3) = a1 + (3-1)d

=> a = a1 + 2d -----------(1)

Also ,

=> a(4) = a1 + (4-1)d

=> b = a1 + 3d ---------(2)

Now ,

Subtracting eq-(1) from eq-(2) , we get ;

=> b - a = (a1 + 3d) - (a1 + 2d)

=> b - a = a1 + 3d - a1 - 2d

=> b - a = d

=> d = b - a

Now ,

Putting d = b - a in eq-(1) , we get ;

=> a = a1 + 2(b - a)

=> a = a1 + 2b - 2a

=> a + 2a - 2b = a1

=> 3a - 2b = a1

=> a1 = 3a - 2b

Now ,

The 10th term of the AP will be given as ;

=> a(10) = a1 + (10-1)d

=> a(10) = a1 + 9d

=> a(10) = 3a - 2b + 9(b - a)

=> a(10) = 3a - 2b + 9b - 9a

=> a(10) = 7b - 6a

Also ,

The general term of the AP will be given as ;

=> a(n) = a1 + (n - 1)d

=> a(n) = 3a - 2b + (n - 1)(b - a)

=> a(n) = 3a - 2b + (n - 1)b - (n - 1)a

=> a(n) = (n - 1)b - 2b - (n - 1)a + 3a

=> a(n) = (n - 1 - 2)b - (n - 1 - 3)a

=> a(n) = (n - 3)b - (n - 4)a

Hence ,

a(10) = 7b - 6a &

a(n) = (n - 3)b - (n - 4)a