Question

The 4th term of an A.P. is 9.The sum of its 6th term and 13th term is 40. What is the expression for the nth term of this A.p.​

Answer 1

Answer:

$\red\bigstar$The expression for the nth term of this AP is an = 2n +1.

SOLUTION

Given That the 4th term of an A.P. is 9.

$\implies$ a + 3d = 9$\longrightarrow$ (1)

Also The  sum of its 6th term and 13th term is 40.

Sixth Term , a₆ = a + 5d

Thirteenth Term , a₁₃ = a + 12d

$\implies$ a + 5d + a + 12d = 40

$\implies$ 2a + 17d = 40 $\longrightarrow$ (2)

$\bigstar$ Let us Multiply (1) by 2

$\implies$ 2a + 6d = 18 $\longrightarrow$ (3)

$\bigstar$ Now, Let us subtract  (3) from (2)

       2a  +  17d  =  40

     (-)2a (-) 6d = ( -)18    

                 11d =  22

∴ Common Difference (d) = 22/11 = 2.

$\bigstar$ Let us substitute d = 2 in (1) to find a

$\implies$ a + 3 × 2 = 9

$\implies$ a + 6 = 9

$\implies$ a = 9 - 6 =3

∴ First Term of AP (a) = 3.

• Now , we have to find the expression for the nth term of this AP.

nth term of an AP ,

$\boxed{an = a + (n - 1) d}$

Here,

a = 3

d = 2

an = 3 + (n-1) 2

an = 3 + 2n - 2

$\blue\bigstar \boxed{\boxed{an = 2n + 1}}$