Question

Find the sum of first 40 terms of the A. P. whose 4 th term is 8 and 6 th term is 14.​

Answer 1

Answer:

The required sum of first 40 terms of the A.P. is 2300

Step-by-step explanation:

Given :

In an A.P.,

• 4th term = 8
• 6th term = 14

To find :

sum of first 40 terms of given A.P.

Solution :

Let 'a' be the first term and 'd' be the common difference.

nth term of an A.P. is given by,

$\underline{\boxed{\bf a_n=a+(n-1)d}}$

4th term :

Put n = 4,

a₄ = a + (4 - 1)d

8 = a + 3d --[1]

6th term :

Put n = 6,

a₆ = a + (6 - 1)d

14 = a + 5d --[2]

Subtract equation [1] from equation [2],

14 - 8 = a + 5d - (a + 3d)

 6 = a + 5d - a - 3d

 6 = 2d

 d = 6/2

 d = 3

∴ Common difference = 3

Substitute d = 3 in equation [1],

8 = a + 3d

8 = a + 3(3)

8 = a + 9

a = 8 - 9

a = -1

∴ First term = -1

Sum of first n terms is given by,

 $\underline{\boxed{\bf S_n=\dfrac{n}{2}[2a+(n-1)d]}}$

Put n = 40,

 $\rm S_{40} = \dfrac{40}{2}[2(-1)+(40-1)(3)] \\\\ \rm S_{40} =20[-2+39(3)] \\\\ \rm S_{40} =20[-2+117] \\\\ \rm S_{40}=20[115] \\\\ \rm S_{40}=2300$

Therefore, the sum of first 40 terms = 2300