Question

It's an arithmetic progression problem. Please help solve this.

Answer 1

Let the first term be a.

Let the common difference be d.

We know that sum of n terms an = a + (n - 1) * d.   ----- (1).

Given that sum of 3rd and 7th term = 6

a + 2d + a + 6d = 6

2a + 8d = 6

a + 4d = 3

a = 3 - 4d   ------ (2)


Given that product of 3rd and 7th term = 6

(a + 2d) * (a + 6d) = 8

a^2 + 6ad + 2ad + 12d^2 = 8

a^2 + 8ad + 12d^2 = 8   ------- (2)


Substitute (1) in (2), we get

We know that (a - b)^2 = a^2 + b^2 - 2ab.

(3 - 4d)^2 + 8(3 - 4d)(d) + 12d^2 = 8

9 + 16d^2 - 24d +  24d - 32d^2 + 12d^2 = 8

9 - 4d^2 = 8

4d^2 = 1

d^2 = 1/4

d = 1/2,-1/2.


Substitute d = 1/2 in (1), we get

a = 3 - 4(1/2)

   = 3 - 2

   = 1


Substitute d = -1/2 in (1), we get

a = 3 - 4(-1/2)

   = 3 + 2

   = 5.



When d = 1/2, a = 1 Then,

The sum of first 20 terms of an AP = (n/2)(2a + (n - 1) * d)

                                                          = (20/2)(2 * 1 + (20 - 1) * 1/2)

                                                          = (10)(2 + 19/2)

                                                          = (10) $\frac{2 * 2 + 19}{2}$

                                                          = $(10)( \frac{23}{2})$

                                                          = 115.


When d = -1/2, a = 5.

Then the sum of first 20 terms of an AP = (n/2)(2a + (n - 1) * d)

                                                                  = (20/2)(2 * 5 + (20 - 1) * (-1/2))

                                                                  = (10)(10 - 19/2)

                                                                  = 10/2

                                                                  = 5.


Therefore the sum of first 20 terms of an AP is 115 (or) 5.


Hope this helps!