The sum up to 10 terms of the series 1.3 + 5.7 + 9.11 + ... is ____________

The sum up to 10 terms of the series 1.3 + 5.7 + 9.11 + ... is ____________

We see that the series is a product of two different APs

1,5,9,.... is one of the AP

3,7,11,.... is another AP

Finding the general term of the first AP.

a = 1, d = 4

1+(n-1)*4 = 4n-3

Similarly, finding the general term of the second AP.

a = 3, d = 4

3 + (n-1)*4 = 4n-1

Therefore, the nth term of the given series

T = (4n-3)*(4n-1)

T = $$16n^2-16n+3$$

Therefore, the sum of the series when the general term is given will be

S = $$16n^2-16n+3$$

S = $$\Sigma\ 16n^2-\Sigma\ 16n+\Sigma\ 3$$

S = $$\dfrac{16n\left(n+1\right)\left(2n+1\right)}{6}-\dfrac{16n\left(n+1\right)}{2}+3n$$

We know the sum of n terms of the square of n terms is $$\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Also, the sum of the first n terms is $$\dfrac{n\left(n+1\right)}{2}$$

S = $$\dfrac{16n\left(n+1\right)\left(2n+1\right)}{6}-\dfrac{16n\left(n+1\right)}{2}+3n$$

We need to find the sum of the first 10 terms. Thus, n = 10

S = $$\dfrac{16\times10\times\ 11\times\ 21}{6}-\dfrac{16\times\ 10\times\ 11}{2}+3\times\ 10$$

S = 6160-880+30 = 5310