Question

Find the sum of first n terms of the AP 4-1/n, 4-2/n, 4-3/n, ....​

Answer 1

Step-by-step explanation:

We have to find:

The sum of 4-1/n, 4-2/n, 4-3/n = ?

solution:

we know that :

1+2+3+4+5+6+. n = n(n+1)/2

and

1+1+1+1+1+. + n = n

Here,

sum of 4-1/n, 4-2/n, 4-3/n up to the nth term

= (4+4+4+4+4+. upto n terms) + (-1/n - 2/n - 3/n-. upto n terms)

= 4 (1+1+1+1. upto n terms) - 1/n (1+2 +3+4. upto n terms)

= 4 n - 1/n x n(n+1)/2

= 4n-(n+1)/2

= [8n-(n+1) ]/2. taking L.C.M

Answer = (7n - 1)/2

$\huge{ son \: prodigal}$