Question

If a,b,and 2a in ap show that 3ab/2(b-a)​

Answer 1

Answer:

$S_n = \frac{3ab}{2 (b-a)}$

Step-by-step explanation:

The correct question is: If the first, second and last term of an AP are a,b and 2a respectively, then show that the sum of all terms of an AP is 3ab/2(b-a)​.

Firstly, as we know that the nth term of an A.P. is given by the following formula;

$a_n=a+(n-1)d$ ,  where a = first term of AP, d = common difference, n = number of terms in an AP and $a_n$ = last term

Since it is given that the first, second and last term of an AP are a,b and 2a respectively, that means;

first term = a

d = second term - first term = b - a

$a_n$ = 2a

So,  $a_n=a+(n-1)d$

       $2a=a+(n-1)(b-a)$

       $2a-a=(n-1)(b-a)$

        $a=(n-1)(b-a)$

        $\frac{a}{b-a} = n - 1$

        $\frac{a}{b-a} +1= n$

        $\frac{a+(b-a)}{b-a} = n$

         $n=\frac{b}{b-a}$    ------------- [equation 1]

Now, the formula for the sum to n terms of an AP when the last term is given to us is;

              $S_n = \frac{n}{2}[\text{first term} + \text{last term}]$

              $S_n = \frac{b}{2\times (b-a)}[a +2a]$      {using equation 1}

              $S_n = \frac{b}{2 (b-a)}[3a]$

              $S_n = \frac{3ab}{2 (b-a)}$

Hence proved.