Answer:
[tex]S_n = \frac{3ab}{2 (b-a)}[/tex]
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;
[tex]a_n=a+(n-1)d[/tex] , where a = first term of AP, d = common difference, n = number of terms in an AP and [tex]a_n[/tex] = 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
[tex]a_n[/tex] = 2a
So, [tex]a_n=a+(n-1)d[/tex]
[tex]2a=a+(n-1)(b-a)[/tex]
[tex]2a-a=(n-1)(b-a)[/tex]
[tex]a=(n-1)(b-a)[/tex]
[tex]\frac{a}{b-a} = n - 1[/tex]
[tex]\frac{a}{b-a} +1= n[/tex]
[tex]\frac{a+(b-a)}{b-a} = n[/tex]
[tex]n=\frac{b}{b-a}[/tex] ------------- [equation 1]
Now, the formula for the sum to n terms of an AP when the last term is given to us is;
[tex]S_n = \frac{n}{2}[\text{first term} + \text{last term}][/tex]
[tex]S_n = \frac{b}{2\times (b-a)}[a +2a][/tex] {using equation 1}
[tex]S_n = \frac{b}{2 (b-a)}[3a][/tex]
[tex]S_n = \frac{3ab}{2 (b-a)}[/tex]
Hence proved.
