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

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Answer:


S_n = (3ab)/(2 (b-a))

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)


(a)/(b-a) = n - 1


(a)/(b-a) +1= n


(a+(b-a))/(b-a) = n


n=(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 = (n)/(2)[\text{first term} + \text{last term}]


S_n = (b)/(2* (b-a))[a +2a] {using equation 1}


S_n = (b)/(2 (b-a))[3a]


S_n = (3ab)/(2 (b-a))

Hence proved.

User Richard Belleville
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