Question

The first second and middle term of an ap are a b c respectively sum of all terms is

Answer 1

Answer:

$Sum=\dfrac{(b+c-2a)(2c+b-a)}{b-a}$

Step-by-step explanation:

First term: a

Second term: b

Common Difference, d = b-a

Middle term = c

$a_n=a+(n-1)d$

$c=a+(\frac{n}{2}-1)(b-a)$

$n=2\dfrac{(b+c-2a)}{b-a}$

Now, we find the sum of nth term of the series.

$S_n=\dfrac{n}{2}(2a+(n-1)d$

$Sum=\dfrac{(b+c-2a)}{b-a}(2a+(2\dfrac{(b+c-2a)}{b-a}-1)(b-a)$

$Sum=\dfrac{(b+c-2a)}{b-a}(2c+b-a)$

Hence, The sum of nth term of the series $\dfrac{(b+c-2a)}{b-a}(2c+b-a)$