Question

show that the sum of an A.P whose first term is 'a' , second term is 'b' and the last term is 'c' is equal to [(a+c)+(b+c-2a)]/2(b-a)​

Answer 1

Step-by-step explanation:

first term = a

second term = b

common difference (d) = (b-a)

Now last term $a_{n}$ = a + (n-1)d

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

Sum up to n terms $S_{n}$

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

Now substitute value

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

we used the value of n-1 from above

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

hence proved ! (MIGHT Help!)

THANKS!