Question

The first term of an A.P. is a, the second term is b and the last term is c. Show that the sum of the A.P. is (+−2)(+)2(−)​

Answer 1

Correct 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)

Solution:

We have

• First term =a
• second term =b
• last term =c

Common difference=(c-b)or (b-a)

We know that ,Genral term of an Ap

$\sf\:t_n=a+(n-1)d$

$\sf\implies\:c=a+(n-1)(c-b)$

$\sf\implies\dfrac{c-a}{c-b}=n-1$

$\sf\implies\dfrac{c-a}{c-b}+1=n$

$\sf\implies\dfrac{c+b-2a}{c-b}=n$

$\sf\dfrac{c+b-2a}{b-a}=n...(1)$

we know sum of n terms

$\sf\:S_n=\dfrac{n}{2}(first term+last term)$

put equation (1)value

$\sf\:S_n=\dfrac{(b+c-2a)}{(b-a)}\times\dfrac{1}{2}\times[a+c]$

$\sf\:S_n=\dfrac{(b+c-2a)(c+a)}{2(b-a)}$