Question

The sum of an A.P whose first term is i+1, second term is j-2 and the last term is k, is equal to

• A.
  $\left(1/2\right)(\frac{k-i-1}{j-i-3}+1)(i+1+k)$
• B.
  $\left(1/3\right)(\frac{k-i-1}{j-i-3}+1)(i+1+k)$
• C.
  $\left(1/2\right)(\frac{k-i-1}{j-i-3}+1)(i-1+k)$
• D.
  $\left(1/2\right)(\frac{k-i-1}{j-i-3}+1)(i+1-k)$

Answer 1

The correct option is A

$\left(1/2\right)(\frac{k-i-1}{j-i-3}+1)(i+1+k)$
First term = i+1
Second term = j-2
Last term = k
As the difference between first term and second term gives common difference,
$d=a_2-a_1$
$d=(j-2)-(i+1)=j-i-3$

In an AP, nth term is equal to,
$a_n=a+(n-1)d$
As k is last term,
$\Rightarrow k=i+1+(n-1)(j-i-3)Thusn=\frac{(k-i-1)}{(j-i-3)}+1$

S= $\frac{n}{2}$(First term + last term)
where S is the sum of the nth terms

Put $n=\left(\frac{k-i-1}{j-i-3}\right)+1$

The Sum of n tems will be as the first term is i+1 and the last term is k
Sum is equal to $\left(\frac{1}{2}\right)(\frac{k-i-1}{j-i-3}+1)(i+1+k)$
Hence, option a is correct.