Question

The sum of an AP whose first term is P, the second term Q and the last term R, is equal to a) (P+Q)(P+R-2Q)/2(Q-R) b)(Q+R)(P=Q-2R)/2(P-Q) c)(P+R)(Q+R-2P)/2(Q-P) d)(P+R)(Q-R+2P0/2(P-Q)

Answer 1

The sum of an AP whose first term is P, the second term Q and the last term R, is equal to $\boxed{\frac{(P+R)(Q+R-2P)}{2(Q-P)}}$

Therefore, option (c) is correct.

Step-by-step explanation:

Given:

First term of AP = P

Second term of AP = Q

last term of AP = R

To find out:

Sum of the AP

Solution:

Let last term be nth term

Common difference = Q - P

We know that if a is the first term of an AP and d is the common difference

Then nth term is given by

$\boxed{a_n=a+(n-1)d}$

Therefore,

$R=P+(n-1)(Q-P)$

$\implies \frac{R-P}{Q-P}=n-1$

$\implies n=\frac{R-P}{Q-P}+1$

$\implies n=\frac{R+Q-2P}{Q-P}$

Again

We know that for n terms in AP if first and last terms are known then sum is given by

$S=\frac{n}{2}[\text{First Term}+\text{Last Term}]$

$\implies S=\frac{R+Q-2P}{2(Q-P)}(P+R)$

$\implies S=\frac{(P+R)(R+Q-2P)}{2(Q-P)}$

Hope this answer is helpful.

Know More:

Q: 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).

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