Question

The p th and qth term of an A.P. is c and d respectively, find rth term.

Answer 1

Answer:

$\frac{pd -cq + rc - rd}{p-q}$

Step-by-step explanation:

Let, first term = a , and differencr between terms = f

So, pth term , a + (p-1)f = c      ... (i)

and qth term , a + (q-1)f = d   ......(ii)

(i) - (ii) ---

► a-a + f(p-1-q+1) = c - d

► f(p-q) = c-d

► f = (c-d) / (p-q)

again , a = c - (p-1)f =   $c-\frac{ (p-1)(c-d)}{p-q}$   =  $\frac{ cp - cq - cp + c +pd -d}{p-q}$ = $\frac{c -cq +pd -d}{p-q}$

Now, rth term = $\frac{c -cq +pd -d}{p-q} + (r-1) \frac{c-d}{p-q}$  $=\frac{c -cq +pd -d+ rc -rd - c +d}{p-q}$ $= \frac{pd -cq + rc - rd}{p-q}$

Answer 2

Answer:

The r th term of the AP is

$\frac{c(r - q) + d(p - r)}{p - q}$

Step-by-step explanation:

Given that,

The p th and q th term of an A.P are c and respectively.Let the first term of the A.P be a,the common difference be k

We know that nth term of A.P is

$t _{n} = a + (n - 1)d$

Therefore p th term is

$a + (p - 1)k= c - - > (1)$

and q th term is

$a + (q - 1)k = d - - > (2)$

subtracting equation 2 from equation 1 we get

$(1) - (2) = > k(p - q) = c - d \\ = > k = \frac{c - d}{p - q}$

Substituting this value of k in equation 1,we get

$a + (p - 1)( \frac{c - d}{p - q} ) = c \\ = > a = \frac{c(p - q) - (p - 1)(c - d)}{p - q} \\ = > a = \frac{cp - cq -( cp - pd - c + d)}{p - q} \\ = > a = \frac{c(1 - q) + d(p - 1)}{p - q}$

Now r th term of the AP is

$a + (r - 1)k$

Substituting values of a and k we get

$= > \frac{c(1 - q) + d(p - 1)}{p - q} + \frac{c - d}{p - q}(r - 1) \\ = \frac{c(1 - q) + d(p - 1) + cr - c - dr + d}{p - q} \\ = \frac{c(1 - q + r - 1) + d(p - 1 - r + 1)}{p - q} \\ = \frac{c(r - q) + d(p - r)}{p - q}$

Therefore, rth term of the AP is

$\frac{c(r - q) + d(p - r)}{p - q}$

#SPJ3