The pth term of an Arithmetic progression is 1/q and qth term is 1/p. The sum of the pqth term is:
a) 1/2(pq+1)
b) 1/2(pq-1)
c) pq+1
d) pq-1

Question

The pth term of an Arithmetic progression is 1/q and qth term is 1/p. The sum of the pqth term is:
a) 1/2(pq+1)
b) 1/2(pq-1)
c) pq+1
d) pq-1

trojanpoem · Answer

$$a_{1} + d(p - 1) = \frac{ 1 }{ q }$$

$$a_{1} + d(q-1) = \frac{ 1 }{ p }$$

r.hamza17 · Answer

We dont need to but the sum of pqth term we just need to check what is the sum of the pqth term.

trojanpoem · Answer

-a1 - d(p-1) = -1/q
a1 + d(q-1) = 1/p

-d(p-1) + d(q-1) = -1/q + 1/p

-dp +d +qd -d = -1/q + 1/p

d(p-q) = p - q /pq

d = 1/pq -> common difference

a1 = 1/q - 1/pq * p + 1/pq
a1 = 1/q - 1/q +1/pq

a1 = 1/pq

trojanpoem · Answer

$$a_{n} = \frac{ 1 }{ pq } + \frac{ 1 }{ pq } (n-1) = \frac{ n }{ pq }$$

trojanpoem · Answer

Now we find the sum.

trojanpoem · Answer

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

trojanpoem · Answer

$$a_{pq} = \frac{ pq }{ 2 } ( \frac{ 2 }{ pq } - \frac{ pq - 1 }{ pq })$$

trojanpoem · Answer

$$\frac{ pq }{ 2 }(\frac{ 2 + pq - 1 }{ pq }) =    1 + pg/2 = 0.5(pg+1)$$

trojanpoem · Answer

Correct the negative sign in the step before this.

r.hamza17 · Answer

Thanx alot mate

trojanpoem · Answer

Any time.