Question

If (p + q)th term of an AP is m and (p – q)th term is n, then pth term is

Answer 1

Answer:-

$\sf{The \ p^{th} \ term \ of \ the \ AP \ is \ \frac{m+n}{2}}$

Given:

• $\sf{t_{(p+q)}=m}$

• $\sf{t_{(p-q)}=n}$

To find:

• $\sf{p^{th} \ term \ of \ the \ AP}$

Solution:

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

$\sf{According \ to \ the \ first \ condition.}$

$\sf{t_{(p+q)}=a+(p+q-1)d}$

$\sf{But, \ t_{(p+q)}=m}$

$\sf{\therefore{a+(p+q-1)d=m...(1)}}$

$\sf{According \ to \ the \ second \ condition}$

$\sf{t_{(p-q)}=a+(p-q-1)d}$

$\sf{But, \ t{(p-q)}=n}$

$\sf{\therefore{a+(p-q-1)d=n...(2)}}$

$\sf{Add \ equation \ (1) \ and \ (2), \ we \ get}$

$\sf{a+(p+q-1)d=m}$

$\sf{+}$

$\sf{a+(p-q-1)d=n}$

_______________________

$\sf{2a+(p+q-1)d+(p-q-1)d=m+n}$

$\sf{\therefore{2a+(p+q-1+p-q-1)d=m+n}}$

$\sf{\therefore{2a+(2p-2)d=m+n}}$

$\sf{\therefore{2a+2(p-1)d=m+n}}$

$\sf{\therefore{2[a+(p-1)d]=m+n}}$

$\sf{Dividing \ both \ sides \ by \ 2, \ we \ get}$

$\boxed{\sf{a+(p-1)d=\frac{m+n}{2}...(3)}}$

$\sf{t_{p}=a+(p-1)d}$

$\sf{...from \ equation (3), \ we \ get}$

$\sf{t_{p}=\frac{m+n}{2}}$

$\sf\purple{\therefore{\tt{The \ p^{th} \ term \ of \ the \ AP \ is \ \frac{m+n}{2}}}}$

Answer 2

$\huge\underline\mathfrak\red{Given}$

• (p + q)th term of an AP is m
• (p - q ) th term of an AP is n

$\huge\underline\mathfrak\color{plum}To \: Find$

• The pth term of AP.

$\huge\underline\mathfrak\color{green}Solution$

$\sf{t _{(p + q)} = m}$-------(1)

$\sf{t _{(p - q)} = n}$--------(2)

We know,

${\boxed{\sf{\pink{t _{n} = a + (n - 1)d}}}}$

Now by first condition,

$\sf{t _{(p + q)} = a + (p + q - 1)d}$

$\sf{\implies a + (p + q - 1)d = m}$--------(3)

By second condition,

$\sf{t _{(p - q)} = a + (p - q - 1)d}$

$\sf{\implies a + (p - q - 1)d = n}$--------(4)

Adding eq (3) and (4) we get,

$\sf{\implies a +( p + q - 1)d + a + (p - q - 1)d = m + n}$

$\sf{\implies 2a + (p + q - 1)d + (p - q - 1)d = m + n}$

$\sf{\implies 2a + (p + q - 1 + p - q - 1)d = m + n}$

$\sf{\implies 2a + 2p - 2)d = m + n}$

$\sf{\implies 2a + 2(p - 1)d = m + n}$

$\sf{\implies 2[a + (p -1 )d ]= m + n}$

Now,by dividing both sides by 2, we get,

$\sf{\implies a + (p - 1)d = \frac{m + n}{2}}$-----(5)

From eq (5),we get,

$\sf{t _{p} = a + (p - 1)d}$

${\boxed{\sf{\color{olive}{t _{p} = \frac{m + n}{2} }}}}$

Hence, the pth term of AP is ${\boxed{\sf{\color{olive}{t _{p} = \frac{m + n}{2} }}}}$

$\huge\underline\mathfrak\blue{ThankYou}$