Question

in an AP, if the pth term is q and the qth term is p, then find its ( p+q) th term.

Answer 1

$\huge{hey}$


$\huge{ \mathfrak{here \: is \: answer}}$

$\bf{see \: in \: pic}$

$\huge \boxed{hope \: it \: helps}$


$\huge{ \mathfrak{THANKS}}$

Answer 2

Hey there !!

➡ Given :-

→ pth term = q.

→ qth term = p.

➡ To prove :-

→ ( p + q )th = 0.

➡ Solution :-

→ pth term = a + ( n - 1 )d.

=> q = a + ( p - 1 )d..........(1).

And,

→ qth term = a + ( n - 1 )d.

=> p = a + ( q - 1 )d..........(2).

▶ Substract equation (1) and (2), we get

a + ( p - 1 )d = q.
a + ( q - 1 )d = p.
(-)...(-)..............(-)
____________

=> pd - d - qd + d = q - p.

=> pd - qd = q - p.

=> d( p - q ) = -( p - q ).

=> d = -1.

▶ Put the value of ‘d’ in equation (1), we get

=> a + ( p - 1 ) (-1) = q.

=> a - p + 1 = q.

=> a = q + p - 1.

▶ Now,

→ $a \tiny{ ( p + q ) }$  = a + ( n - 1 )d.

=> $a \tiny{ ( p + q ) }$  = p + q - 1 + ( p + q - 1 ) (-1).

=> $a \tiny{ ( p + q ) }$  = p + q - 1 - p - q + 1.


$\huge \boxed{ \boxed{ \bf => a \tiny{ ( p + q ) } \large = 0. }}$

✔✔ Hence, it is proved ✅✅.

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THANKS

#BeBrainly.