Question

If the pth term of an AP is q and its qth term is p then show that its ( p + q )th is zero.

Answer 1

HEY THERE!!!



Given Arithmetic Sequence or Expression:-



*the pth term of an AP is q *



Tp=q



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



Again,



its qth term is p



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

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Subtractaction from 1 from 2


=> a+(q-1)d -[a+(p-1)d]=p-q


=> a+qd-d-[a+pd-d]=P-q

$\implies{a+qd-d-a-pd+d=P-q} \\ \\ \implies \cancel{a}+qd \: \: \cancel{ - d}- \cancel{a}-pd \: \: \cancel{ + d}=P-q \\ \\ \implies \: qd - pd = p - q \\ \\ \implies \: - d(p - q) = p - q \\ \\ \implies \: - d \cancel{(p - q) = p - q} \\ \\ \\ \implies{d = - 1}$



Here, Putting the value of common Difference in Equation;-



a+(p-1)d=q

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

=> a-p+1=q

=>a=p+q-1

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According to the Question;-

T(p+q)=?

=> T(p+q)=p+q-1+[p+q -1]-1

=> T(p+q)=p+q-1-p-q+1

$\implies \: T(p+q)= \cancel{p+q-1-p-q+1} \\ \\ \\ \implies T(p+q)=0$

Hence , It's Proved!!

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.