Question

If the pth term of an AP is q and the qth term is p,proff that its nth term is p+q_n

Answer 1

Answer:

Step-by-step explanation:

Given,

ap=a+(p-1)d=q........›1

aq=a+(q-1)d=p.........›2

Now solve 1&2

We get d=-1

And a=p+q-1

Then ,an=a+(n-1)d

an=p+q-1+(n-1)(-1)

an=p+q-1-(n-1)

an=p+q-1-n+1

an=p+q-n

So an=p+q-n

Hence proved.........

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Answer 2

Correct Question :-

If the pth term of an AP is q and the qth term is p,then prove that the nth term is (p+q-n)

Given :-

• Pth term of an AP is q and the qth term is p

To find :-

• The nth term is (p+q-n)

Solution :-

Let a be the first term and d the common difference of the given AP.

Then,

$\rm\:T_p=a+(p-1)d\:\:and\:T_q=a+(q-1)d$

Now,

$\rm\:T_p=q\:and\:T_q=p\:\:\:\:(given)$

Now we have,

$\rm\:a+(p-1)d=q\:...........................(i)$

$\rm\:a+(q-1)d=p\:...........................(ii)$

On subtracting eq(i) from eq(ii), we get

$\rm\:(q-p)d=(p-q)$

$\rm\longrightarrow\:d=-1$

putting d = -1 in eq (i) ,we get

$\rm\:a=(p+q-1)$

$\rm\:Thus,\:a=(p+q-1)\:and\:d=-1$

$\rm\therefore\:nth\:term=a+(n-1)d$

$\rm\longrightarrow\:(p+q-1)+(n-1)\times\:(-1)$

$\rm\longrightarrow\:(p+q-n)$

$\bf\:Hence,nth\:term=(p+q-n)$