In an A.P. if Tₚ = q and Tq = p, then find the rth term.
Please solve this problem

Formula:

We know that

nth term of the AP is Tn = a+(n-1)d

Where, a = First term

d = Common difference

n = number of terms

Now,

Given that

pth term of the A.P. = Tp = q

⇛a + (p-1)d = q

⇛(p-1)d = q - a

⇛d = (q-a)/(p-1) →→→→(i)

and

qth term of the A.P. = Tq = p

⇛a + (q-1) d = p

⇛ (q-1)d = p-a

⇛ d = (p-a)/(q-1) →→→→(ii)

From Eqn(i) and Eqn(ii)

⇛(q-a)/(p-1) = (p-a)/(q-1)

On applying cross multiplication then

⇛(q-a)×(q-1) = (p-a)×(p-1)

⇛q²-q-aq+a = p²-p-ap+a

⇛q²-q-aq = p²-p-ap

⇛ap-aq = p²-p+q-q²

⇛a(p-q) = (p²-q²)-(p-q)

⇛a(p-q) = (p+q)(p-q)-(p-q)

⇛ a(p-q) = (p-q){(p+q)-1}

On cancelling (p-q) both sides then

⇛a = (p+q-1) →→→→Eqn(iii)

On Substituting the value of a in Eqn(ii) then

⇛ d = [p-(p+q-1)]/(q-1)

⇛d = (p-p-q+1)/(q-1)

⇛d = (-q+1)/(q-1)

⇛d = -(q-1)/(q-1)

⇛d = -1

We have,

a = p+q-1 and d = -1

Now, rth term of the AP

⇛ Tr = a + (r-1)d

⇛Tr = (p+q-1)+(r-1)(-1)

⇛ Tr = p+q-1+(-r+1)

⇛ Tr = p+q-1-r+1

⇛ Tr = p+q-r

Therefore, Tr = p+q-r

Answer: rth term of the given A.P. is p+q-r

see explanation

Step-by-step explanation:

In an arithmetic sequence the common difference d is

d = a₂ - a₁ = 10 - 8 = 2

To obtain the next term in the sequence add d to the previous term, that is

a₅ = 14 + 2 = 16

a₆ = 16 + 2 = 18

a₇ = 18 + 2 = 20

The next 3 terms in the sequence are 16, 18, 20

The n th term equation for an arithmetic sequence is

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 8 and d = 2, thus

= 8 + 2(n - 1) = 8 + 2n - 2 = 2n + 6