Question

p times the pth term of an AP is q times the qth term. Find the (p+q)th term.​

Answer 1

$\large\underline{\sf{Solution-}}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

↝ pᵗʰ term is,

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

and

↝ qᵗʰ term is,

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

According to statement

$\rm \: p \: a_p \: = \: q \: a_q \:$

So, on substituting the values from above, we get

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

$\rm \: ap + d{p}^{2} - dp = aq + {dq}^{2} - dq$

$\rm \: ap + d{p}^{2} - dp - aq - {dq}^{2} + dq = 0$

$\rm \: (ap - aq) + ( {dp}^{2} - {dq}^{2}) - (dp - dq) = 0$

$\rm \: a(p - q) + d( {p}^{2} - {q}^{2}) - d(p - q) = 0$

$\rm \: a(p - q) + d(p - q)(p + q) - d(p - q) = 0$

$\rm \:(p - q)[a + d(p + q) - d] = 0$

$\rm \:a + d(p + q) - d = 0 \: \: \: \{ \: as \: p \: \ne \: q \}$

$\rm \:a + [p + q - 1]d = 0 \:$

$\rm\implies \:a_{p + q} \: = \: 0$

$\rule{190pt}{2pt}$

Additional Information :-

↝ Sum of n  terms of an arithmetic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

Answer 2

$\underline{\huge \tt{ \red{A} \blue{n} {S} \green{w} \pink{E} \orange{r}}}$

nth term of A.P :-

$\bigstar \boxed{\sf \: a_{n} = a + (n - 1)d}$

Given :-

• p times of pth term = q times of qth term.

$\\ \sf \: p \{a + (p - 1)d \} = q \{ a + (q - 1)d\} \\ \sf \: p \{a + pd - d \} = q\{a + qd - d \} \\ \sf \: pa + {p}^{2} d - pd = qa + {q}^{2} d - qd \\ \sf \: pa - qa + {p}^{2}d - {q}^{2}d - pd + qd =0 \\ \sf \: a(p - q) + d( {p}^{2} - {q}^{2} - p + q) = 0 \\ \sf \: a(p - q) + d \{ (p - q)(p + q) - p + q\} = 0 \\ \sf \: a(p - q) + d(p - q)(p + q - 1) = 0$

$\sf \: dividing \: both \: sides \: by \: (p - q) \: we \: get.. \\ \\ \sf \: a + d(p + q - 1) = 0 \: \: \: \: - - - (i)$

Now ,

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

From equation (i) ,

$\dagger \underline{ \boxed{ \sf \red{ \: a_{p + q} = 0}}}$