Question

(◍•ᴗ•◍)❤ Please solve this (◍•ᴗ•◍)❤​

Answer 1

Question :–

• The pth , qth and rth terms of A.P. are a , b and c respectively. Prove that a(q - r) + b(r - p) + c(p - q) = 0.

ANSWER :–

GIVEN :–

• Pth term = a

• qth term = b

• rth term = c

TO FIND :–

• Prove that a(q - r) + b(r - p) + c(p - q) = 0.

SOLUTION :–

• We know that nth term of A.P. is –

$\\ \longrightarrow \large { \boxed{ \rm \: T_{n} = a_{1} + (n - 1)d}}$

• Here –

$\\ \rm \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: a_{1} \: = \: first \: \: term \: \:of \: \: A.P.$

$\\ \rm \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: d\: = \: common \: \: difference \: \:of \: \: A.P.$

$\\ \rm \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: n \: = \: total \: \: number \: \: of \: \: terms \: \:in \: \: A.P.$

$\\ \rm \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: T_{n} \: = \: n \th \: \: term \: \: of \ \: \: A.P.$

• According to the first condition –

$\\ \longrightarrow \rm \: T_{p} = a$

$\\ \implies \rm \: a_{1} + (p- 1)d = a ---eq.(1)$

• According to the second condition –

$\\ \longrightarrow \rm \: T_{q} = b$

$\\ \implies \rm \: a_{1} + (q- 1)d = b ---eq.(2)$

• According to the third condition –

$\\ \longrightarrow \rm \: T_{r} = c$

$\\ \implies \rm \: a_{1} + (r- 1)d = c ---eq.(3)$

• Now Let's take L.H.S. –

$\\ = \rm \: a(q - r) + b(r - p) + c(p - q)$

• Now put the values of a , b and c from eq.(1) , eq.(2) and eq.(3) –

$\\ = \rm \:( a_{1} + (p- 1)d)(q - r) + (a_{1} + (q- 1)d)(r - p) +( a_{1} + (r- 1)d)(p - q)$

$\\ = \rm \ ( a_{1} + p.d- d)(q - r) + (a_{1} + q.d- d)(r - p) +( a_{1} + rd- d)(p - q)$

$\\ = \rm \ a_{1} .q+ p.d.q- d.q - a_{1} .r - p.d.r + d.r +a_{1}.r + q.d.r- d.r - a_{1}.p - q.d.p + d.r.p +a_{1} .p+ r.d.p- d.p -a_{1} .q - r.d.q + d.q$

• All term are cancelled , so –

$\\ = \rm \ 0$

$\\ = \rm \ R.H.S. \: \: \: \: \: \: \: \: \: (Hence \: \: \: proved)$

$\\ \rule{220}{2}$

Answer 2

$\large\bf\underline{Question:-}$

The pth ,qth and rth terms of AP are a,b and c respectively.

then prove that a(q-r) +b(r-p)+c(p-q) = 0

$\large\bf\underline{Given:-}$

• the pth , qth and rth terms of an AP a,b and c.

$\large\bf\underline {To \: prove:-}$

• a(q-r) +b(r-p)+c(p-q) = 0

$\huge\bf\underline{Solution:-}$

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

we know that ,

$\bf \: a_n = a+(n-1)d$

$\rm \: T_P = x+(p-1)d \\ </p><p></p><p> \rm \: T_q = x +(q-1)d \\ </p><p></p><p> \rm \: T_r = x +(r-1)d \:$

Then,

• x + (p-1)d = a ........(i)
• x + (q-1)d = b ......(ii)
• x + (r-1)d = c .......(iii)

On Multiplying :-

eq.(i) by (q-r) , eq. (ii) by( r-p) and eq.(iii) by (p-q)

we get,

$\rm \: a(q-r) +b(r-p)+c(p-q)$

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

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

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

$\small\rm\:=a[0+0+0]+d[qp-q-rp+rq-r-pq+p+pr-p-qr+q]$

$\small\rm\:=0+d(o)$

$\small\rm\:=0$

LHS = RHS

Hence, proved