Maths / Arithmetic Progression / General Term Of an AP

QUESTION

If pth ,qth and rth terms of an AP are a, b and c respectively, show that a(q-r)+b(r-p)+c(p-q)=0.

EXPLANATION
Explain TypeExplanation Content
Text

$\dpi{120} \fn_jvn \large a_p=a$

$\dpi{120} \fn_jvn \large x-(p-1)d=a$

Multiplying both sides with (q-r)

$\dpi{120} \fn_jvn \large x(q-r)-(p-1)(q-r)d=a(q-r)$

$\dpi{120} \fn_jvn \large xq-xr-(pq-pr-q+r)d=a(q-r)$                    (1)

$\dpi{120} \fn_jvn \large a_q=b$

$\dpi{120} \fn_jvn \large x-(q-1)d=b$

Multiplying both sides with (r-p)

$\dpi{120} \fn_jvn \large x(r-p)-(q-1)(r-p)d=b(r-p)$

$\dpi{120} \fn_jvn \large xr-xp-(qr-qp-r+p)d=b(r-p)$                    (2)

$\dpi{120} \fn_jvn \large a_r=c$

$\dpi{120} \fn_jvn \large x-(r-1)d=c$

Multiplying both sides with (p-q)

$\dpi{120} \fn_jvn \large x(p-q)-(r-1)(p-q)d=c(p-q)$

$\dpi{120} \fn_jvn \large xp-xq-(rp-rq-p+q)d=c(p-q)$                    (3)

adding 1, 2 and 3 we get

$\dpi{120} \fn_jvn \large xq-xr-(pq-pr-q+r)d+xr-xp-(qr-qp-r+p)d+xp-xq-(rp-rq-p+q)d=a(q-r)+b(r-p)+c(p-q)$

$\dpi{120} \fn_jvn \large (xq-xr+xr-xp+xp-xq)-(pq-pr-q+r+qr-qp-r+p+rp-rq-p+q)d=a(q-r)+b(r-p)+c(p-q)$

$\dpi{120} \fn_jvn \large (0)-(0)d=a(q-r)+b(r-p)+c(p-q)$

$\dpi{120} \fn_jvn \large a(q-r)+b(r-p)+c(p-q) =0$

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