Question

please help q39 part 2...........

Answer 1

▶ Question :-

→ If the pth, qth and rth terms of an AP be a, b, c respectively, then show that

==> ( a - b )r + ( b - c )p + ( c - a )q = 0.

▶ Explanation :-

▶ Given :-

→ Pth term = a .

→ qth term = b .

→ rth term = c .

→ a( q - r ) + b( r - p ) + c( p - q ) = 0 . [ From part 1 ] .

▶ Solution :-

→ Solving LHS .

°•° ( a - b )r + ( b - c )p + ( c - a )q.

= ar - br + bp - cp + cq - aq .

= ar - aq - br + bq - cp + cq .

= -(aq - ar) - (br - bp) - (cp - cq) .

= - [a(q - r)+ b(r - p) + c(p - q)] .

$\huge \boxed{ \pink { = 0 . }}$

[ Given ] .

✔✔ Hence, it is proved ✅✅.

Answer 2

here is your answer ☺☺☺☺☺☺☺☺

1) question


Here is the answer to your question.

Let A and D be the first term and common difference of A.P.
ap = a ⇒A + (p– 1) D = a … (1)
aq = b ⇒A + (q – 1) D = b … (2)
ar = c ⇒ A + (r – 1) D = c … (3)

1. a (q – r) + b (r – p) + c (p – q)
= [A + (p– 1) D] (q – r) + [A + (q – 1) D] (r – p) + [A + (r – 1) D] (p – q)]
= A (q – r) + (p – 1) (q – r)D + A (r – p) + (q – 1) (r – p) D + A (p – q) + (r – 1) (p – q) D
= A × (q – r + r – p + p – q) + D × (pq – pr – q + r + qr – pq – r + p + pr – rq – p + q)
= A × 0 + D × 0
= 0

2. a (q – r) + b (r – p) + c (p – q) = 0 (Proved)
⇒ aq – ar + br – bp + cp – cq = 0
⇒ r (–a + b) + p (–b + c) + q (–c + a) = 0
⇒ – r (a – b) – p (b – c) – q (c – a) = 0
⇒ r (a – b) – p(b – c) + q (c – a) = 0