Question

Question :-
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☆ Sum of first p, q and r terms of an A.P. are a, b and c, respectively.
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→ Prove that
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$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {\dfrac{a}{p}(q - r) + \dfrac{b}{q}(r - p) + \dfrac{c}{r}(p - q) = 0}$
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Answer 1

Solution

Given :-

• Sum of p terms = a
• Sum of q terms = b
• Sum of r terms = c

Prove That:-

• (q-r)a/p + (r-p)b/q + (p-q)c/r = 0

Proof :-

Let,

• F is first terms for all .

• Common Defference of all terms = D

Using Formula

$\dag\boxed{\underline{\tt{\blue{\:S_{n}\:=\:\dfrac{n}{2}\times\big(2F+(n-1)D\big)}}}}$

Now, First Calculate sum of p terms

➠ Sp = p/2[2F+ (p-1)D] = a

➠ a/p = [2F + (p-1)D]/2 ___________(1)

Again, Calculate sum of q terms,

➠ Sq = q/2[2F + (q-1)D] = b

➠b/q = [2F + (q - 1)D]/2 ____________(2)

Again, Calculate sum of r terms

➠ Sr = r/2[2F + (r-1)D] = c

➠ c/r = [2F + (r-1)D]/2_____________(3)

Now, Multiply by (q - r) in equ(1)

➠ (q-r)a/p = (q-r)[2F+(p-1)D]/2 _______(4)

Similarly,

➠(r-b)b/q = (r-b)[2F+(q-1)D]/2________(5)

And,

➠(p-q)c/r = (p-q)[2F+(r-1)D]/2________(6)

Now, adding equ(4) , (5) & (6)

➠ (q-r)a/p + (r-b)b/q + (p-q)c/r = F(q-r+r-b+p-q) + D/2(pq - qr -q + r + rq - pq + p - r + rp - rq - p -q )

➠ (q-r)a/p + (r-b)b/q + (p-q)c/r = F(0) + D/2(0)

➠ (q-r)a/p + (r-b)b/q + (p-q)c/r = 0 + 0

➠ (q-r)a/p + (r-b)b/q + (p-q)c/r = 0

That's Proved.

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