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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 25 of 106
Marks: +1, -0
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that ap\frac{a}{p} (q - r) + bq\frac{b}{q} (r - p) + cr\frac{c}{r} (p - q) = 0
Solution:  
Let the first term be A & common difference be D. We have
SpS_p = p2\frac{p}{2} [2A + (p - 1) D] = a ⇒ A + (p - 1) D2\frac{D}{2} = ap\frac{a}{p} , SqS_q = q2\frac{q}{2} [2A + (q - 1) D] = b
⇒ A + (q - 1) D2\frac{D}{2} = bq\frac{b}{q} , SrS_r = r2\frac{r}{2} [2A + (r - 1) D] = c ⇒ A + (r - 1) D2\frac{D}{2} = cr\frac{c}{r}
Now, ap(q−r)+\frac{a}{p} (q - r) +b/q$(r−p)+(r - p) +c/r(p−q)=(p - q) =(A + (p-1) D/2)(q-r)++(A+(q-1) D/2)(r-p)++(A+r-1)D/2)(p-q)++D/2=A[(q−r)+(r−p)+(p−q)]+= A [(q - r) + (r - p) + (p - q)] +D/2[(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q)]=A(q−r+r−p+p−q)+[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)] = A (q - r + r - p + p - q) +D/2[pq−rp−q+r+qr−pq−r+p+rp−qr−p+q]=A(0)+[pq - rp - q + r + qr - pq - r + p + rp - qr - p + q] = A(0) +q/p(0)=0+0=0 Hence (0) = 0 + 0 = 0 \text{ Hence }b/q(q−r)+(q - r) +c/r$ (p - q) = 0
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