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Arithmetic Progressions
CH05 Question Bank
CBSE_CH05_QUESTION_BANK
Grade 10

Question:

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

Step-by-Step Solution

Key Concept: Express a, b, c using the nth-term formula in terms of the AP's first term A and common difference D, then substitute and simplify.
Let the AP have first term $A$ and common difference $D$. Then $a=A+(p-1)D,\ b=A+(q-1)D,\ c=A+(r-1)D$. [1.0 Mark]

$a(q-r)+b(r-p)+c(p-q)=A[(q-r)+(r-p)+(p-q)]+D[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]$. [1.0 Mark]

The coefficient of $A$ is $(q-r)+(r-p)+(p-q)=0$. For the $D$ term: expanding $(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)$, the '$-1$' parts give $-(q-r)-(r-p)-(p-q)=0$, and the remaining parts $p(q-r)+q(r-p)+r(p-q)=pq-pr+qr-pq+rp-rq=0$ as well. So the whole expression is $0$. Hence proved. [1.0 Mark]

Correct Answer:
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