Question

4. If p™term of an A.P is q, q™ term is p, prove that its n™term is (p + q-n).​

Answer 1

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: { \orange{ given}}\\ { \pink{ \boxed{ \green{ \therefore ap = q}}}} \\ { \pink{ \boxed{ \green{ \therefore aq = p}}}} \\ \\ \: \: \: \: \: \: \: \: \: { \blue {to \: prove}} \\ {\purple{ \boxed {\red{nth \: term = (p + q - n)}}}}$

☆ PROOF:

$\to ap = q \\ \to a + (p - 1)d = q - - - - - (1) \\ \\ \to aq = p \\ \to a + (q - 1)d = p - - - - - (2)$

☆ Subtracting (2) from (1)

$\to a + (p - 1)d - (a + (q - 1)d = q - p \\ \to a + (p - 1)d - a - (q - 1)d = q - p \\ \to (p - 1 - q + 1)d = q - p\\ \to (p - q)d = q - p \\ \to d = \frac{q - p}{ p - q} \\ \to d = \frac{ - (p - q)}{p - q} \\ { \boxed {\to d = - 1}}$

☆ Putting value of d in (1)

$\to \: a + (p - 1) \times ( - 1) = q \\ \to a + ( - p + 1) = q \\ \to a =p - 1 + q \\ \\ \to nth \: term = a + (n - 1)( - 1) \\ \to \: p - 1 + q + (n - 1)d \\ \to p - 1 + q - n + 1 \\ {\green{ \boxed{\to \: (p + q - n )}}}$

$\huge{\pink{\boxed{\green{\underline{\sf{PROVED}}}}}}$

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