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Standard VI

Mathematics

Equivalent Fractions

For an A.P., ...

Question

For an A.P., $T_{p} = q$ and $T_{q} = p$ , then $T_{p + q} =$ ..........

0

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p - q

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p + q

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p + q - 1

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Solution

The correct option is C $0$
$T_{p} = a + (p - 1) d = q$
$T_{q} = a + (q - 1) d = p$
solving the above two equations,
$(p - q) d = (q - p), d = - 1$
$a = p + q - 1$
$T_{p + q} = a + (p + q - 1) (- 1) = a - a = 0$

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For an A.P., Tp=q and Tq=p , then Tp+q=..........

The correct option is C 0Tp=a+(p−1)d=qTq=a+(q−1)d=psolving the above two equations,(p−q)d=(q−p),d=−1a=p+q−1Tp+q=a+(p+q−1)(−1)=a−a=0

For an A.P., $T_{p} = q$ and $T_{q} = p$ , then $T_{p + q} =$ ..........

The correct option is C $0$
$T_{p} = a + (p - 1) d = q$
$T_{q} = a + (q - 1) d = p$
solving the above two equations,
$(p - q) d = (q - p), d = - 1$
$a = p + q - 1$
$T_{p + q} = a + (p + q - 1) (- 1) = a - a = 0$