If Tp , Tq,Tr are pth, qth and rth terms of an A.P then |Tp&Tq&Tr p&q&r 1&1&1| is equal to

Question

If Tp , Tq,Tr are pth, qth and rth terms of an A.P then |Tp&Tq&Tr p&q&r 1&1&1| is equal to (A) 1 (B) -1 (C) 0 (D) p + q + r.

Tardigrade Admin · Accepted Answer

Tp = a+ (p-1) d; Tq = a + (q-1) d; Tr =a + (r-1)d ∴ the given determinant |a+(p-1)d&a+(q-1)d&a+(r-1)d p&q&r 1&1&1| Operate R1 - (R2 - R3)d, we get the determinant = |a&a&a p&q&r 1&1&1| = 0

Q. If $T_{p}, T_{q}, T_{r}$ are $p$ th, $q$ th and $r$ th terms of an A.P then $∣ ∣ T_{p} p 1 T_{q} q 1 T_{r} r 1 ∣ ∣$ is equal to

1

-1

0

p + q + r.

$T_{p} = a + (p - 1) d; T_{q} = a + (q - 1) d;$
$T_{r} = a + (r - 1) d$
$∴$ the given determinant
$∣ ∣ a + (p - 1) d p 1 a + (q - 1) d q 1 a + (r - 1) d r 1 ∣ ∣$
Operate $R_{1} - (R_{2} - R_{3}) d$ ,
we get the determinant = $∣ ∣ a p 1 a q 1 a r 1 ∣ ∣ = 0$

Q. If Tp​,Tq​,Tr​ are pth, qth and rth terms of an A.P then ∣∣​Tp​p1​Tq​q1​Tr​r1​∣∣​ is equal to

1

-1

0

p + q + r.

Tp​=a+(p−1)d;Tq​=a+(q−1)d; Tr​=a+(r−1)d ∴ the given determinant ∣∣​a+(p−1)dp1​a+(q−1)dq1​a+(r−1)dr1​∣∣​ Operate R1​−(R2​−R3​)d, we get the determinant = ∣∣​ap1​aq1​ar1​∣∣​=0

Q. If $T_{p}, T_{q}, T_{r}$ are $p$ th, $q$ th and $r$ th terms of an A.P then $∣ ∣ T_{p} p 1 T_{q} q 1 T_{r} r 1 ∣ ∣$ is equal to

$T_{p} = a + (p - 1) d; T_{q} = a + (q - 1) d;$
$T_{r} = a + (r - 1) d$
$∴$ the given determinant
$∣ ∣ a + (p - 1) d p 1 a + (q - 1) d q 1 a + (r - 1) d r 1 ∣ ∣$
Operate $R_{1} - (R_{2} - R_{3}) d$ ,
we get the determinant = $∣ ∣ a p 1 a q 1 a r 1 ∣ ∣ = 0$