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(a) not in A.P./G.P./H.P.

(b) in A.P.

(C) in G.P.

(d) in H.P.

Answer

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Hint: Try to obtain the required terms from given terms using properties of sequences.

Since ,it is given that $a,b,c,d$ are in A.P., then $d,c,b,a$ are also in A.P.

A series of terms is known as a H.P. series when the reciprocals of elements are

in arithmetic progression or A.P.

So,

$ \Rightarrow \dfrac{1}{d},\dfrac{1}{c},\dfrac{1}{b},\dfrac{1}{a}$ are in H.P.

Thus, after multiplying the above terms with $abcd$,

We get,

$ \Rightarrow \dfrac{{abcd}}{d},\dfrac{{abcd}}{c},\dfrac{{abcd}}{b},\frac{{abcd}}{a}$ are in H.P.

$ \Rightarrow abc, abd, acd, bcd$ are in H.P.

Hence, the required answer is (d) in H.P.

Note: To solve these types of questions, perform the specific manipulations and obtain the required solution.

Since ,it is given that $a,b,c,d$ are in A.P., then $d,c,b,a$ are also in A.P.

A series of terms is known as a H.P. series when the reciprocals of elements are

in arithmetic progression or A.P.

So,

$ \Rightarrow \dfrac{1}{d},\dfrac{1}{c},\dfrac{1}{b},\dfrac{1}{a}$ are in H.P.

Thus, after multiplying the above terms with $abcd$,

We get,

$ \Rightarrow \dfrac{{abcd}}{d},\dfrac{{abcd}}{c},\dfrac{{abcd}}{b},\frac{{abcd}}{a}$ are in H.P.

$ \Rightarrow abc, abd, acd, bcd$ are in H.P.

Hence, the required answer is (d) in H.P.

Note: To solve these types of questions, perform the specific manipulations and obtain the required solution.