If a1, a2 , a3 , a4 are in A.P., then (1/√a1+√a2)+(1/√a2+√a3)+(1/√a3+√a4)=

Question

If a1, a2 , a3 , a4 are in A.P., then (1/√a1+√a2)+(1/√a2+√a3)+(1/√a3+√a4)= (A) (√a4-√a1/a3-a2) (B) (a4-a1/a3-a2) (C) (a3-a2/√a4-√a1) (D) (a1-a4/a3-a2)

Tardigrade Admin · Accepted Answer

(1/√a1+√a2)+(1/√a2+√a3)+(1/√a3+√a4) =(√a1-√a2/(√a1+√a2)(√a1-√a2)) +(√a2-√a3/(√a2+√a3)(√a2-√a3))+(√a3-√a4/(√a3+√a4)(√a3-√a4)) [by rationalisation] =(√a1-√a2/a1-a2)+(√a2-√a3/a2-a3)+(√a3-√a4/a3-a4) ∵ a1, a2, a3 and a4 are in AP. ∴ a2-a1=a3-a2=a4-a3 or a1-a2=a2-a3=a3-a4 Thus, (1/√a1+√a2)+(1/√a2+√a3)+(1/√a3+√a4) =(√a1-√a2/a1-a2)+(√a2-√a3/a1-a2)+(√a3-√a4/a1-a2) =((√a1-√a2)+(√a2-√a3)+(√a3-√a4)/(a1-a2))=(-√a4+√a1/a1-a2) =(√a4-√a1/a2-a1)=(√a4-√a1/a3-a2)

Q. If $a_{1}, a_{2}, a_{3}, a_{4}$ are in A.P., then $\frac{1}{a _{1} + a _{2}} + \frac{1}{a _{2} + a _{3}} + \frac{1}{a _{3} + a _{4}} =$

$\frac{a _{4} - a _{1}}{a _{3 - a_{2}}}$

$\frac{a _{4} - a _{1}}{a _{3} - a _{2}}$

$\frac{a _{3} - a _{2}}{a _{4} - a _{1}}$

$\frac{a _{1} - a _{4}}{a _{3} - a _{2}}$

$\frac{a _{5} - a _{0}}{a _{1} - a _{4}}$

Q. If a1​,a2​,a3​,a4​ are in A.P., then a1​​+a2​​1​+a2​​+a3​​1​+a3​​+a4​​1​=

a3−a2​​a4​​−a1​​​

a3​−a2​a4​−a1​​

a4​​−a1​​a3​−a2​​

a3​−a2​a1​−a4​​