BITSAT 2019 Paper

(c) As a1,a2,a3 ,...... , an , are in A.P. we get,
a2−a1 = a3−a2 = ........... = an−an−1 = d (say)
Now,

1

√a1+√a2

=

√a1−√a2

a1−a2

=

√a1−√a2

−d

Similarly,

1

√a2+√a3

=

√a2−√a3

−d

, ....... ,

1

√an−1+√an

=

√an−1−√an

−d

∴

1

√a1+√a2

+

1

√a2+√a3

+ ....... +

1

√an−1+√an

=
=

√a1−√an

−d

= −

1

d

[

a1−an

√a1+√an

]
= −

1

d

[

a1−{a1+(n+1)d}

√a1+√an

]
[formula for nth term]
= −

1

d

[

−(n−1)d

√a1+√an

] = [

(n−1)

√a1+√an

]