KEAM 2011 Math Paper

Let S1=a,S2=a+d,S3=a+2d...
S101=a+100d are in AP
Given, ⇒‌‌‌

1

a

−‌

1

a+100d

=‌

d

6

⇒‌‌S101=‌

600

a

....(i)
Given that, ‌‌S1+S101=50
a+‌

600

a

=50
⇒‌‌a2−50a+600=0
⇒‌‌a2−30a−20a+600=0
⇒‌‌a(a−30)−20(a−30)=0
⇒‌‌(a−30)(a−20)=0
⇒‌‌a=20,30
So, ‌‌S1=20 or S1=30
and ‌‌S101=a+100d=‌

600

a

[from Eq. (i)]
When a=20,20+100d=‌

600

20

=30
100d=10⇒d=‌

1

10

When a=30,30+100d=‌

600

30

=20
d=−‌

1

10

So, when S1=20, then, S101=a+100d
=20+‌

100

10

=30
when S1=30, then S101=a+100d
=30−‌

100

10

=20
Hence,
(i) When S1=20,S101=30
|S1−S101|=|20−30|=10
(ii) When S1=30,S101=20
|S1−S101|=|30−20|=10