Let S1=a,S2=a+d,S3=a+2d... S101=a+100d are in AP Given,
‌
1
S1S2
+‌
1
S2S3
+...+‌
1
S100S101
=‌
1
6
⇒
1
a(a+d)
+‌
1
(a+d)(a+2d)
+...+‌
1
(a+99d)(a+100d)
=‌
1
6
⇒‌
1
d
{‌
1
a
−‌
1
a+d
}+‌
1
d
{‌
1
a+d
−‌
1
a+2d
}+...+‌
1
d
{‌
1
a+99d
−‌
1
a+100d
}=‌
1
6
⇒‌‌‌
1
a
−‌
1
a+100d
=‌
d
6
⇒‌‌‌
100d
a(a+100d)
=‌
d
6
⇒600=a(a+100d)
⇒‌‌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