(i) a 1 = a2 = 2, an = a-1, n > 2
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1
Answer:
Step-by-step explanation:
Since, a
1
,a
2
,a
3
,.....a
2n
form an AP.
Therefore,
a
2
−a
1
=a
4
−a
3
=....a
2n
−a
2n−1
=d
Here, a
1
2
−a
2
2
+a
3
2
−a
4
2
+....+a
2n−1
2
−a
2n
2
=(a
1
−a
2
)(a
1
+a
2
)+(a
3
−a
4
)(a
3
+a
4
)+....+(a
2n−1
−a
2n
)⋅(a
2n−1
+a
2n
)
=−d(a
1
+a
2
+....+a
2n
)=−d(
2
2n
(a
1
+a
2n
)
Also, we know a
2n
=a
1
+(2n−1)d
⇒d=
2n−1
a
2n
−a
1
⇒−d=
2n−1
a
1
−a
2n
∴Therefore, the sum is
=
2n−1
n(a
1
−a
2n
)⋅(a
1
+a
2n
)
=
2n−1
n
⋅(a
1
2
−a
2n
2
).
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