I an A.P. if S1 = T1 + T2 + T3 + T4 + ...........+ Tn ( n is odd ) and S2 = T2 + T4 + T6 + .... + T(n-1), then find the value of s1/S2 in terms of n.
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Step-by-step explanation:
ANSWER
S
n
=
2
n
[2a+(n−1)d]
S
1
=
2
n
[2a+(n−1)d]...(1)
S
2
=
2
2
n−1
{2(a+d)+[
2
(n−1)
−1]2d}
S
2
=
4
(n−1)
[2a+2d+(n−1)d−2d]
S
2
=
4
(n−1)
[2a+(n−1)d]...(2)
(1)÷(2) gives
S
2
S
1
=
4
(n−1)
[2a+(n−1)d]
2
n
[2a+(n−1)d]
S
2
S
1
=
2
n
×
(n−1)
4
S
2
S
1
=
n−1
2n
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