Question

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.​

Answer 1

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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