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If S1, S2, S3 be the respectively the sum of terms of n, 2n, 3n an A.P. the value of S 1 ÷ (S 2 -S 3 ) is given by?
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If S1, S2, S3 be the respectively the sum of terms of n, 2n, 3n an A.P...
Question: If S1, S2, S3 be the respectively the sum of terms of n, 2n, 3n an A.P. the value of S1 ÷ (S2 -S3) is given by?

Solution:

To solve this problem, we need to first find the formulas for S1, S2, and S3:

Formula for Sum of n terms of an A.P:

Sn = n/2[2a + (n-1)d]

Where,

Sn = Sum of n terms
a = First term
d = Common difference

Formula for S1:

S1 = n/2[2a + (n-1)d]

Formula for S2:

S2 = 2n/2[2a + (2n-1)d]

Simplifying this equation, we get:

S2 = n[2a + (2n-1)d]

Formula for S3:

S3 = 3n/2[2a + (3n-1)d]

Simplifying this equation, we get:

S3 = (3n/2)[2a + (3n-1)d]

Now, substituting the formulas of S1, S2, and S3 in the expression S1 ÷ (S2 -S3), we get:

S1 ÷ (S2 -S3) = [n/2[2a + (n-1)d]] / {[n[2a + (2n-1)d]] - [(3n/2)[2a + (3n-1)d]]}

Simplifying this expression, we get:

S1 ÷ (S2 -S3) = [2/3(2n-1)]

Therefore, the value of S1 ÷ (S2 -S3) is [2/3(2n-1)].

Conclusion:

Thus, the solution to the given problem is [2/3(2n-1)].
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If S1, S2, S3 be the respectively the sum of terms of n, 2n, 3n an A.P. the value of S 1 ÷ (S 2 -S 3 ) is given by?
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