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I've hardly give-up on mathematics whenever i remember and sight this question: the sum of the first twenty-one terms of a linear sequence(A.P.) is 28, and the sum of the first twenty-eight terms is 21. Find which term of the sequence is 0 and also the sum of the terms preceding it. sir, i believe that you can convert my dreppression to happiness by solving the question effectively. may God bless you, Sir.

What you call a linear sequence is also know as an arithmetic sequence which has the form

An+1 = An + d where An+1 is the (n+1)th term of the sequence and is related to the nth term, An, by adding the fixed constant d. Note that

An = A1 + (n-1)d

The sum of the first n terms is

Sn = sum(n=1,N){An} = A1 + A2 + ... + AN.

This can also be written Sn = (n/2)(A1 + AN).

For the current problem, we have

(21/2)(A1 + A21) = (21/2)(A1 + 20d) = 28 and

(28/2)(A1 + A28) = (28/2)(A1 + 27d) = 21.

These are 2 independent equations in 2 unknowns, namely A1 and d. A little algebra gives

A1 = 3.03

d = -0.17.

We also want to find which term in the sequence is 0 (or close to it) so

Aj = 0 = A1 + (j-1)(-0.17) => j = 19.

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