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4. (a) If \( T_{7}=-4 \) and \( S_{16}=24 \) of an arithmetic scrics, determine the first term and the constant difference of the series. The fifth term of an arithmetic sequence is 0 and thirteenth term is (b) 12. Determine the sum of the first 21 terms of sequence. (c) The 1 st term of an arithmetic sequence is 6 and the sum of the first five terms is 250 . Calculate the 12 th term of the sequence. 5. The first term and the last term of an arithmetic series is 5 and 61 respectively while the sum of all the terms is 957 . Determine the number of terms in the series. The sum of the first 10 terms of an arithmetic series is 145 and the sum of its fourth and ninth term is five times the third term. Determine the first terin and constant difference. Given is the series \( 1+2+3+4+5+\ldots+n \) (a) Show that \( S_{n}=\frac{n(n+1)}{2} \). (b) Find the sum of the first 1001 terms excluding all multiples of 7 . (betermine the first three terms of each of the following geometric (sequences of which: (a) the 6 th term is 28 and 11 th term is 896 . (b) the 2 nd term is 3 and the 4 th term is \( 6 \frac{3}{4} \).

Ask by Ortega Jimenez. in South Africa
Feb 03,2025

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Answer

First term (\( a \)) is -26 and the common difference (\( d \)) is \( \frac{11}{3} \).

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Mind Expander

An arithmetic sequence can be traced back to its roots, where ancient mathematicians used these patterns in various forms, way before our calculators could chime in! The great Greek philosopher, Euclid, studied properties of such sequences in his mathematical works. It was later refined during the Renaissance as scholars applied algebraic techniques, paving the way for modern mathematical principles. So, next time you work with these sequences, you can imagine how their elegance has evolved through centuries! When solving problems with arithmetic sequences, one common mistake is jumping directly into calculations without laying out the basic formulas first. Always remember your sequence's first term (\(a\)) and common difference (\(d\)). A simple pitfall is to confuse the formula for the \(n\)th term \(T_n = a + (n-1)d\) with the sum of the first \(n\) terms \(S_n = \frac{n}{2} [2a + (n - 1)d]\). Double-checking these formulas before you begin will save you from losing those tricky points in assessments!

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