If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Step-by-Step Solution
Key Concept: For an arithmetic progression, the sum of the first n terms $S_n$ is related to the individual terms by $a_n = S_n - S_{n-1}$ (with $S_0 = 0$). Hence the first term $a_1 = S_1$, the second term $a_2 = S_2 - S_1$, etc. By substituting the given expression $S_n = 4n - n^2$ we can obtain each required term.
1. Given: $S_n = 4n - n^2$.
2. First term $a_1$ (or $S_1$):
$$a_1 = S_1 = 4(1) - 1^2 = 4 - 1 = 3.$$
3. Sum of first two terms $S_2$:
$$S_2 = 4(2) - 2^2 = 8 - 4 = 4.$$
4. Second term $a_2$:
$$a_2 = S_2 - S_1 = 4 - 3 = 1.$$
5. Third term $a_3$:
$$S_3 = 4(3) - 3^2 = 12 - 9 = 3,$$
$$a_3 = S_3 - S_2 = 3 - 4 = -1.$$
6. Tenth term $a_{10}$:
$$S_{10} = 4(10) - 10^2 = 40 - 100 = -60,$$
$$S_{9} = 4(9) - 9^2 = 36 - 81 = -45,$$
$$a_{10} = S_{10} - S_{9} = -60 - (-45) = -15.$$
7. General $n^{th}$ term $a_n$:
$$a_n = S_n - S_{n-1}$$
$$=\big[4n - n^2\big] - \big[4(n-1) - (n-1)^2\big]$$
$$= 4n - n^2 - \big[4n - 4 - (n^2 - 2n + 1)\big]$$
$$= 4n - n^2 - (4n - 4 - n^2 + 2n - 1)$$
$$= 4n - n^2 - (6n - 5 - n^2)$$
$$= -2n + 5.$$
Hence, $a_n = 5 - 2n$.
8. Verification: For $n=1,2,3,10$ the formula gives $a_1=3$, $a_2=1$, $a_3=-1$, $a_{10}=-15$, matching the values obtained earlier.
Correct Answer: First term $a_1 = 3$.
Sum of first two terms $S_2 = 4$.
Second term $a_2 = 1$.
Third term $a_3 = -1$.
Tenth term $a_{10} = -15$.
General $n^{th}$ term $a_n = 5 - 2n$.