Question

If the sum of the first n terms of an AP is $4n - {n^2}$, what is the first term? What is the sum of the first two terms? also find the ${2^{nd}},{3^{rd}},{10^{th}}$and the \[{n^{th}}\] terms of AP.

Answer 1

Hint: First, we will know what AP is.
An arithmetic progression can be given by $a,(a + d),(a + 2d),(a + 3d),...$where $a$is the first term and $d$is a common difference.
$a,b,c$are said to be in arithmetic progression if the common difference between any two-consecutive number of the series is the same that is $b - a = c - b \Rightarrow 2b = a + c$
Given that the sum of the terms are $4n - {n^2}$and we need to find the arithmetic progression for many values.
 Formula used: ${a_n} = a + (n - 1)d$ where $d$is the common difference, $a$is the first term, since we know that difference between consecutive terms is constant in any A.P

Complete step by step answer:
Since we need to find some terms for the given AP is$4n - {n^2}$, take this as ${S_n} = 4n - {n^2}$(generalized form).
Hence starting with the first term put $n = 1$then we get${S_n} = 4n - {n^2} \Rightarrow {S_1} = 4(1) - {(1)^2}$.
Further solving this we get, ${S_1} = 4(1) - {(1)^2} \Rightarrow {S_1} = 3$(since it is the first term, we can also able to say this is $a$in the given AP), thus we get${S_1} = a = 3$.
Similarly, for the second term, we have to put$n = 2$; then we get${S_n} = 4n - {n^2} \Rightarrow {S_2} = 4(2) - {(2)^2}$, further solving this we get${S_2} = 4(2) - {(2)^2} \Rightarrow 4$.
The second question is about the sum of the first two terms; thus we get, ${S_2} = {a_1} + {a_2} = 4$(where one is the first term). Further solving we get${S_2} = {a_1} + {a_2} = 4 \Rightarrow 3 + {a_2} = 4 \Rightarrow {a_2} = 1$.
Now for the third term as follows, (put n as three)${S_n} = 4n - {n^2} \Rightarrow {S_3} = 4(3) - {(3)^2} \Rightarrow 12 - 9 \Rightarrow {S_3} = 3$.
For the next tenth term we need to find the common difference and thus using the third term, ${a_1} + {a_2} + {a_3} = 3$where a is the only unknown, substitute for the all other we get, $3 + 1 + {a_3} = 3 \Rightarrow {a_3} = - 1$
Hence the common difference is$d = {a_3} - {a_2} \Rightarrow - 1 - 1 = - 2$.
Now for the tenth term, we have, generalized from the ${a_n} = a + (n - 1)d \Rightarrow {a_{10}} = a + 9d \Rightarrow 3 + 9( - 2) \Rightarrow - 15$is the tenth term.
Finally, for the nth term we have ${a_n} = a + (n - 1)d \Rightarrow {a_n} = 3 + (n - 1)( - 2) \Rightarrow 5 - 2n$
Hence, we get the solution as ${S_1} = a = 3$
 ${S_2} = 4$
${a_2} = 1$
${S_3} = 3$
${a_3} = - 1$
${a_{10}} = - 15$
and ${a_n} = 5 - 2n$

Note: In AP the terms are generalized as,
3 terms in AP; $(a - d),a,(a + d)$
4 terms; $(a - 3d),(a - d),a,(a + d),(a + 3d)$
The resulting sequence also will be in AP. In an arithmetic progression, the sum of terms from beginning and end will be constant.
We have found the AP for first two terms and then generalized for the sum of the first two terms by ${S_2} = {a_1} + {a_2} = 4$