# Ex.5.3 Q11 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

If the sum of the first \(n\) terms of an AP is \(4n - {n^2}\), what is the first term (that is \(S_1\))?

What is the sum of first two terms?

What is the second term? Similarly find the \(3\rm{rd,}\) the\(10\rm{th}\) and the \(n\rm{th}\) terms.

## Text Solution

**What is Known?**

\({S_n} = 4{n^2} - {n^2}\)

**What is Unknown?**

\({S_1},{\rm{ }}{S_2},{\rm{ }}{a_2},{\rm{ }}{a_3},{\rm{ }}{a_{10}}\) and \({a_n}\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(nth\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\)is the common difference and \(n\) is the number of terms and \(l\) is the last term.

**Steps:**

Given,

Sum of first \(n\) terms, \({S_n} = 4n - {n^2}\)

Therefore,

Sum of first term,

\[a = {S_1} = 4 \times 1 - {1^2} = 4 - 1 = 3\]

Sum of first two terms,

\[{S_2} = 4 \times 2 - {2^2} = 8 - 4 = 4\]

Sum of first three terms,

\[{S_3} = 4 \times 3 - {3^2} = 12 - 9 = 3\]

Second term,

\[{a_2} = {S_2} - {S_1} = 4 - 3 = 1\]

Third term,

\[{a_3} = {S_3} - {S_2} = 3 - 4 = - 1\]

Tenth term,

\[{a_{10}} = {S_{10}} - {S_9}\]

\[\begin{align}& = \left( {4 \times 10 - {{10}^2}} \right) - \left( {4 \times 9 - {9^2}} \right)\\ &= \left( {40 - 100} \right) - \left( {36 - 81} \right)\\& = - 60 + 45\\ &= - 15\end{align}\]

\(n\rm{th}\) term, \({a_n} = {S_n} - {S_{n - 1}}\)

\[\begin{align}&\!=\!\left[ {4n - {n^2}} \right]\!-\!\left[\!{4\left( {n\!-\!1} \right)\!-\!{{\left( {n\!-\!1} \right)}^2}} \!\right]\\ &= 4n - {n^2} - 4n + 4 + {\left( {n - 1} \right)^2}\\ &= 4 - {n^2} + {n^2} - 2n + 1\\ &= 5 - 2n\end{align}\]

Hence, the sum of first two terms is \(4.\)

The second term is \(1.\)

\(3\rm{rd},\; 10\rm{th},\) and \(n\rm{th}\) terms are \(−1, −15,\) and \((5 - 2n)\) respectively.