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

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## 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.

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 11

## 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.

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