# Ex.5.3 Q10 Arithmetic Progressions Solution - NCERT Maths Class 10

Go back to  'Ex.5.3'

## Question

Show that $${a_1},{\rm{ }}{a_2},...{\rm{ }},{\rm{ }}{a_n},{\rm{ }}...$$form an AP where $${a_n}$$ is defined as below

(i) $${a_n} = 3 + 4n$$

(ii) $${a_n} = 9 - 5n$$

Also find the sum of the first $$15$$ terms in each case.

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 10

## Text Solution

(i) $${a_n} = 3 + 4n$$

What is Known?

$$\,{a_n} = 3 + 4n$$

What is Unknown?

Whether $${a_1},{\rm{ }}{a_2},...{\rm{ }},{\rm{ }}{a_n},{\rm{ }}...$$ form an AP

Reasoning:

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

Steps:

Given,

• $$nth$$ term, $${a_n} = 3 + 4n$$

\begin{align}{a_1} &= 3 + 4 \times 1\\&= 7\\{a_2} &= 3 + 4 \times 2 = 3 + 8\\&= 11\\{a_3} &= 3 + 4 \times 3 = 3 + 12\\&= 15\\{a_4} &= 3 + 4 \times 4 = 3 + 16\\&= 19\end{align}

It can be observed that

\begin{align}{a_2} - {a_1} &= 11 - 7 \\&= 4\\{a_3} - {a_2} &= 15 - 11\\& = {\rm{ }}4\\{a_4} - {a_3} &= 19 - 15\\& = {\rm{ }}4\end{align}

i.e., the difference of $${a_n}$$ and $${a_{n - 1}}$$ is constant.

Therefore, this is an AP with common difference as $$4$$ and first term as $$7.$$

Sum of $$n$$ terms,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{15}} &= \frac{{15}}{2}\left[ {2 \times 7 + \left( {15 - 1} \right)4} \right]\\ &= \frac{{15}}{2}\left[ {14 + 14 \times 4} \right]\\ &= \frac{{15}}{2} \times 70\\ &= 15 \times 35\\ &= 525\end{align}

(ii) $${a_n} = 9 - 5n$$

What is Known?

$${a_n} = 9 - 5n$$

What is Unknown?

Whether $${a_1},{\rm{ }}{a_2},...{\rm{ }},{\rm{ }}{a_n},{\rm{ }}...$$ form an AP

Reasoning:

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

Steps:

Given,

• $$nth$$ term, $${a_n} = 9 - 5n$$

\begin{align}{a_1} &= 9 - 5 \times 1\\&= 9 - 5 \\&= 4\\{a_2} &= 9 - 5 \times 2\\&= 9 - 10\\&= - 1\\{a_3} &= 9 - 5 \times 3 \\&= 9 - 15\\&= - 6\\{a_4}&= 9 - 5 \times 4\\&= 9 - 20\\&= - 11\end{align}

It can be observed that

\begin{align}{a_2} - {a_1} &= ( - 1{\rm{ )}} - {\rm{ }}4 \\&= - 5\\{a_3} - {a_2}& = ( - 6) - ( - 1)\\& = - 5\\{a_4} - {a_3} &= ( - 11) - ( - 6) \\&= - 5\end{align}

i.e., the difference of $${a_n}$$ and $${a_{n - 1}}$$ is constant.

Therefore, this is an A.P. with common difference as $$(−5)$$ and first term as $$4.$$

Sum of n terms of AP,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{15}} &= \frac{{15}}{2}\left[ {2 \times 4 + \left( {15 - 1} \right)\left( { - 5} \right)} \right]\\ &= \frac{{15}}{2}\left[ {8 + 14\left( { - 5} \right)} \right]\\ &= \frac{{15}}{2}\left[ {8 - 70} \right]\\ &= \frac{{15}}{2}\left[ { - 62} \right]\\& = - 465\end{align}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school