# Ex.5.2 Q3 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

In the following AP, find the missing terms in the boxes:

i) \begin{align}2, \boxed{\;\;}, 26 \end{align}

ii) \begin{align}\boxed{\;\;}13, \boxed{\;\;}, 3\end{align}

iii) \begin{align}5, \boxed{\;\;}, \boxed{\;\;}, 9 \frac{1}{2}\end{align}

iv) \begin{align}-4, \boxed{\;\;}, \boxed{\;\;}, \boxed{\;\;}, \boxed{\;\;} \end{align}

v) \begin{align}\boxed{\;\;}, 38, \boxed{\;\;}, \boxed{\;\;}, \boxed{\;\;},-22 \end{align}

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 3

## Text Solution

Reasoning:

\begin{align}{a_n} = a + \left( {n - 1} \right)d\end{align}

i) What is Known?

The AP with missing terms.

What is Unknown?

Second term.

Steps:

Let common difference be $$d.$$

First term \begin{align}a = 2 \end{align}

\begin{align} \text { Third term } &=a+2d \\ &=2+2d \end{align}

\begin{align} \text { Third term } &=26(\text { Given }) \\ 2+2d&=26 \\ 2d&=26-2 \\ 2d&=24 \\ d&=12 \end{align}

\begin{align} \text { Second term } &=a+d \\ &=2+12 \\ &=14 \end{align}

The missing term in the box is $$14.$$

ii) What is Known?

The AP with missing terms.

What is Unknown?

First and the third term.

Steps:

\begin{align}{a_n} = a + (n - 1)d\\{a_2} = 13,\,\,\,{a_4} = 3\end{align}

Second term $$= 13$$

\begin{align}a + (2 - 1)d &= 13\\a + d &= 13 \qquad\ldots (1)\end{align}

Fourth term $$= 3$$

$$\begin{array}{l} a + (4 - 1)d = 3\\ a + 3d = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots (2) \end{array}$$

Solving $$(1)$$ and $$(2)$$ for $$a$$ and $$d$$

Equation $$(2)$$ – Equation $$(1)$$ gives

\begin{align}{3d-d}&=3-13 \\ {2d}&=-10 \\ {d}&=-5\end{align}

Putting $$d$$ in equation $$(1)$$

\begin{align}{a + (-5)}&=13 \\ {a-5}&=13 \\ {a} &= 18\end{align}

\begin{align} \text { Third term } &=18-10 \\ &=8 \end{align}

Hence the missing terms in the boxes are $$18$$ and $$8.$$

iii) What is Known?

The AP with missing terms.

What is Unknown?

Missing terms in the boxes.

Steps:

First term $$a = 5$$

Fourth term \begin{align}{a_4} = \frac{{19}}{2}\end{align}

\begin{align} \text { Fourth term }{a_4}&= a +(4-1) d \\ &=5+3 d \\ 5+3 d &=\frac{19}{2} \\ 6 d &=9 \\ d &=\frac{3}{2} \end{align}

\begin{align}{\text { Second term }{a_2}}&=a+d \\&=5+\frac{3}{2} \\ &=\frac{13}{2}\end{align}

\begin{align}{\text{Third term }{a_3}}&= a + 2d\\ &= 5 + 2 \times \frac{3}{2}\\ &= 5 + 3\\ &= 8\end{align}

The missing terms in the boxes are \begin{align}\frac{{13}}{2}\end{align} and $$8.$$

iv) What is Known?

The AP with missing terms.

What is Unknown?

Missing terms in the boxes.

Steps:

First term \begin{align} a = -4 \end{align}

\begin{align} {\text { Sixth term } a_{6}}&=6 \\ a+5d &=6 \\-4+5d&=6 \\ 5d&=10 \\d&=2 \end{align}

\begin{align} {\text{Second term}}\!=\!a + d \!=\!\!-\! 4 + 2 =\,{\rm{ - 2}}\\{\text{Third term = a + 2d = }}{\rm{ - 4 + 4 = 0}}\\{\text{Fourth term = a + 3d = }}{\rm{ - 4 + 6 = 2}}\\{\text{Fifth term = a + 4d = }}\,\,{\rm{ - 4 + 8 = 4}} \end{align}

Hence the missing terms are \begin{align}-2,0,2,4. \end{align}

v) What is Known?

The AP with missing terms.

What is Unknown?

Missing terms in the boxes.

Steps:

Let the first term be \begin{align}= a. \end{align}

Common difference \begin{align}= d \end{align}

Second term \begin{align}= 38 \end{align} (Given)

\begin{align} a+d = 38 \quad…(1) \end{align}

Sixth term \begin{align} = -22 \end{align} (Given)

Sixth term \begin{align} = a + 5d = -22\quad \dots(2) \end{align}

Solving $$(1)$$ and $$(2)$$for $$a$$ and $$d$$

\begin{align}{4d}&= - 60 \\ {d}&=-15\end{align}

Put the $$d$$ value in …$$(1)$$

\begin{align}{a-15}&=38 \\ {a}&=53\end{align}

\begin{align} \text { Third term } &=a+2d \\ &=53-30=23 \end{align}

\begin{align} \text { Fourth term } &=a+3 d \\ &=53-45=8 \end{align}

\begin{align} \text { Fifth term } &=a+4d \\ &=53-60=-7 \end{align}

Hence the missing terms are \begin{align} 53, 23, 8, -7. \end{align}

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