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

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 \(\begin{align} a = -4 \end{align}\)

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

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

 

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 \(\begin{align} 53, 23, 8, -7. \end{align}\)