# Arithmetic Progression Practice Questions: how to find missing terms of AP?

CBSE NCERT Solutions Chapter 5 Arithmetic Progressions Exercise 5.2 Question 3

In the following AP's find the missing terms:

(i) 2, __ , 26

(ii) __, 13, __, 3

(iii) 5, __, __, $9\frac{1}{2}$

(iv) -4. __, __, __, __, 6

(v) __, 38, __, __, __, -22

Solution (i)

2, __ , 26

We know that difference between consecutive terms is constant in any Arithmetic progression.   (1)

Let the missing term be $x$.

Using property (1), we can say that

$x-2 = 26 - x$

$\Rightarrow 2x = 28$

$\Rightarrow x = \frac{28}{2}$

$\Rightarrow x = 14$

Therefore, missing term is $14$.

Solution (ii)

__, 13, __, 3

Let missing terms be $x$ and $y$.

The sequence becomes:  $x, 13, y, 3$

We know that difference between consecutive terms is constant in any Arithmetic progression.   (1)

Using property (1), we can say that

$y- 13 = 3 - y$

$\Rightarrow 2y = 16$

$\Rightarrow y = \frac{16}{2}=8$

Also, using property (1), we can say that

$13 - x = y - 13$

$\Rightarrow x + y = 26$

But, we have $y = 8$, Putting value of $y$ in the above equation, we get

$x + 8 = 26$

$\Rightarrow x = 18$

Therefore, missing terms are 18 and 8.

Solution (iii)

We can also use the following method to solve similar problems. It is also easy. The first two questions can also be solved by the same method.

5, __, __, $9\frac{1}{2}$

Here, first term = $a = 5$

And, 4th term =${{a}_{4}}=\frac{19}{2}$

Using formula ${{a}_{n}}=a+(n-1)d$,   to find nth term of arithmetic progression, we get

${{a}_{4}}=5+(4-1)d$

$\Rightarrow\frac{19}{2} = 5 + 3d$

$\Rightarrow3d = \frac{19}{2} -5$

$\Rightarrow3d =\frac{9}{2}$
$\Rightarrow d =\frac{3}{2}$

Therefore, we get common difference = d = $\frac{3}{2}$

Second term = a+ d =  $5 +\frac{3}{2} =\frac{13}{2}$

Third term = second term + d = $\frac{13}{2} +\frac{3}{2} = 8$

Therefore, missing terms are $\frac{13}{2}$ and $8$

Solution (iv)

-4. __, __, __, __, 6

Here, First term = a = -4

6th term = $a_6$ = 6

Using formula ${{a}_{n}}=a+(n-1)d$,   to find nth term of arithmetic progression, we get

${{a}_{6}}=-4+(6-1)d$

$\Rightarrow 6 = -4 + 5d$

$\Rightarrow 5d = 10$

$\Rightarrow d = \frac{10}{5} = 2$

Therefore, common difference = d = 2

Second term = first term + d = a+ d = -4 + 2 = -2

Third term = second term + d = -2 + 2 = 0

Fourth term = third term + d = 0 + 2 = 2

Fifth term = fourth term + d = 2 + 2 = 4

Therefore, missing terms are -2, 0, 2 and 4.

Solution (v)

__, 38, __, __, __, -22

In this problem, we are given 2nd and 6th term.

Using formula ${{a}_{n}}=a+(n-1)d$,   to find nth term of arithmetic progression, we get

${{a}_{2}} = a+(2-1)d$     And      ${{a}_{6}} = a+(6-1)d$

$\Rightarrow 38 = a + d$   And    $-22 = a + 5d$

These are equations in two variables, we can solve them using any method. Lets solve them using substitution method.

Using equation ($38 = a + d$), we can say that $a = 38 -d$. Putting value of $a$ in equation ($-22 = a + 5d$),  we get

$-22 = 38 -d + 5d$

$\Rightarrow 4d = -60$

$\Rightarrow d = \frac{-60}{4} = -15$

Using this value of $d$ and putting this in equation $38 = a + d$, we get

$38 = a -15$

$\Rightarrow a = 53$

Therefore, we get $a = 53$ and $d = -15$

First term = a = 53

Third term = second term + d = 38 - 15 = 23

Fourth term = third term + d = 23 - 15 = 8

Fiftht term = fourth term + d =  8 - 15 = -7

Therefore, missing terms are 53, 23, 8 and -7.

Posted in : Arithmetic Progressions, NCERT Solutions

Tags :