Question

In the following APs, find the missing terms in the boxes:
A) $2,\;\quad \boxed{},\quad 26$
B) $\boxed{},\;\quad 13,\quad \boxed{},\quad 3$
C) $5,\quad \boxed{},\;\quad \boxed{},\quad 9\dfrac{1}{2}$
D) $ - 4,\quad \boxed{},\;\quad \boxed{},\quad \boxed{},\quad \boxed{},\quad 6$
E) $\boxed{},\quad 38,\;\quad \boxed{},\quad \boxed{},\quad \boxed{},\quad - 2$

Answer 1

Hint: An arithmetic progression can be given by a, (a+d), (a+2d), (a+3d), ……
a, (a+d), (a+2d), (a+3d),….. where a = first term, d = common difference.
a,b,c are said to be in AP if the common difference between any two consecutive number of the series is same ie \[b - a = c - b \Rightarrow 2b = a + c\]
Formula to consider for solving these questions
an=a+ (n− 1)d
Where d -> common difference
A -> first term
n-> term
an -> nth term

Complete step-by-step answer:
A) $2,\;\quad \boxed{},\quad 26$
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
\[\begin{array}{*{20}{l}}
  { \Rightarrow x - 2 = 26 - x} \\
  { \Rightarrow 2x = 28} \\
  { \Rightarrow x = \dfrac{{28}}{2}} \\
  { \Rightarrow x = 14}
\end{array}\]
Therefore, the missing term is 14.
 B) $\boxed{},\;\quad 13,\quad \boxed{},\quad 3$
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
\[\begin{array}{*{20}{l}}
  { \Rightarrow y - 13 = 3 - y} \\
  { \Rightarrow 2y = 16} \\
  { \Rightarrow y = \dfrac{{16}}{2} = 8}
\end{array}\]
Also, using property (1), we can say that
\[\begin{array}{*{20}{l}}
  { \Rightarrow 13 - x = y - 13} \\
  { \Rightarrow x + y = 26}
\end{array}\]
But, we have y=8, Putting value of y in the above equation, we get
\[
  \; \Rightarrow x + 8 = 26 \\
   \Rightarrow x = 18 \\
 \]
Therefore, missing terms are 18 and 8.
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.
C) $5,\quad \boxed{},\;\quad \boxed{},\quad 9\dfrac{1}{2}$
Here, first term = a=5
And, 4th term \[ = {a_4} = \dfrac{{19}}{2}\]
Using formula an=a+(n−1)d, to find nth term of arithmetic progression, we get
\[
  \;\; \Rightarrow {a_4} = 5 + \left( {4 - 1} \right)d\;\; \\
   \Rightarrow \dfrac{{19}}{2} = 5 + 3d\;\; \\
   \Rightarrow 3d = \dfrac{{19}}{2} - 5\; \\
   \Rightarrow 3d = \dfrac{9}{2} \\
   \Rightarrow d = \dfrac{3}{2} \\
 \]
Therefore, we get common difference = \[d{\text{ }} = \;\dfrac{3}{2}\]
Second term = \[a + {\text{ }}d{\text{ }} = {\text{ }}\;5 + \dfrac{3}{2} = \dfrac{{13}}{2}\]
Third term = second term + d \[ = \;\dfrac{{13}}{2} + \dfrac{3}{2} = 8\]
Therefore, missing terms are \[\dfrac{{13}}{2}\] and 8
D) $ - 4,\quad \boxed{},\;\quad \boxed{},\quad \boxed{},\quad \boxed{},\quad 6$
Here, First term = a = -4
6th term = a6 = 6
      Using formula an=a+(n−1)d, to find nth term of arithmetic progression, we get
      \[\begin{array}{*{20}{l}}
  { \Rightarrow {a_6} = - 4 + \left( {6 - 1} \right)d} \\
  { \Rightarrow 6 = - 4 + 5d} \\
  { \Rightarrow 5d = 10} \\
  { \Rightarrow d = \dfrac{{10}}{5} = 2}
\end{array}\]
      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.
 E) $\boxed{},\quad 38,\;\quad \boxed{},\quad \boxed{},\quad \boxed{},\quad - 2$
In this problem, we are given 2nd and 6th term.
Using formula an=a+(n−1)d, to find nth term of arithmetic progression, we get
\[\begin{array}{*{20}{l}}
  {{a_2} = a + \left( {2 - 1} \right)d\;\;{\text{ }}\;{\text{ }}and{\text{ }}\;{\text{ }}\;{\text{ }}\;{a_6} = a + \left( {6 - 1} \right)d} \\
  { \Rightarrow 38 = a + d\;\;{\text{ }}and{\text{ }}\;{\text{ }}\; - 22 = a + 5d}
\end{array}\]
These are equations in two variables, we can solve them using any method. Lets solve them using a 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
\[\begin{array}{*{20}{l}}
  { \Rightarrow - 22 = 38 - d + 5d} \\
  { \Rightarrow 4d = - 60} \\
  { \Rightarrow d = - \dfrac{{60}}{4} = - 15}
\end{array}\]
Using this value of d and putting this in equation 38=a+d, we get
\[\begin{array}{*{20}{l}}
  { \Rightarrow 38 = a - 15} \\
  { \Rightarrow a = 53}
\end{array}\]
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
Fifth term = fourth term + d = 8 - 15 = -7
Therefore, missing terms are 53, 23, 8 and -7.

Note: To solve most of the problems related to AP, the terms can be conveniently taken as
3 terms: (a−d),a,(a+d)
4 terms: (a−3d),(a−d),(a+d),(a+3d)
5 terms: (a−2d),(a−d),a,(a+d),(a+2d)
${t_n}={S_n}-{S_{n-1}}$
If each term of an AP is increased, decreased, multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.
In an AP, the sum of terms equidistant from beginning and end will be constant.