Question

In the following A.P’s, find the missing terms in the blank spaces:
i) 2,___, 26
ii) __, 13 ,___, 4
iii) 5, __, ___, $ 9\dfrac{1}{2} $
iv) -4, __,__,__, __. 6
v) __, 38 ,__, __, __, -22

Answer 1

Hint: A sequence is a set of terms in a definite order with a rule of obtaining its terms. One of such sequences is arithmetic progression. A sequence of numbers is said to be A.P when the difference between any two consecutive numbers is always the same.

Complete step-by-step answer:
If a is the first term and d is the common difference, the A.P can be written as a, a+d, a+2d…… and so on. Nth term in an A.P is given by $ {t_n} = a + (n - 1)d $ . Using this we can find all the missing terms in the given A.P series above.
The concept of nth term in an A.P can be used to find the missing terms be it the second, third or fourth term by changing the value of n accordingly.
I.Solving the first part we get,
a(given as the first term)=2
 we are given with third term by,
 $
\Rightarrow {a_3} = 26 = a + (3 - 1)d = a + 2d \\
  26 = 2 + 2d \\
  24 = 2d \\
  d = 12 \;
 $
We have to calculate the second term, we have the value of d(common difference)=12
So
  $
\Rightarrow {a_2} = a + (2 - 1)d = a + d \\
\Rightarrow {a_2} = 2 + 12 = 14 \\
 $
Now let us solve the second part,
II.The first term here and $ {a_3} $ the last term is not given. We have $ {a_2} = 13 = a + d $ and $ {a_4} = 3 = a + (4 - 1)d = a + 3d $
We can find the value of d by
  $
\Rightarrow {a_4} - {a_2} = a + 3d - a - d \\
  {a_4} - {a_2} = 2d \\
  3 - 13 = 2d \\
   - 10 = 2d \;
  d = - 5 \;
 $
And similarly a can be found out by substituting the values for fourth or second term
 $
\Rightarrow 3 = a + 3d \\
\Rightarrow 3 = a + 3( - 5) \\
\Rightarrow 3 = a - 15 \\
\Rightarrow a = 15 + 3 = 18 \;
 $
Third term can be calculated as
 $
\Rightarrow {a_3} = a + 2d = 18 + 2( - 5) \\
\Rightarrow {a_3} = 18 - 10 = 8 \;
 $
The third part of the question can be calculated as,

III.We have been given with the first term a=5 and fourth term from which we can calculate the value of d by resolving the mixed fraction and taking the L.C.M.
 $
\Rightarrow {a_4} = a + (4 - 1)d = 5 + 3d \\
  9\dfrac{1}{2} = 5 + 3d \\
  9\dfrac{1}{2} - 5 = 3d \\
  \dfrac{{19}}{2} - 5 = 3d \\
  \dfrac{{19 - 10}}{2} = \dfrac{9}{2} = 3d \\
\Rightarrow d = \dfrac{3}{2} \;
 $
Second and third term can be calculated as,
 $
\Rightarrow {a_2} = a + d = 5 + \dfrac{3}{2} = \dfrac{{10 + 3}}{2} = \dfrac{{13}}{2} \\
\Rightarrow {a_3} = a + 2d = 5 + 2\left( {\dfrac{3}{2}} \right) = 5 + 3 = 8 \;
 $

IV.The fourth part of the question has given the value for a=-4 and sixth term as 6.therefore it can be solved just like the part above
 $
\Rightarrow {a_6} = a + (6 - 1)d = a + 5d = - 4 + 5d \\
\Rightarrow 6 = - 4 + 5d \\
\Rightarrow 10 = 5d \\
\Rightarrow d = 2 \;
 $
Therefore all the missing terms can be calculated now since we have the value of both and a by,
 $
\Rightarrow {a_2} = a + d = - 4 + 2 = - 2 \\
\Rightarrow {a_3} = a + 2d = - 4 + 2(2) = 0 \\
\Rightarrow {a_4} = a + 3d = - 4 + 3(2) = 2 \\
\Rightarrow {a_5} = a + 4d = - 4 + 4(2) = 4 \;
$

V.In the last part of the question we have the value of second term and the sixth term
 $
\Rightarrow {a_2} = 38 = a + d \\
\Rightarrow {a_6} = - 22 = a + 5d \;
 $
We can find the value of d by
 $
\Rightarrow {a_6} - {a_2} = - 22 - 38 = a + 5d - a - d \\
   - 60 = 4d \\
\Rightarrow d = \dfrac{{ - 60}}{4} = - 15 \;
 $
Value of a can be found out by substituting the value of d in second term
 $
\Rightarrow {a_2} = 38 = a + ( - 15) \\
\Rightarrow a = 38 + 15 = 53 \;
 $
Similarly the other missing terms can be found out as
 $
\Rightarrow {a_3} = a + 2d = 53 + 2( - 15) = 53 - 30 = 23 \\
\Rightarrow {a_4} = a + 3d = 53 + 3( - 15) = 53 - 45 = 8 \\
\Rightarrow {a_5} = a + 4d = 53 + 4( - 15) = 53 - 60 = - 7 \;
 $

Note: If a fixed number is added to or subtracted from each term of a given A.P, then the resulting sequence is also an A.P. with the same common difference as that of the given A.P. same is with multiplication also.