Question

Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

	a	d	n	an
	7	3	8	….
	-18	….	10	0
	….	-3	18	-5
	-18.9	2.5	….	3.6
	3.5	0	105	….

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⇒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:
It is given that
i) $a = 7,d = 3,n = 8$
Now, we know that
\[
  {a_n} = a + {\text{ }}\left( {n - {\text{ }}1} \right)d \\
  {a_8} = 7 + {\text{ }}\left( {8 - {\text{ }}1} \right)3 = 28 \\
 \]
Therefore,
\[{a_8} = 28\]
ii) It is given that
$a = - 18,n = 10,{a_{10}} = 0$
Now, we know that
\[
  {a_n} = a + {\text{ }}\left( {n - {\text{ }}1} \right)d \\
  {a_{10}} = - 18 + {\text{ }}\left( {10 - {\text{ }}1} \right)d = - 18 + 9d \\
   \Rightarrow d = \dfrac{{18}}{9} = 2 \\
 \]
Therefore,
\[d = 2\]
iii) It is given that
$d = - 3,n = 18,{a_{18}} = - 5$
Now, we know that
\[
  {a_n} = a + {\text{ }}\left( {n - {\text{ }}1} \right)d \\
  {a_{18}} = a + {\text{ }}\left( {18 - {\text{ }}1} \right)( - 3) = - 51 + a \\
   \Rightarrow a = 51 + {a_{18}} = 51 - 5 = 46 \\
 \]
Therefore,
\[a = 46\]
iv) It is given that
$a = - 18.9,d = 2.5,{a_n} = 3.6$
Now, we know that
\[
  {a_n} = a + {\text{ }}\left( {n - {\text{ }}1} \right)d \\
  3.6 = - 18.9 + {\text{ }}\left( {n - {\text{ }}1} \right)(2.5) = - 22.5 + 2.5n \\
   \Rightarrow n = \left( {\dfrac{{3.6 + 22.5}}{{2.5}}} \right) = 10 \\
 \]
Therefore,
\[n = 10\]
v) It is given that
$a = 3.5,d = 0,n = 105$
Now, we know that
\[
  {a_n} = a + {\text{ }}\left( {n - {\text{ }}1} \right)d \\
  {a_{105}} = 3.5 + {\text{ }}\left( {105 - {\text{ }}1} \right)0 = 3.5 \\
 \]
Therefore,
\[{a_{105}} = 3.5\]

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.