Question

Choose the correct choice in the following and justify:
 1) 30th term of the AP: 10,7,4,..., is
A) 97
B) 77
C) −77
D) 87
2) 11th term of the AP: −3,$ - \dfrac{1}{2}$ 2,...., is
A) 28
B) 22
C) 38
D) \[ - 48\dfrac{1}{2}\]​

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:
Given that
A.P. 10,7,4,…
First term, a=10
Common difference, d=a2​-a1​=7−10=−3
We know that
\[
  {a_n} = a + {\text{ }}\left( {n - {\text{ }}1} \right)d \\
  {a_{30}} = 10 + {\text{ }}\left( {30 - {\text{ }}1} \right)( - 3) = 10 + (29)( - 3) = - 77 \\
 \]
Therefore the 30th term of the AP is -77.

Hence, the correct answer is option C.

Given that
A.P. is −3,2−1​,2,......
First term a=−3
Common difference, \[d = {a_2} - {a_1} = 2 - 1 - \left( { - 3} \right) = \dfrac{5}{2}\] ​
We know that
\[
  {a_n} = a + {\text{ }}\left( {n - {\text{ }}1} \right)d \\
  {a_{11}} = - 3 + {\text{ }}\left( {11 - {\text{ }}1} \right)\left( {\dfrac{5}{2}} \right) = - 3 + 25 = 22 \\
 \]
Therefore the 11th term of the AP is 22.

Hence, the correct answer is option B.

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.