Question

For what value of n, are the ${n^{th}}$ terms of two APs $63, 65, 67,...$ and $3, 10, 17, …$ equal?

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
${a_n}=a+ (n− 1)d$
Where d -> common difference
A -> first term
n-> term
${a_n} -> {n^{th}}$ term

Complete step-by-step answer:
Lets first consider AP 63, 65, 67…
First term = a = 63, Common difference = d = 65 – 63 = 2
Using formula , to find nth term of arithmetic progression,
${a_n} = 63 + (n − 1) (2)$… (1)
Now, consider second AP 3, 10, 17…
First term = a = 3, Common difference = d = 10 – 3 = 7
Using formula , to find nth term of arithmetic progression,
${a_n} = 3 + (n − 1) (7)$… (2)
According to the given condition:
\[\begin{array}{*{20}{l}}
  {\left( 1 \right){\text{ }} = {\text{ }}\left( 2 \right)} \\
  { \Rightarrow 63{\text{ }} + {\text{ }}(n - 1){\text{ }}\left( 2 \right){\text{ }} = {\text{ }}3{\text{ }} + {\text{ }}(n - 1){\text{ }}\left( 7 \right)} \\
  { \Rightarrow 63{\text{ }} + {\text{ }}2n-2{\text{ }} = {\text{ }}3{\text{ }} + {\text{ }}7n - 7} \\
   \Rightarrow 65{\text{ }} = {\text{ }}5n \\
   \Rightarrow n{\text{ }} = {\text{ }}13 \\
  \end{array}\]

Therefore, the ${13^{th}}$ terms of both the AP’s are equal.

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.