Question

(a) Write the algebraic form of the arithmetic sequence 1, 4, 7, 10…
(b) Is 100 a term of the sequence? Why?
(c) Prove that square of any term of this sequence is also a term of it.

Answer 1

Hint: We find the common difference using terms of the given sequence. Form algebraic form of AP using general form of nth term of AP. Put nth term as 100 and check if it satisfies the equation of nth term of AP. For part (c) we take the square of the general term obtained in part (a) and use the factorization method to reduce it into smaller forms.

Complete step-by-step answer:
(a)
We are given the arithmetic sequence 1, 4, 7, 10…
Here first term is 1
\[ \Rightarrow a = 1\]
Since second term is 4, we can find the common difference by subtracting first term from second term
\[ \Rightarrow d = 4 - 1\]
\[ \Rightarrow d = 3\]
We know any arithmetic sequence has nth term given by the formula \[{a_n} = a + (n - 1)d\]
Here we substitute the value of \[a = 1,d = 3\]
\[ \Rightarrow {a_n} = 1 + (n - 1)3\]
Multiplying the terms in RHS
\[ \Rightarrow {a_n} = 1 + 3n - 3\]
\[ \Rightarrow {a_n} = 3n - 2\]................… (1)
\[\therefore \]Algebraic form of arithmetic series is \[{a_n} = 3n - 2\]
(b)
We have to check if the number 100 is a term of the AP
Substitute the value of \[{a_n} = 100\]in equation (1)
\[ \Rightarrow 100 = 3n - 2\]
Shift all constant values to one side of the equation
\[ \Rightarrow 3n = 100 + 2\]
\[ \Rightarrow 3n = 102\]
Divide both sides by 3
\[ \Rightarrow \dfrac{{3n}}{3} = \dfrac{{102}}{3}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow n = 34\]
Since we got n as a whole number, we can say 100 is the 34th term of the AP.
(c)
We have to prove that square of any term of this sequence is also a term of it
Since the general term in equation (1) I.e. \[{a_n} = 3n - 2\] is the term of the sequence.
Square both sides of the term
\[ \Rightarrow {\left( {{a_n}} \right)^2} = {\left( {3n - 2} \right)^2}\]
Use the identity \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]in RHS
\[ \Rightarrow {\left( {{a_n}} \right)^2} = {(3n)^2} + {(2)^2} - 2(3n)(2)\]
\[ \Rightarrow {\left( {{a_n}} \right)^2} = 9{n^2} - 12n + 4\]
Now we factorize this quadratic equation
We can write \[4 = 6 - 2\]
\[ \Rightarrow {\left( {{a_n}} \right)^2} = 9{n^2} - 12n + 6 - 2\]
Take 3 common from first three terms in RHS
\[ \Rightarrow {\left( {{a_n}} \right)^2} = 3\left[ {3{n^2} - 4n + 2} \right] - 2\]
We can write \[3\left[ {3{n^2} - 4n + 2} \right] = 3\left[ {3{n^2} - 4n + 2 + 2 - 2} \right]\]
i.e. \[3\left[ {3{n^2} - 4n + 2} \right] = 3\left[ {3{n^2} - 4n + 4 - 2} \right]\]
i.e. \[3\left[ {3{n^2} - 4n + 2} \right] = 3\left[ {{{(3n - 2)}^2} - 2} \right]\]
\[ \Rightarrow {\left( {{a_n}} \right)^2} = 3\left[ {{{(3n - 2)}^2} - 2} \right] - 2\]
Put \[\left[ {{{(3n - 2)}^2} - 2} \right] = k\]
\[ \Rightarrow {\left( {{a_n}} \right)^2} = 3k - 2\]
So, the square of the term is a term of the arithmetic progression as it is of the type \[{a_n} = 3n - 2\]
Hence proved

Note: * An arithmetic progression is a sequence of terms having common differences between them. If ‘a’ is the first term of an AP, ‘d’ is the common difference, then the $n^{th}$ term of an AP can be found as \[{a_n} = a + (n - 1)d\] .