Chapter ISG, Problem 7.10.4P

a.

To determine

To identify: the type of the sequence 1, 2, 3….100 and explore a way to find its sum based on grouping pairs of numbers from each end of the sequence.

Answer to Problem 7.10.4P

Sequence is Arithmetic.

The sum of n term of the given sequence is $S_{100}=5050$ .

Explanation of Solution

Given information: Given sequence is 1, 2, 3….100

Calculation:

Given sequence is 1, 2, 3….100.

To identify the type of the given sequence, first subtract each term from the term that follows it.

  $2-1=3-2=4-3=5-4=1.$

There is a common difference of d= 1 so, the given sequence is arithmetic

Use the formula for an arithmetic sequence.

  $\begin{array}{l}{a}_n=a_{n-1}+d\\ a_n=a_{n-1}+1\end{array}$

The first term is 1 and $a_n=a_{n-1}+1$$,2\le n\le 100$ .

Then $\begin{array}{l}{a}_1=1\\ a_2=1+1\\ a_3=1+2\\ \mathrm{.......}\\ \mathrm{.......}\\ a_{100}=1+99\end{array}$

The sum of n term of the given sequence is:

  $\begin{array}{l}{S}_{100}=a_1+a_2+\mathrm{........}{a}_{99}+a_{100}\\ S_{100}=1+(1+1)+(1+2)\mathrm{.............}(1+98)+(1+99)\mathrm{......}(1)\\ \text{By}\text{writing}\text{the}\text{term}\text{in}\text{reverse}\text{order}\text{.}\\ S_{100}=(1+98)+(1+99)\mathrm{............}(1+2)+(1+1)+\mathrm{1...}(2)\\ Adding(1)and(2).\\ 2S_{100}=(2+99)+(2+98)\mathrm{..........}(2+1)+2\\ 2S_{100}=100(2+99).\\ 2S_{100}=10100.\\ S_{100}=5050\end{array}$

b.

To determine

To find : an explicit formula for the sum S of n terms of an athematic sequence whose first term is $a_1$ and whose nth or last term is $a_n$

Answer to Problem 7.10.4P

S = $\frac{n}{2}$ [2 $a_1$ + ( n - 1)d].

Explanation of Solution

Given information: Suppose, $a_1,a_2,a_3\mathrm{..........}{a}_n$ be an Arithmetic Progression whose first term is $a_1$ and common difference is d.

Calculation:

Then,

a1 = $a_1$

a2 = $a_1$ + d

a3 = $a_1$ + 2d

a4 = $a_1$ + 3d

………..

………..

  $a_n$ = $a_1$ + (n - 1)d

Now,

S = a1 + a2 + a3 + ………….. + an-1 + an

S = $a_1$ + ( $a_1$ + d) + ( $a_1$ + 2d) + ( $a_1$ + 3d) + ……….. + { $a_1$ + (n - 2)d} + { $a_1$ + (n - 1)d} ……………….. (i)

By writing the terms of S in the reverse order, we get,

S = { $a_1$ + (n - 1)d} + { $a_1$ + (n - 2)d} + { $a_1$ + (n - 3)d} + ……….. + ( $a_1$ + 3d) + ( $a_1$ + 2d) + ( $a_1$ + d) + $a_1$

Adding the corresponding terms of (i) and (ii), we get

2S = {2 $a_1$ + (n - 1)d} + {2 $a_1$ + (n - 1)d} + {2 $a_1$ + (n - 1)d} + ………. + { $a_1$ + (n - 2)d}

2S = n[2 $a_1$ + (n -1)d]

⇒S = $\frac{n}{2}$ [2 $a_1$ + (n - 1)d].