Question

Consider the arithmetic sequence 9, 15, 21.
(a). Write the algebraic form of this sequence.
(b). Find the twenty-fifth term of this sequence
(c). Find the sum of terms from twenty-fifth to fiftieth of this sequence
(d). Can the sum of some terms of this sequence be 2015? Why?

Answer 1

Hint: To find each of the answers, we apply the formulae of the sum of n terms in a sequence and the ${n^{th}}$term of the sequence. The algebraic form is expressing the sequence in form of variables. To find if the sum is 2015, we equate the sum of n terms equal to 2015 and solve for n.

Complete step-by-step solution:
Given Arithmetic sequence: 9, 15, 21
The first term a, of this sequence is a = 9 and
The common difference d is,$ d = 15 – 9 = 21 – 15 = 6$.
(a). Algebraic form
We know for a natural number n, the ${n^{th}}$term of a sequence is given by ${n^{th}}$ term = $a + (n-1)d$
i.e. ${X_n}$ = $9 + (n-1)6 = 6n + 3$, where n = 1, 2, 3…..
Hence algebraic form of the sequence 9, 15, 21 is ${X_n}$= 6n + 3
(b). Twenty fifth term of the sequence
We know the ${n^{th}}$ term of the sequence is given by ${n^{th}}$ term = $a + (n-1) d$
${25^{th}}$ term = $9 + (25 – 1)6 = 153$
(c). Sum of terms from twenty fifth to fiftieth.
We know the sum of first n terms of a sequence is given by, ${S_n} = \dfrac{1}{2}n\left[ {{X_1} + {X_n}} \right]$, where ${X_1}$ is the first term of the sequence and ${X_n}$ is the ${n^{th}}$ term of the sequence.
(Sum of terms from twenty-fifth to fiftieth, Sum of first fifty terms subtracted by the sum of first twenty-four terms.
The ${24^{th}}$term of the sequence is given by, $9 + (24 – 1)6 = 147$
The ${50^{th}}$term of the sequence is given by, $9 + (50 – 1)6 = 303$
Sum of the first 24 terms, ${S_{24}} = \dfrac{{24}}{2}\left[ {{X_1} + {X_{24}}} \right] = 12\left[ {9 + 147} \right] = 1872$
Sum of the first 50 terms, ${S_{50}} = \dfrac{{50}}{2}\left[ {{X_1} + {X_{50}}} \right] = 25\left[ {9 + 303} \right] = 7800$
Sum of the terms from twenty fifth to fiftieth
= ${S_{50}} - {S_{24}}$
= $7800 – 1872$
= $5928.$
(d). Sum of some terms of this sequence be 2015?
Let the sum of the first n terms of the sequence be 2015, ${S_n} = \dfrac{1}{2}n\left[ {{X_1} + {X_n}} \right]$
$ \Rightarrow 2015 = \dfrac{1}{2}n\left[ {9 + 6n + 3} \right]$
$ \Rightarrow 2015 = 3{n^2} + 6n$
$ \Rightarrow 3{n^2} + 6n - 2015$ ……… (i)
Now, using the formula for finding the roots of a quadratic equation $a{x^2} + bx + c = 0$ is given by, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, provided ${b^2} - 4ac \geqslant 0$
Comparing equation (i), with this, we get a = 3, b = 6 and c = -2015.
$n = \dfrac{{ - 6 \pm \sqrt {{6^2} - 4 \times 3 \times \left( { - 2015} \right)} }}{{2 \times 3}}$
$n = \dfrac{{ - 6 \pm \sqrt {24216} }}{6}$
$n = \dfrac{{ - 6 \pm 2\sqrt {6054} }}{6}$
Here, we can see that the value of n for equation (i) is not a natural number.
Hence 2015 cannot be a sum of some terms in this sequence.

Note: In order to solve this type of question the key is to know the formulae of the sum of n terms and the ${n^{th}}$ term of a sequence. It is important to observe that sum of terms from twenty-fifth to fiftieth is the sum of the first fifty terms subtracted by the sum of the first twenty-four terms.
Also, in the formulae of sequences n can take only whole number values, i.e. 0, 1, 2 ……