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Q3(vi):
In an AP: (vi) given $a = 2, d = 8, S_n = 90$, find $n$ and $a_n$.

Solution :

Given:

First term of the Arithmetic Progression ($a$) = $2$

Common difference ($d$) = $8$

Sum of $n$ terms ($S_n$) = $90$

To Find:

Number of terms ($n$) and the $n^{th}$ term ($a_n$).

Step 1: Formulating the equation for $n$ using the Sum formula

The formula for the sum of the first $n$ terms of an Arithmetic Progression is given by:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

Substituting the given values into the formula:

$90 = \frac{n}{2} [2(2) + (n - 1)8]$

Step 2: Simplifying the algebraic expression

$90 = \frac{n}{2} [4 + 8n - 8]$

$90 = \frac{n}{2} [8n - 4]$

Factor out $2$ from the bracket:

$90 = \frac{n}{2} \cdot 2(4n - 2)$

$90 = n(4n - 2)$

$90 = 4n^2 - 2n$

Step 3: Solving the quadratic equation

Rearrange the equation into the standard form $an^2 + bn + c = 0$:

$4n^2 - 2n - 90 = 0$

Divide the entire equation by $2$ to simplify:

$2n^2 - n - 45 = 0$

Using the splitting the middle term method, we look for two numbers that multiply to $(2 \times -45) = -90$ and add to $-1$:

The numbers are $-10$ and $9$.

$2n^2 - 10n + 9n - 45 = 0$

$2n(n - 5) + 9(n - 5) = 0$

$(2n + 9)(n - 5) = 0$

This gives two possible values for $n$:

$2n + 9 = 0 \implies n = -\frac{9}{2}$

$n - 5 = 0 \implies n = 5$

[Since the number of terms $n$ must be a positive integer, we reject $n = -\frac{9}{2}$]

Therefore, $n = 5$.

Step 4: Finding the $n^{th}$ term ($a_n$)

The formula for the $n^{th}$ term of an AP is:

$a_n = a + (n - 1)d$

Substitute $a = 2$, $n = 5$, and $d = 8$:

$a_5 = 2 + (5 - 1)8$

$a_5 = 2 + (4 \times 8)$

$a_5 = 2 + 32$

$a_5 = 34$

Final Answer: The number of terms $n = 5$ and the $n^{th}$ term $a_n = 34$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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