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Q9:
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first $n$ terms.

Solution :

Given:

The sum of the first $7$ terms of an Arithmetic Progression (AP), denoted as $S_7 = 49$.

The sum of the first $17$ terms of the same AP, denoted as $S_{17} = 289$.

To Find:

The sum of the first $n$ terms of the AP, denoted as $S_n$.

Formulae Used:

The sum of the first $n$ terms of an AP is given by the formula:

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

Where $a$ is the first term and $d$ is the common difference.

Step 1: Formulating the equation for the first 7 terms

Using the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$ for $n = 7$:

$49 = \frac{7}{2} [2a + (7 - 1)d]$

$49 = \frac{7}{2} [2a + 6d]$

Divide both sides by $7$:

$7 = \frac{1}{2} [2a + 6d]$

Multiply by $2$:

$14 = 2a + 6d$

Divide the entire equation by $2$ to simplify:

$7 = a + 3d$ --- (Equation 1)

Step 2: Formulating the equation for the first 17 terms

Using the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$ for $n = 17$:

$289 = \frac{17}{2} [2a + (17 - 1)d]$

$289 = \frac{17}{2} [2a + 16d]$

Divide both sides by $17$ (since $17^2 = 289$):

$17 = \frac{1}{2} [2a + 16d]$

Multiply by $2$:

$34 = 2a + 16d$

Divide the entire equation by $2$ to simplify:

$17 = a + 8d$ --- (Equation 2)

Step 3: Solving the system of linear equations

Subtract Equation 1 from Equation 2:

$(a + 8d) - (a + 3d) = 17 - 7$

$a - a + 8d - 3d = 10$

$5d = 10$

$d = 2$

Substitute $d = 2$ into Equation 1:

$7 = a + 3(2)$

$7 = a + 6$

$a = 7 - 6$

$a = 1$

Step 4: Finding the sum of the first $n$ terms ($S_n$)

Substitute $a = 1$ and $d = 2$ into the general formula $S_n = \frac{n}{2} [2a + (n - 1)d]$:

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

$S_n = \frac{n}{2} [2 + 2n - 2]$

$S_n = \frac{n}{2} [2n]$

$S_n = n \times n$

$S_n = n^2$

Final Answer: The sum of the first $n$ terms is $n^2$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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