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Q3(ii):
In an AP: (ii) given $a = 7, a_{13} = 35$, find $d$ and $S_{13}$.

Solution :

Given:

In an Arithmetic Progression (AP):

  • First term, $a = 7$
  • The $13^{th}$ term, $a_{13} = 35$

To Find:

  • Common difference, $d$
  • Sum of the first 13 terms, $S_{13}$

Step 1: Finding the common difference ($d$)

The formula for the $n^{th}$ term of an Arithmetic Progression is given by:

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

Substituting the given values for the $13^{th}$ term ($n = 13$):

$a_{13} = a + (13 - 1)d$

$35 = 7 + 12d$ [Substituting $a_{13} = 35$ and $a = 7$]

Subtracting 7 from both sides:

$35 - 7 = 12d$

$28 = 12d$

Dividing both sides by 12:

$d = \frac{28}{12}$

Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, 4:

$d = \frac{7}{3}$

Step 2: Finding the sum of the first 13 terms ($S_{13}$)

The formula for the sum of the first $n$ terms of an AP when the first and last terms are known is:

$S_n = \frac{n}{2}(a + a_n)$

Here, $n = 13$, $a = 7$, and $a_{13} = 35$. Substituting these values into the formula:

$S_{13} = \frac{13}{2}(7 + 35)$

$S_{13} = \frac{13}{2}(42)$

Performing the division inside the expression:

$S_{13} = 13 \times \left(\frac{42}{2}\right)$

$S_{13} = 13 \times 21$

Calculating the product:

$S_{13} = 273$

Final Answer: The common difference $d = \frac{7}{3}$ and the sum of the first 13 terms $S_{13} = 273$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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