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Q3(iv):
In an AP: (iv) given $a_3 = 15, S_{10} = 125$, find $d$ and $a_{10}$.

Solution :

Given:

In an Arithmetic Progression (AP):

1. The 3rd term, $a_3 = 15$

2. The sum of the first 10 terms, $S_{10} = 125$

To find:

1. The common difference, $d$

2. The 10th term, $a_{10}$

Step 1: Formulating equations based on the general term of an AP

The formula for the $n^{th}$ term of an AP is given by: $a_n = a + (n - 1)d$, where $a$ is the first term and $d$ is the common difference.

For $n = 3$:

$a_3 = a + (3 - 1)d$

$15 = a + 2d$ --- (Equation 1)

Step 2: Formulating equations based on the sum of $n$ terms of an AP

The formula for the sum of the first $n$ terms of an AP is given by: $S_n = \frac{n}{2} [2a + (n - 1)d]$.

For $n = 10$ and $S_{10} = 125$:

$125 = \frac{10}{2} [2a + (10 - 1)d]$

$125 = 5 [2a + 9d]$

Divide both sides by 5:

$25 = 2a + 9d$ --- (Equation 2)

Step 3: Solving the system of linear equations

From Equation 1, express $a$ in terms of $d$:

$a = 15 - 2d$

Substitute this expression for $a$ into Equation 2:

$25 = 2(15 - 2d) + 9d$

$25 = 30 - 4d + 9d$

$25 = 30 + 5d$

Subtract 30 from both sides:

$25 - 30 = 5d$

$-5 = 5d$

$d = -1$

Step 4: Finding the first term $a$

Substitute $d = -1$ back into the expression for $a$:

$a = 15 - 2(-1)$

$a = 15 + 2$

$a = 17$

Step 5: Calculating the 10th term $a_{10}$

Using the formula $a_n = a + (n - 1)d$ for $n = 10$:

$a_{10} = a + (10 - 1)d$

$a_{10} = 17 + 9(-1)$

$a_{10} = 17 - 9$

$a_{10} = 8$

Final Answer: The common difference $d = -1$ and the 10th term $a_{10} = 8$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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