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Q10:
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution :

Given: An Arithmetic Progression (AP) where the 17th term ($a_{17}$) exceeds the 10th term ($a_{10}$) by 7.

To find: The common difference ($d$) of the Arithmetic Progression.

Step 1: Defining the general term of an Arithmetic Progression

The $n^{th}$ term of an Arithmetic Progression is given by the formula:
$a_n = a + (n - 1)d$
where:
$a$ = the first term of the AP
$d$ = the common difference
$n$ = the position of the term in the sequence

Step 2: Expressing the 17th and 10th terms using the formula

For the 17th term ($n = 17$):
$a_{17} = a + (17 - 1)d$
$a_{17} = a + 16d$ --- (Equation 1)

For the 10th term ($n = 10$):
$a_{10} = a + (10 - 1)d$
$a_{10} = a + 9d$ --- (Equation 2)

Step 3: Formulating the equation based on the given condition

The problem states that the 17th term exceeds the 10th term by 7. Mathematically, this is expressed as:
$a_{17} - a_{10} = 7$

Step 4: Substituting the expressions into the equation

Substitute Equation 1 and Equation 2 into the condition established in Step 3:
$(a + 16d) - (a + 9d) = 7$

Step 5: Solving for the common difference ($d$)

Distribute the negative sign across the terms in the second parenthesis:
$a + 16d - a - 9d = 7$

Group the like terms:
$(a - a) + (16d - 9d) = 7$

Simplify the expression:
$0 + 7d = 7$
$7d = 7$

Divide both sides by 7 to isolate $d$:
$d = \frac{7}{7}$
$d = 1$

Final Answer: The common difference of the Arithmetic Progression is 1.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2


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