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Q5(ii):
Find the number of terms in each of the following APs : (ii) 18, $15\frac{1}{2}$, 13, . . . , – 47

Solution :

Given: An Arithmetic Progression (AP) sequence: $18, 15\frac{1}{2}, 13, \dots, -47$.

To find: The number of terms ($n$) in the given AP.

Step 1: Identify the parameters of the Arithmetic Progression.

An Arithmetic Progression is defined by its first term ($a$) and its common difference ($d$).

The first term is $a = 18$.

The second term is $a_2 = 15\frac{1}{2} = \frac{31}{2}$.

The common difference ($d$) is calculated as $d = a_2 - a_1$:

$d = \frac{31}{2} - 18$

$d = \frac{31}{2} - \frac{36}{2}$ [Converting 18 to a fraction with denominator 2]

$d = -\frac{5}{2}$

Step 2: State the General Term Formula.

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

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

Where:

  • $a_n$ is the last term (given as $-47$)
  • $a$ is the first term ($18$)
  • $n$ is the number of terms
  • $d$ is the common difference ($-\frac{5}{2}$)

Step 3: Substitute the values into the formula and solve for $n$.

$-47 = 18 + (n - 1)\left(-\frac{5}{2}\right)$

Subtract 18 from both sides:

$-47 - 18 = (n - 1)\left(-\frac{5}{2}\right)$

$-65 = (n - 1)\left(-\frac{5}{2}\right)$

Multiply both sides by $-\frac{2}{5}$ to isolate $(n - 1)$:

$-65 \times \left(-\frac{2}{5}\right) = n - 1$

$\frac{130}{5} = n - 1$ [Since $65 \div 5 = 13$]

$26 = n - 1$

Add 1 to both sides:

$n = 26 + 1$

$n = 27$

Step 4: Conclusion.

Since $n$ represents the count of terms, it must be a positive integer. Our result $n = 27$ satisfies this condition.

Final Answer: The number of terms in the given AP is 27.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2


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