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Q3(v):
In an AP: (v) given $d = 5, S_9 = 75$, find $a$ and $a_9$.

Solution :

Given:

The common difference of the Arithmetic Progression (AP), $d = 5$.

The sum of the first 9 terms of the AP, $S_9 = 75$.

To Find:

The first term ($a$) and the ninth term ($a_9$).

Step 1: Identifying the relevant formulas

To solve for $a$, we use the formula for the sum of the first $n$ terms of an AP:

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

To solve for $a_9$, we use the formula for the $n^{th}$ term of an AP:

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

Step 2: Substituting the given values into the sum formula

Given $n = 9$, $d = 5$, and $S_9 = 75$, we substitute these into the sum formula:

$75 = \frac{9}{2} [2a + (9 - 1)5]$

[Substituting $n=9, d=5, S_9=75$]

Step 3: Solving for $a$

$75 = \frac{9}{2} [2a + (8)5]$

[Simplifying the term inside the parenthesis: $9-1 = 8$]

$75 = \frac{9}{2} [2a + 40]$

[Multiplying $8 \times 5 = 40$]

$75 = 9(a + 20)$

[Factoring out 2 from the bracket: $\frac{9}{2} \times 2(a + 20) = 9(a + 20)$]

$75 = 9a + 180$

[Distributing 9 into the parenthesis]

$9a = 75 - 180$

[Transposing 180 to the left side]

$9a = -105$

$a = \frac{-105}{9}$

$a = \frac{-35}{3}$

[Dividing both numerator and denominator by 3]

Step 4: Calculating the ninth term ($a_9$)

Using the formula $a_n = a + (n - 1)d$ for $n = 9$:

$a_9 = a + (9 - 1)d$

$a_9 = \left(\frac{-35}{3}\right) + 8(5)$

[Substituting $a = \frac{-35}{3}$ and $d = 5$]

$a_9 = \frac{-35}{3} + 40$

$a_9 = \frac{-35 + 120}{3}$

[Finding a common denominator: $40 = \frac{120}{3}$]

$a_9 = \frac{85}{3}$

Final Answer:

The first term $a = -\frac{35}{3}$ and the ninth term $a_9 = \frac{85}{3}$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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