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Q3(x):
In an AP: (x) given $l = 28, S = 144$, and there are total 9 terms. Find $a$.

Solution :

Given:

The last term of the Arithmetic Progression (AP), denoted by $l$ (or $a_n$), is $28$.

The sum of the terms of the AP, denoted by $S_n$, is $144$.

The total number of terms in the AP, denoted by $n$, is $9$.

To find:

The first term of the AP, denoted by $a$.

Step 1: Identifying the relevant formula

For an Arithmetic Progression where the first term ($a$), the last term ($l$), and the total number of terms ($n$) are known, the sum of the terms ($S_n$) is given by the formula:

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

[This formula is derived from the general sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ by substituting $l = a + (n-1)d$].

Step 2: Substituting the given values into the formula

Given $S_n = 144$, $n = 9$, and $l = 28$, we substitute these into the equation:

$144 = \frac{9}{2}(a + 28)$

Step 3: Solving for $a$

To isolate $a$, we first multiply both sides of the equation by $2$ to eliminate the denominator:

$144 \times 2 = 9(a + 28)$

$288 = 9(a + 28)$

Next, divide both sides by $9$:

$\frac{288}{9} = a + 28$

[Performing the division: $288 \div 9 = 32$]

$32 = a + 28$

Finally, subtract $28$ from both sides to solve for $a$:

$a = 32 - 28$

$a = 4$

Step 4: Verification (Optional)

If $a = 4$ and $l = 28$ with $n = 9$, we can find the common difference $d$ using $l = a + (n-1)d$:

$28 = 4 + (9-1)d$

$24 = 8d$

$d = 3$

Checking the sum: $S_9 = \frac{9}{2}[2(4) + (9-1)3] = \frac{9}{2}[8 + 24] = \frac{9}{2}[32] = 9 \times 16 = 144$. The result is consistent.

Final Answer: The first term $a$ is 4.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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