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Q10(i):
Show that $a_1, a_2, . . ., a_n, . . .$ form an AP where $a_n$ is defined as below : (i) $a_n = 3 + 4n$ Also find the sum of the first 15 terms in each case.

Solution :

Given: The $n^{th}$ term of a sequence is defined by the expression $a_n = 3 + 4n$.

To Find:
1. Show that the sequence forms an Arithmetic Progression (AP).
2. Calculate the sum of the first 15 terms ($S_{15}$).

Step 1: Generating the terms of the sequence
To determine if the sequence is an AP, we calculate the first few terms by substituting $n = 1, 2, 3, ...$ into the given formula $a_n = 3 + 4n$.

For $n = 1$: $a_1 = 3 + 4(1) = 3 + 4 = 7$
For $n = 2$: $a_2 = 3 + 4(2) = 3 + 8 = 11$
For $n = 3$: $a_3 = 3 + 4(3) = 3 + 12 = 15$
For $n = 4$: $a_4 = 3 + 4(4) = 3 + 16 = 19$

Step 2: Verifying the Arithmetic Progression
A sequence is an AP if the difference between consecutive terms is constant. This constant is called the common difference ($d$).

Calculate the differences:
$a_2 - a_1 = 11 - 7 = 4$
$a_3 - a_2 = 15 - 11 = 4$
$a_4 - a_3 = 19 - 15 = 4$

[Since the difference $a_n - a_{n-1} = 4$ is constant for all $n$, the sequence is an AP with first term $a = 7$ and common difference $d = 4$.]

Step 3: Calculating the sum of the first 15 terms
The formula for the sum of the first $n$ terms of an AP is given by:
$S_n = \frac{n}{2} [2a + (n - 1)d]$

Here, $n = 15$, $a = 7$, and $d = 4$. Substituting these values into the formula:

$S_{15} = \frac{15}{2} [2(7) + (15 - 1)4]$

Step 4: Simplifying the expression
$S_{15} = \frac{15}{2} [14 + (14)4]$
$S_{15} = \frac{15}{2} [14 + 56]$
$S_{15} = \frac{15}{2} [70]$
$S_{15} = 15 \times 35$ [Dividing 70 by 2]

$S_{15} = 525$

Final Answer: The sequence forms an AP with a common difference of 4, and the sum of the first 15 terms is 525.


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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