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Q10(ii):
Show that $a_1, a_2, . . ., a_n, . . .$ form an AP where $a_n$ is defined as below : (ii) $a_n = 9 – 5n$ 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 formula $a_n = 9 - 5n$.

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 $n = 1, 2, 3, \dots$ into the given formula $a_n = 9 - 5n$.

For $n = 1$: $a_1 = 9 - 5(1) = 9 - 5 = 4$
For $n = 2$: $a_2 = 9 - 5(2) = 9 - 10 = -1$
For $n = 3$: $a_3 = 9 - 5(3) = 9 - 15 = -6$
For $n = 4$: $a_4 = 9 - 5(4) = 9 - 20 = -11$

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

Calculate the differences:
$d_1 = a_2 - a_1 = -1 - 4 = -5$
$d_2 = a_3 - a_2 = -6 - (-1) = -6 + 1 = -5$
$d_3 = a_4 - a_3 = -11 - (-6) = -11 + 6 = -5$

[Since $a_{n} - a_{n-1} = -5$ for all $n$, the difference is constant.]

Therefore, the sequence $4, -1, -6, -11, \dots$ forms an AP with the first term $a = 4$ and common difference $d = -5$.

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, $n = 15$, $a = 4$, and $d = -5$. Substituting these values into the formula:

$S_{15} = \frac{15}{2} [2(4) + (15 - 1)(-5)]$
$S_{15} = \frac{15}{2} [8 + (14)(-5)]$
$S_{15} = \frac{15}{2} [8 - 70]$
$S_{15} = \frac{15}{2} [-62]$

Step 4: Final Arithmetic Simplification
$S_{15} = 15 \times \left(\frac{-62}{2}\right)$
$S_{15} = 15 \times (-31)$
$S_{15} = -465$

Final Answer: The sequence forms an AP with a common difference of $-5$, and the sum of the first 15 terms is $-465$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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