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Q8:
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution :

Given:

The second term of the Arithmetic Progression ($a_2$) = $14$.

The third term of the Arithmetic Progression ($a_3$) = $18$.

The number of terms to be summed ($n$) = $51$.

To Find:

The sum of the first $51$ terms ($S_{51}$) of the Arithmetic Progression.

Step 1: Defining the variables and formulas

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

The formula for the $n^{th}$ term of an AP is given by: $a_n = a + (n - 1)d$.

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

Step 2: Formulating equations based on given terms

Using the $n^{th}$ term formula for the second term ($n=2$):

$a_2 = a + (2 - 1)d$

$14 = a + d$ --- (Equation 1)

Using the $n^{th}$ term formula for the third term ($n=3$):

$a_3 = a + (3 - 1)d$

$18 = a + 2d$ --- (Equation 2)

Step 3: Solving for $a$ and $d$

Subtract Equation 1 from Equation 2 to eliminate $a$:

$(a + 2d) - (a + d) = 18 - 14$

$a - a + 2d - d = 4$

$d = 4$

Substitute the value of $d = 4$ into Equation 1 to find $a$:

$a + 4 = 14$

$a = 14 - 4$

$a = 10$

Step 4: Calculating the sum of the first 51 terms

We use the sum formula $S_n = \frac{n}{2} [2a + (n - 1)d]$ with $n = 51$, $a = 10$, and $d = 4$:

$S_{51} = \frac{51}{2} [2(10) + (51 - 1)4]$

$S_{51} = \frac{51}{2} [20 + (50)4]$

$S_{51} = \frac{51}{2} [20 + 200]$

$S_{51} = \frac{51}{2} [220]$

$S_{51} = 51 \times 110$

$S_{51} = 5610$

Final Answer: The sum of the first 51 terms of the Arithmetic Progression is 5610.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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