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Q2:
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Solution :
Given: An Arithmetic Progression (AP) where the sum of the third term ($a_3$) and the seventh term ($a_7$) is $6$, and their product is $8$.
To Find: The sum of the first sixteen terms ($S_{16}$) of the AP.
Step 1: Defining the variables and formulas
Let the first term of the AP be $a$ and the common difference be $d$.
The $n^{th}$ term of an AP is given by the formula: $a_n = a + (n - 1)d$.
The sum of the first $n$ terms of an AP is given by: $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Step 2: Expressing the given conditions algebraically
Using the formula for the $n^{th}$ term:
$a_3 = a + (3 - 1)d = a + 2d$
$a_7 = a + (7 - 1)d = a + 6d$
Condition 1: Sum of the terms is $6$
$(a + 2d) + (a + 6d) = 6$
$2a + 8d = 6$
Dividing by $2$: $a + 4d = 3 \implies a = 3 - 4d$ --- (Equation 1)
Condition 2: Product of the terms is $8$
$(a + 2d)(a + 6d) = 8$ --- (Equation 2)
Step 3: Solving for $a$ and $d$
Substitute Equation 1 into Equation 2:
$((3 - 4d) + 2d)((3 - 4d) + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
Using the algebraic identity $(x - y)(x + y) = x^2 - y^2$:
$3^2 - (2d)^2 = 8$
$9 - 4d^2 = 8$
$-4d^2 = 8 - 9$
$-4d^2 = -1$
$d^2 = \frac{1}{4} \implies d = \pm \frac{1}{2}$
Case 1: If $d = \frac{1}{2}$
$a = 3 - 4(\frac{1}{2}) = 3 - 2 = 1$
Case 2: If $d = -\frac{1}{2}$
$a = 3 - 4(-\frac{1}{2}) = 3 + 2 = 5$
Step 4: Calculating $S_{16}$ for both cases
The formula for $S_{16}$ is $S_{16} = \frac{16}{2} [2a + (16 - 1)d] = 8[2a + 15d]$.
For Case 1 ($a=1, d=1/2$):
$S_{16} = 8[2(1) + 15(1/2)] = 8[2 + 7.5] = 8[9.5] = 76$
For Case 2 ($a=5, d=-1/2$):
$S_{16} = 8[2(5) + 15(-1/2)] = 8[10 - 7.5] = 8[2.5] = 20$
Final Answer: The sum of the first sixteen terms is either 76 or 20.
More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.4
- Q1: Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find $n$ for $a_n < 0$]
- Q3: A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = $\frac{250}{25} + 1$]
- Q4: The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$. [Hint : $S_{x-1} = S_{49} - S_x$]
- Q5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = $\frac{1}{4} \times \frac{1}{2} \times 50$ m$^3$]
CBSE Solutions for Class 10 Mathematics Arithmetic Progression
Chapters in CBSE - Class 10 Mathematics
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