MP Board · Class 10 · Mathematics · Arithmetic ProgressionsIn an Arithmetic Progression, the sum of the 4th and 8th terms is 24, and the sum of the 6th and 10th terms is 44. Find the first term, the common difference, and the first 20 terms of this Arithmetic Progression. Show complete step-by-step working.
Step 1: Write down the general formula for the $n$-th term of an AP\nThe formula for the $n$-th term ($a_n$) of an Arithmetic Progression is:
$$a_n = a + (n - 1)d$$\nwhere $a$ is the first term and $d$ is the common difference.
Step 2: Form equations based on the given conditions\nAccording to the first condition, the sum of the 4th ($a_4$) and 8th ($a_8$) terms is 24:
$$a_4 + a_8 = 24$$\nUsing the $n$-th term formula: $$[a + (4 - 1)d] + [a + (8 - 1)d] = 24$$ $$(a + 3d) + (a + 7d) = 24$$ $$2a + 10d = 24$$\nDividing the entire equation by 2: $$a + 5d = 12 \quad \text{--- (Equation 1)}$$ \nAccording to the second condition, the sum of the 6th ($a_6$) and 10th ($a_{10}$) terms is 44: $$a_6 + a_{10} = 44$$\nUsing the $n$-th term formula: $$[a + (6 - 1)d] + [a + (10 - 1)d] = 44$$ $$(a + 5d) + (a + 9d) = 44$$ $$2a + 14d = 44$$\nDividing the entire equation by 2: $$a + 7d = 22 \quad \text{--- (Equation 2)}$$
Step 3: Solve the simultaneous linear equations\nSubtract Equation 1 from Equation 2:
$$(a + 7d) - (a + 5d) = 22 - 12$$ $$2d = 10$$ $$d = \frac{10}{2} = 5$$ \nNow, substitute the value of $d = 5$ into Equation 1: $$a + 5(5) = 12$$ $$a + 25 = 12$$ $$a = 12 - 25 = -13$$ \nThus, the first term $a = -13$ and the common difference $d = 5$.
Step 4: Find the sum of the first 20 terms ($S_{20}$)\nThe formula for the sum of the first $n$ terms is:
$$S_n = \frac{n}{2} [2a + (n - 1)d]$$\nSubstitute $n = 20$, $a = -13$, and $d = 5$ into the formula: $$S_{20} = \frac{20}{2} [2(-13) + (20 - 1)5]$$ $$S_{20} = 10 [-26 + (19)5]$$ $$S_{20} = 10 [-26 + 95]$$ $$S_{20} = 10 [69]$$ $$S_{20} = 690$$
Final Answer
- First term ($a$) = $-13$
- Common difference ($d$) = $5$
- Sum of the first 20 terms ($S_{20}$) = $690$