1. **Problem Statement:** Find the sums of the given arithmetic progressions (APs) in Q4, Q5, Q6, Q7, Q9, Q10, and Q12. --- 2. **Formula for sum of n terms of an AP:** $$S_n = \frac{n}{2} [2a + (n-1)d]$$ where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms. --- ### Q4: Find the sum of the following APs. (i) $1, 5, 9, \ldots$ to 25 terms - $a=1$, $d=5-1=4$, $n=25$ - $$S_{25} = \frac{25}{2} [2(1) + (25-1)4] = \frac{25}{2} [2 + 96] = \frac{25}{2} \times 98 = 25 \times 49 = 1225$$ (ii) $75, 66, 57, \ldots$ to 15 terms - $a=75$, $d=66-75=-9$, $n=15$ - $$S_{15} = \frac{15}{2} [2(75) + (15-1)(-9)] = \frac{15}{2} [150 - 126] = \frac{15}{2} \times 24 = 15 \times 12 = 180$$ (iii) $1, 2, 3, 4, \ldots$ to 100 terms - $a=1$, $d=1$, $n=100$ - $$S_{100} = \frac{100}{2} [2(1) + (100-1)1] = 50 [2 + 99] = 50 \times 101 = 5050$$ (iv) $1, 3, 5, 7, \ldots$ to 75 terms - $a=1$, $d=2$, $n=75$ - $$S_{75} = \frac{75}{2} [2(1) + (75-1)2] = \frac{75}{2} [2 + 148] = \frac{75}{2} \times 150 = 75 \times 75 = 5625$$ (v) $t, 3t, 5t, \ldots$ to 25 terms - $a=t$, $d=3t - t = 2t$, $n=25$ - $$S_{25} = \frac{25}{2} [2t + (25-1)2t] = \frac{25}{2} [2t + 48t] = \frac{25}{2} \times 50t = 25 \times 25t = 625t$$ --- ### Q5: Find sum to $n$ terms of AP $7, 10, 13, 16, \ldots$ and then find $S_5$. - $a=7$, $d=3$ - Sum to $n$ terms: $$S_n = \frac{n}{2} [2(7) + (n-1)3] = \frac{n}{2} [14 + 3n - 3] = \frac{n}{2} (3n + 11)$$ - For $n=5$: $$S_5 = \frac{5}{2} (3 \times 5 + 11) = \frac{5}{2} (15 + 11) = \frac{5}{2} \times 26 = 5 \times 13 = 65$$ - Calculator check: $7 + 10 + 13 + 16 + 19 = 65$ --- ### Q6: Find indicated sums for APs. (i) $1.2, 1.7, 2.2, \ldots$ - $a=1.2$, $d=1.7 - 1.2 = 0.5$ - $$S_7 = \frac{7}{2} [2(1.2) + (7-1)0.5] = \frac{7}{2} [2.4 + 3] = \frac{7}{2} \times 5.4 = 7 \times 2.7 = 18.9$$ - $$S_{15} = \frac{15}{2} [2(1.2) + (15-1)0.5] = \frac{15}{2} [2.4 + 7] = \frac{15}{2} \times 9.4 = 15 \times 4.7 = 70.5$$ (ii) $4, 9, 14, 19, \ldots$ - $a=4$, $d=5$ - $$S_8 = \frac{8}{2} [2(4) + (8-1)5] = 4 [8 + 35] = 4 \times 43 = 172$$ - $$S_{18} = \frac{18}{2} [2(4) + (18-1)5] = 9 [8 + 85] = 9 \times 93 = 837$$ --- ### Q7: Find the 12th term of AP: $5, 2, -1, \ldots$ - $a=5$, $d=2-5 = -3$ - $n=12$ - Formula for $n$th term: $$a_n = a + (n-1)d$$ - $$a_{12} = 5 + (12-1)(-3) = 5 + 11 \times (-3) = 5 - 33 = -28$$ --- ### Q9: $a_n = 3n + 2$. Find $S_n$, $S_{10}$, and $S_{15}$. - First term $a = a_1 = 3(1) + 2 = 5$ - Common difference $d = a_2 - a_1 = (3 \times 2 + 2) - 5 = 8 - 5 = 3$ - Sum to $n$ terms: $$S_n = \frac{n}{2} [2a + (n-1)d] = \frac{n}{2} [2 \times 5 + (n-1)3] = \frac{n}{2} (10 + 3n - 3) = \frac{n}{2} (3n + 7)$$ - $$S_{10} = \frac{10}{2} (3 \times 10 + 7) = 5 \times 37 = 185$$ - $$S_{15} = \frac{15}{2} (3 \times 15 + 7) = \frac{15}{2} \times 52 = 15 \times 26 = 390$$ --- ### Q10: Sum of all integers between 100 and 200 which are multiples of 7. - First multiple of 7 greater than 100 is $105$ ($7 \times 15$) - Last multiple of 7 less than 200 is $196$ ($7 \times 28$) - Number of terms: $$n = 28 - 15 + 1 = 14$$ - AP: $105, 112, 119, \ldots, 196$ - $a=105$, $d=7$ - Sum: $$S_{14} = \frac{14}{2} [2(105) + (14-1)7] = 7 [210 + 91] = 7 \times 301 = 2107$$ --- ### Q12: First term $a=2$, 8th term equals three times 3rd term. Find sum of 20 terms. - Let common difference be $d$ - $a_8 = a + 7d$, $a_3 = a + 2d$ - Given: $a_8 = 3 a_3$ - Substitute: $$2 + 7d = 3(2 + 2d)$$ $$2 + 7d = 6 + 6d$$ $$7d - 6d = 6 - 2$$ $$d = 4$$ - Sum of 20 terms: $$S_{20} = \frac{20}{2} [2(2) + (20-1)4] = 10 [4 + 76] = 10 \times 80 = 800$$