Question

Find the sum of the following APs:

1. \(2, 7, 12, .......,\) to \(10\) terms.
2. \(–37, –33, –29,.......,\) to \(12\) terms.
3. \(0.6, 1.7, 2.8, ......., \)to \(100\) terms.
4. \(\frac {1}{15}, \frac {1}{12}, \frac {1}{10}, .......,\) to \(11\) terms.

Answer 1

(i) The sum of the first 10 terms of the arithmetic progression (A.P.) \(2, 7, 12 ,…\) is calculated as follows: The first term \(a = 2\). The common difference \(d = a_2 - a_1 = 7 - 2 = 5\). The number of terms \(n = 10\). The formula for the sum of an A.P. is \(S_n = \frac n2[2a+(n-1)d]\). Substituting the values, \(S_{10} = \frac {10}{2}[2\times 2+(10-1)5]\) \(S_{10} = 5[4+9\times5]\) \(S_{10} = 5[4+45]\) \(S_{10} = 5\times 49\) \(S_{10} = 245\)

(ii) The sum of the first 12 terms of the A.P. \(−37, −33, −29 ,…\) is calculated as follows: The first term \(a = −37\). The common difference \(d = a_2 - a_1 = (−33) - (−37) = − 33 + 37 = 4\). The number of terms \(n = 12\). Using the formula \(S_n = \frac n2[2a+(n-1)d]\): \(S_{12} = \frac {12}{2}[2(-37)+(12-1)4]\) \(S_{12} = 6[-74+11\times4]\) \(S_{12} = 6[-74+44]\) \(S_{12} = 6\times(-30)\) \(S_{12} = -180\)

---

(iii) The sum of the first 100 terms of the A.P. \(0.6, 1.7, 2.8 ,…\) is calculated as follows: The first term \(a = 0.6\). The common difference \(d = a_2 - a_1 = 1.7 - 0.6 = 1.1\). The number of terms \(n = 100\). Using the formula \(S_n = \frac n2[2a+(n-1)d]\): \(S_{100} = \frac {100}{2}[2(0.6)+(100-1)1.1]\) \(S_{100} = 50[1.2+99\times1.1]\) \(S_{100} = 50[1.2+108.9]\) \(S_{100} = 50[110.1]\) \(S_{100} = 5505\)

---

(iv) For the A.P. \(\frac {1}{15} , \frac {1}{12} , \frac {1}{10} ,………,\) to 11 terms: The first term \(a = \frac {1}{15}\). The number of terms \(n = 11\). The common difference \(d = a_2 - a_1\). \(d = \frac {1}{12}-\frac {1}{15}\) To find a common denominator, which is 60: \(d = \frac {5}{60}-\frac {4}{60}\) \(d = \frac {1}{60}\) Using the formula for the sum of an A.P., \(S_n = \frac n2[2a+(n-1)d]\): \(S_{11} = \frac {11}{2}[2(\frac {1}{15})+(11-1)\frac {1}{60}]\) \(S_{11} = \frac {11}{2}[\frac {2}{15}+\frac {10}{60}]\) Simplify \(\frac {10}{60}\) to \(\frac 16\): \(S_{11} = \frac {11}{2}[\frac {2}{15}+\frac 16]\) To add the fractions inside the bracket, find a common denominator, which is 30: \(S_{11} = \frac {11}{2}[\frac {4}{30}+\frac {5}{30}]\) \(S_{11} = \frac {11}{2}[\frac {9}{30}]\) Now multiply the fractions: \(S_{11} = \frac {11 \times 9}{2 \times 30}\) \(S_{11} = \frac {99}{60}\) Simplify the fraction by dividing both numerator and denominator by 3: \(S_{11} = \frac {33}{20}\)