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) \(2, 7, 12 ,…, \)to \(10\) terms
For this A.P., \(a = 2\), \(d = a_2 − a_1 = 7 − 2 = 5\) and \(n = 10\)
We know that,
\(S_n = \frac n2[2a+(n-1)d]\)

\(S_{10} = \frac {10}{2}[2\times 2+(10-1)5]\)
\(S_{10} = 5[4+9\times5]\)
\(S_{10} = 5\times 49\)
\(S_{10} = 245\)

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(ii) \(−37, −33, −29 ,…,\) to \(12\) terms
For this A.P., \(a = −37\), \(d = a_2 − a_1 = (−33) − (−37) = − 33 + 37 = 4,  n = 12\)
We know that,
\(S_n = \frac n2[2a+(n-1)d]\)

\(S_{12} = \frac {12}{2}[2(-37)+(12-1)4]\)

\(S_{12} = 6[-74+11\times4]\)
\(S_{12} = 6[-74+44]\)
\(S_{12} = 6\times(-30)\)
\(S_{12} = -180\)

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(iii) \(0.6, 1.7, 2.8 ,…,\) to \(100\) terms
For this A.P., \(a = 0.6, d = a_2 − a_1 = 1.7 − 0.6 = 1.1\) and \(n = 100\)
We know that,
\(S_n = \frac n2[2a+(n-1)d]\)

\(S_{100}\) \(= \frac {100}{2}[2(0.6)+(100-1)1.1]\)
\(S_{100}\)\(= 50[1.2+99\times1.1]\)
\(S_{100}\)\(= 50[1.2+108.9]\)
\(S_{100}\)\(= 50[1.2+99\times1.1]\)
\(S_{100}\)\(= 50[110.1]\)
\(S_{100}\) \(= 5505\)

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(iv) \(\frac {1}{15} , \frac {1}{12} , \frac {1}{10} ,………,\) to 11 terms

For this A.P.,
\(a = \frac {1}{15}\)
\(n = 11\)
\(d = a_2-a_1\)

\(d = \frac {1}{12}-\frac {1}{15}\)

\(d = \frac {5-4}{60}\)

\(d = \frac {1}{60}\)

We know that,

\(S_n = \frac n2[2a+(n-1)d]\)

\(S_{11} = \frac {11}{2}[2(\frac {1}{15})+(11-1)\frac {1}{60}]\)

\(S_{11}\) \(= \frac {11}{2}[\frac {2}{15}+\frac {10}{60}]\)

\(S_{11}\) \(= \frac {11}{2}[\frac {2}{15}+\frac 16]\)

\(S_{11}\) \(= \frac {11}{2}[\frac {4+5}{30}]\)

\(S_{11}\) \(= \frac {11}{2} \times \frac {9}{30}\)

\(S_{11}\) \(= \frac {33}{20}\)