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Arithmetic Progressions
EXERCISE 5.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
Find the sum of the following APs: (i) 2, 7, 12, . . ., to 10 terms. (ii) –37, –33, –29, . . ., to 12 terms. (iii) 0.6, 1.7, 2.8, . . ., to 100 terms. (iv) 1 1 1 , , 15 12 10 , . . ., to 11 terms.
Step-by-Step Solution
Key Concept: For an arithmetic progression (AP) with first term \(a\), common difference \(d\) and \(n\) terms, the sum is given by \[ S_n = \frac{n}{2}\bigl[2a+(n-1)d\bigr] \] or equivalently \[ S_n = \frac{n}{2}(a + l) \] where \(l = a+(n-1)d\) is the last term. The method is to identify \(a\), \(d\) and \(n\) for each series, compute the last term (if convenient) and then apply the formula.
1. Series (i) - First term \(a = 2\). - Common difference \(d = 7-2 = 5\). - Number of terms \(n = 10\). - Last term \(l = a+(n-1)d = 2+9\times5 = 47\). - Sum \(S_{10}=\frac{10}{2}(a+l)=5\times(2+47)=5\times49=\mathbf{245}\).
2. Series (ii) - First term \(a = -37\). - Common difference \(d = -33-(-37)=4\). - Number of terms \(n = 12\). - Last term \(l = -37+11\times4 = 7\). - Sum \(S_{12}=\frac{12}{2}(a+l)=6\times(-37+7)=6\times(-30)=\mathbf{-180}\).
3. Series (iii) - First term \(a = 0.6\). - Common difference \(d = 1.7-0.6 = 1.1\). - Number of terms \(n = 100\). - Last term \(l = 0.6+99\times1.1 = 0.6+108.9 = 109.5\). - Sum \(S_{100}=\frac{100}{2}(a+l)=50\times(0.6+109.5)=50\times110.1=\mathbf{5505}\).
4. Series (iv) (interpreted as the AP \(1,\;\frac{1}{5},\;\frac{1}{12},\;\frac{1}{22},\dots\) with first term \(a=1\) and common difference \(d=\frac{1}{5}-1=-\frac{4}{5}\)) - First term \(a = 1\). - Common difference \(d = \frac{1}{5}-1 = -\frac{4}{5}\). - Number of terms \(n = 11\). - Sum \(S_{11}=\frac{11}{2}\bigl[2a+(11-1)d\bigr]=\frac{11}{2}\bigl[2\times1+10\times(-\frac{4}{5})\bigr] =\frac{11}{2}\bigl[2-8\bigr]=\frac{11}{2}\times(-6)=\mathbf{-33}\).
Thus the required sums are: - (i) \(245\) - (ii) \(-180\) - (iii) \(5505\) - (iv) \(-33\)
Correct Answer:(i) 245, (ii) -180, (iii) 5505, (iv) -33
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