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Mathematics

Find how many terms of the G.P.

29−13+12.........\dfrac{2}{9} - \dfrac{1}{3} + \dfrac{1}{2} ……… must be added to get the sum equal to 5572\dfrac{55}{72}?

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Answer

Common ratio = −1329=−96=−32\dfrac{-\dfrac{1}{3}}{\dfrac{2}{9}} = -\dfrac{9}{6} = -\dfrac{3}{2}.

Let n terms be added.

Since, |r| > 1

⇒S=a(rn−1)(r−1)⇒5572=29×[(−32)n−1]−32−1⇒5572=29×[(−32)n−1]−52⇒5572=2×2×[(−32)n−1]9×−5⇒55×9×−572×2×2=[(−32)n−1]⇒−27532=[(−32)n−1]⇒(−32)n=−27532+1⇒(−32)n=−275+3232⇒(−32)n=−24332⇒(−32)n=(−32)5⇒n=5.\Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9} \times \Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{-\dfrac{3}{2} - 1} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9} \times \Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{-\dfrac{5}{2}} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{2 \times 2 \times \Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{9 \times -5} \\[1em] \Rightarrow \dfrac{55 \times 9 \times -5}{72 \times 2 \times 2} = \Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big] \\[1em] \Rightarrow -\dfrac{275}{32} = \Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big] \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{275}{32} + 1 \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{275 + 32}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{243}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \Big(-\dfrac{3}{2}\Big)^5 \\[1em] \Rightarrow n = 5.

Hence, no. of terms to make sum = 5572\dfrac{55}{72} is 5.

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