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Sum of how many terms of the G.P. 29−13+12−...... is 5572?\dfrac{2}{9} - \dfrac{1}{3} + \dfrac{1}{2} - ……\text{ is } \dfrac{55}{72} ?

AP GP

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Answer

Common ratio (r) = −1329=−92×3=−32\dfrac{-\dfrac{1}{3}}{\dfrac{2}{9}} = -\dfrac{9}{2 \times 3} = -\dfrac{3}{2}.

Let no. of terms be n.

By formula,

Sum of G.P. (when |r| > 1 ) = a(rn−1)(r−1)\dfrac{a(r^n - 1)}{(r - 1)}

Substituting value we get :

⇒29[(−32)n−1](−32−1)=5572⇒29[(−32)n−1](−3−22)=5572⇒29[(−32)n−1](−52)=5572⇒29[(−32)n−1]=5572×−52⇒[(−32)n−1]=5572×−52×92⇒[(−32)n−1]=−27532⇒(−32)n=−27532+1⇒(−32)n=−275+3232⇒(−32)n=−24332⇒(−32)n=(−32)5⇒n=5.\Rightarrow \dfrac{\dfrac{2}{9}\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{\Big(-\dfrac{3}{2} - 1\Big)} = \dfrac{55}{72} \\[1em] \Rightarrow \dfrac{\dfrac{2}{9}\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{\Big(\dfrac{-3 - 2}{2}\Big)} = \dfrac{55}{72} \\[1em] \Rightarrow \dfrac{\dfrac{2}{9}\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{\Big(-\dfrac{5}{2}\Big)} = \dfrac{55}{72} \\[1em] \Rightarrow \dfrac{2}{9}\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big] = \dfrac{55}{72} \times -\dfrac{5}{2} \\[1em] \Rightarrow \Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big] = \dfrac{55}{72} \times -\dfrac{5}{2} \times \dfrac{9}{2} \\[1em] \Rightarrow \Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big] = -\dfrac{275}{32} \\[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, sum of 5 terms = 5572\dfrac{55}{72}.

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