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Find the maximum sum of the AP 21, \(20\frac{2}{5},\;19\frac{4}{5}, \ldots .\)1). ∞2). 3783). \(431\frac{5}{7}\)4). 448 |
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Answer» Summation of AP $(= \;\frac{n}{2}\left( {{a_0} + {a_n}} \right))$ Where, n is number of TERMS, a0 is first term and an is the last term. It is a decreasing AP. Thus, maximum sum will be OBTAINED TILL the last term is positive. GIVEN AP, 21, $(20\frac{2}{5},\;19\frac{4}{5}, \ldots .)$ C.d. = 20.4 – 21 = -0.6 an = a0 + (n – 1)d = 21 – 0.6(n – 1) For it to be positive : 21 – 0.6(n – 1) > 0 ⇒ 21 > 0.6(n – 1) ⇒ n < 36 ∴ n = 35 an = 21 – 0.6 × 34 = 0.6 ∴ Maximum sum of the given AP $(= \;\frac{{35}}{2}\left( {21 + 0.6} \right) = \;378)$ |
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