1.

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

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 $(= \;\frac{{35}}{2}\left( {21 + 0.6} \right) = \;378)$


Discussion

No Comment Found