First slide

Arithmetic progression

Question

The minimum number of terms from the beginning of the series $20 + 22 \frac{2}{3} + 25 \frac{1}{3} + \dots$ so that the sum may
exceed 1568, is

Moderate

A

25
B

27
C

28
D

29

Solution

It is in A.P. for which $a = 20, d = 2 \frac{2}{3} = \frac{8}{3}$
Now, $S_{n} > 1568$
$\begin{array}{l} \Rightarrow \frac{n}{2} [40 + (n - 1) \frac{8}{3}] > 1568 \\ \Rightarrow \frac{n}{2} \times \frac{112 + 8 n}{3} > 1568 \\ \Rightarrow n^{2} + 14 n > \frac{6}{8} \times 1568 = 1176 \\ \Rightarrow n^{2} + 14 n - 1176 > 0 \\ or (n + 42) (n - 28) > 0 \end{array}$
As n is positive, n – 28 > 0 i.e., n > 28 $∴$ Minimum value of n = 29.

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