First slide

Arithmetic progression

Question

The number of terms of the A.P. 1, 4, 7, . . . that must be taken to obtain a sum of 715
is

Moderate

A

24
B

23
C

22
D

21

Solution

Here $a = 1, d = 3$ and $S_{n} = 715$ We have
$\begin{array}{l} 715 = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [2 (1) + (n - 1) (3)] \\ \Rightarrow 1430 = 3 n^{2} - n or 3 n^{2} - n - 1430 = 0 \\ n = \frac{1}{6} (1 \pm \sqrt{1 + 3 \times 4 \times 1430}) = \frac{1}{6} (1 \pm 131) \\ = 22, - 65 / 3 \end{array}$
As $n$ must be a natural number $n$ = 22

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