Question

Find the value of n if
i) 1 + 4 + 7 + 10 +...... to n terms = 590
ii) 50 + 46 + 42 + 38 + ...to n terms = 336

Answer 1

Solution:

i) 1 + 4 + 7 + 10 +...... to n terms = 590

Here given series is AP,

first term a= 1

common difference d = 3

Sum of n terms is given by
$S_{n} = \frac{n}{2} \Big[2a + (n - 1)d\Big] \\ \\ 590 = \frac{n}{2} (2 + (n - 1)3) \\ \\ 1180 = n(2 + 3n - 3) \\ \\ 1180 = n(3n - 1) \\ \\ 3 {n}^{2} - n - 1180 = 0 \\ \\ now \: solve \: this \: quadratic \: eq \\ \\ n_{1,2} = \frac{1 ±\sqrt{1 + 14160} }{6} \\ \\ n_{1,2} = \frac{1 ± 119}{6} \\ \\ n_{1} = \frac{1 + 119}{6} \\ \\ n_{1}= \frac{120}{6} = 20 \\ \\ n_{2} = \frac{1 - 119}{6} \\ \\ n_{2} = \frac{ - 118}{6} = \frac{ - 59}{3}$
n can't be negative and fractional,so discard second value.

Thus n = 20

2) 50 + 46 + 42 + 38 + ...to n terms = 336

here,a= 50

d = -4

$S_{n} = \frac{n}{2} \bigg[2a + (n - 1)d\bigg] \\ \\ 336 = \frac{n}{2} (100 - (n - 1)4) \\ \\ 672 = n(100 - 4n + 4) \\ \\ 672 = n( - 4n + 104) \\ \\ - 4{n}^{2} + 104n - 672= 0 \\ \\ 2 {n}^{2} - 52n + 336 = 0 \\ \\ now \: solve \: this \: quadratic \: eq \\ \\ n_{1,2} = \frac{52 ±\sqrt{2704 - 2688} }{4} \\ \\ n_{1,2} = \frac{52 ±4}{4} \\ \\ n_{1} = \frac{52 + 4}{4} \\ \\ n_{1}= \frac{56}{4} = 14 \\ \\ n_{2} = \frac{52- 4}{4} \\ \\ n_{2} = \frac{ 48}{4} = 12$
Hope it helps you.