Question

Solve:
(i)25 + 22 + 19 + 16 + ... + x = 115
(ii) 1 + 4 + 7 + 10 + ... + x = 590

Answer 1

(i) 25 + 22 + 19 + 16 + ... + x = 115
Here, a = 25, d = -3, $S_n=115$
We know:
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 115=\frac{n}{2}[2\times 25+(n-1)\times (-3\left)\right]\Rightarrow 115\times 2=n[50-3n+3]\Rightarrow 230=n(53-3n)\Rightarrow 230=53n-3n^2\Rightarrow 3n^2-53n+230=0$
By quadratic formula:
$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Substituting a = 3, b = -53 and c = 230, we get:
$n=\frac{53\pm \sqrt{(53{)}^2-4\times 3\times 230}}{2\times 3}=\frac{46}{6},10\Rightarrow n=10,asn\ne \frac{46}{6}\therefore a_n=a+(n-1)d\Rightarrow x=25+(10-1)(-3)\Rightarrow x=25-27=-2$

(ii) 1 + 4 + 7 + 10 + .... + x = 590
Here, a = 1, d = 3
We know:
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 590=\frac{n}{2}[2\times 1+(n-1)\times (3\left)\right]\Rightarrow 590\times 2=n[2+3n-3]\Rightarrow 1180=n(3n-1)\Rightarrow 1180=3n^2-n\Rightarrow 3n^2-n-1180=0$
By quadratic formula:
$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Substituting a = 3, b = -1 and c = - 1180, we get:
$\Rightarrow n=\frac{1\pm \sqrt{(1{)}^2+4\times 3\times 1180}}{2\times 3}=\frac{-118}{6},20\Rightarrow n=20,asn\ne \frac{-118}{6}\therefore a_n=x=a+(n-1)d\Rightarrow x=1+(20-1)\left(3\right)\Rightarrow x=1+60-3=58$