find the value of:
i^n+100+i^n+50+in+48+i^n+46
Answers
............................
Answer:
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
\begin{gathered}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} \\ \\\end{gathered}
S
n
=
2
n
[2a+(n−1)d]
590=
2
n
(2+(n−1)3)
1180=n(2+3n−3)
1180=n(3n−1)
3n
2
−n−1180=0
nowsolvethisquadraticeq
n
1,2
=
6
1±
1+14160
n
1,2
=
6
1±119
n
1
=
6
1+119
n
1
=
6
120
=20
n
2
=
6
1−119
n
2
=
6
−118
=
3
−59
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
\begin{gathered}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 \\ \\\end{gathered}
S
n
=
2
n
[2a+(n−1)d]
336=
2
n
(100−(n−1)4)
672=n(100−4n+4)
672=n(−4n+104)
−4n
2
+104n−672=0
2n
2
−52n+336=0
nowsolvethisquadraticeq
n
1,2
=
4
52±
2704−2688
n
1,2
=
4
52±4
n
1
=
4
52+4
n
1
=
4
56
=14
n
2
=
4
52−4
n
2
=
4
48
=12
Hope it helps you.
i hope thish answer is your right answer