step by step by usingsequence and series

Question

find the 20th term and the sum of the first 20 term of the sequence

1. 7, 4, 1, -2, ..........................
2. given a=5 an = 17 Sn =44  find d and n

find the sum of thefirst 10 term of the sequence 4, 12, 36, 108.

Answer 1

1) 7, 4, 1, -2. First term is a=7 and common difference is d=-3.

Series is a, a+d, a+2d, ..., a+(n-1)d where n is the term number, starting at n=1.

Sum of the series is S_n=a+(a+d)+(a+2d)+...+(a+(n-1)d).

There are n a's so S_n=an+d(1+2+3+...+(n-1)).

We can pair terms in the series in brackets, which is the sum of the natural numbers up to n-1:

1+2+3+...+n-1=(1+n-1)+(2+n-2)+(3+n-3)+...=n+n+n+...

There are n-1 terms so there are (n-1)/2 pairs and each pair sums to n so the sum of the natural numbers up to n-1 is n(n-1)/2, because each pair sums to n. By the way, it doesn't matter whether n is odd or even, the formula still works.

So the sum of the given series is an+dn(n-1)/2=n(a+d(n-1)/2). Now put a=7, n=20 and d=-3:

S₂₀=140-60×19/2=140-570=-430.

Let's check if the formula is correct by summing the given 4 terms: 7+4+1-2=10. Now put n=4:

S₄=28-12(3)/2=28-18=10. So the formula is correct.

20th term=7+19(-3)=7-57=-50.

2) Since the nth term is a+(n-1)d

a=5, a_n=17, S_n=44.

a+(n-1)d=17=5+(n-1)d ①; an+dn(n-1)/2=44=5n+dn(n-1)/2=n(5+d(n-1)/2) ②.

We have two equations and two unknowns so we can find the unknowns.

We know from ① that 5+(n-1)d=17, so (n-1)d=12 and (n-1)d/2=6, which can be substituted in ②

n(5+6)=44, 11n=44, n=4. We can now use ① to find d: 5+3d=17, 3d=12, d=4.

The series is 5, 9, 13, 17 (a_n=a₄=17); S_n=S₄=5+9+13+17=44.

3) Geometric series 4, 12, 36, 108, ... a=4, common ratio r=3.

s₁₀=a(rⁿ-1)/(r-1)=4(3¹⁰-1)/2=2(59049-1)=118096.