Answer:
First term: 6
Common difference: 15/4
Step-by-step explanation:
General form of arithmetic progression:
[tex]a_n=a+(n-1)d[/tex]
(where [tex]a[/tex] is the initial term and [tex]d[/tex] is the common difference between terms)
Question (a)
[tex]\implies a_3=a+(3-1)d=a+2d[/tex]
[tex]\implies a_9=a+(9-1)d=a+8d[/tex]
[tex]\implies a_{25}=a+(25-1)d=a+24d[/tex]
Question (b)
General form of a geometric progression: [tex]a_n=ar^{n-1}[/tex]
(where [tex]a[/tex] is the initial term and [tex]r[/tex] is the common ratio)
Therefore, the first three terms of a geometric series are:
[tex]a_1=ar^0=a[/tex]
[tex]a_2=ar^1=ar[/tex]
[tex]a_3=ar^2[/tex]
To find the common ratio:
[tex]r=\dfrac{ar^2}{ar}=\dfrac{ar}{a} \implies \dfrac{a_3}{a_2}=\dfrac{a_2}{a_1}[/tex]
If the 3rd 9th and 25th terms of an arithmetic progression form the first three consecutive terms of a geometric series, then inputting these into the above formula for r:
[tex]\implies \dfrac{a+24d}{a+8d}=\dfrac{a+8d}{a+2d}[/tex]
[tex]\implies (a+24d)(a+2d)=(a+8d)(a+8d)[/tex]
[tex]\implies a^2+26ad+48d^2=a^2+16ad+64d^2[/tex]
[tex]\implies 10ad=16d^2[/tex]
[tex]\implies 10a=16d[/tex]
The 6th and 7th terms of the arithmetic progression are:
[tex]a_6=a+5d[/tex]
[tex]a_7=a+6d[/tex]
If the sum of the 7th and twice the 6th term of the arithmetic progression is 78, then:
[tex]\implies a_7+2(a_6)=78[/tex]
[tex]\implies (a+6d)+2(a+5d)=78[/tex]
[tex]\implies 3a+16d=78[/tex]
Substituting [tex]10a=16d[/tex] into [tex]3a+16d=78[/tex] and solving for a:
[tex]\implies 3a+10a=78[/tex]
[tex]\implies 13a=78[/tex]
[tex]\implies a=6[/tex]
Substituting [tex]a=6[/tex] into [tex]3a+16d=78[/tex] and solving for d:
[tex]\implies 3(6)+16d=78[/tex]
[tex]\implies 16d=60[/tex]
[tex]\implies d=\dfrac{15}{4}[/tex]
