.
Let "a" be the first term of the AP, and let "d" be its common difference.
Then from the condition you have these two equations
3*(a+d) = a + 4d (1) (which means = )
a + 2d = 10 (2) (which means = 10 )
From equation (1)
3a + 3d = a + 4d
2a = d (3)
From equation (2)
2a + 4d = 20,
and substituting (replacing) here 2a = d from (3), you get
d + 4d = 20,
5d = 20,
d = 20/5 = 4.
Now from equation (2)
a = 10-3d = 10-2*4 = 2.
So, the progression has the first term a= 2 and the common difference d= 4.
In particular, the 20-th term is
= a + 19*d = 2 + 19*4 = 78. ANSWER
Solved.
------------------
On arithmetic progressions, see the lessons
- Arithmetic progressions
- The proofs of the formulas for arithmetic progressions
- Problems on arithmetic progressions
- Word problems on arithmetic progressions
- Chocolate bars and arithmetic progressions
- One characteristic property of arithmetic progressions
- Solved problems on arithmetic progressions
- Calculating partial sums of arithmetic progressions
- Finding number of terms of an arithmetic progression
- Advanced problems on arithmetic progressions
- Problems on arithmetic progressions solved MENTALLY
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.