Arithmetic progression 2

The 3rd term of an Arithmetic progression is 10 more than the first term while the fifth term is 15 more than the second term. Find the sum of the 8th and 15th terms of the Arithmetic progression if the 7th term is 7 times the first term.

Correct result:

s1 =  115

Solution:

a3=10+a1 a5=15+a2  a1+2d=10+a1 d=10/2=5  a7=7a1=a1+6d   7 a1=a1+6 d 7 a1=a1+6 5  6a1=30  a1=5  a8=a1+7 d=5+7 5=40 a15=a1+14 d=5+14 5=75  s1=a8+a15=40+75=115



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